Solve 2/sqrt{2x-1}+v2+1/2x^2-1=v2+1/sqrt{2x+1}

$$\frac { 2 } { \sqrt { 2 } x - 1 } + \frac { v 2 + 1 } { 2 x ^ { 2 } - 1 } = \frac { v 2 + 1 } { \sqrt { 2 } x + 1 }$$

Solve for v_2

$$v_{2}=-\frac{x+2\sqrt{2}}{-x+\sqrt{2}}$$

$$x\neq \sqrt{2}\text{ and }|x|\neq \frac{\sqrt{2}}{2}$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2x^{2}-1.

$$\left(2x^{2}-1\right)\left(2^{\frac{1}{2}}x-1\right)^{-1}\times 2+v_{2}+1=\left(2x^{2}-1\right)\left(2^{\frac{1}{2}}x+1\right)^{-1}\left(v_{2}+1\right)$$

Use the distributive property to multiply 2x^{2}-1 by \left(2^{\frac{1}{2}}x-1\right)^{-1}.

$$\left(2x^{2}\left(2^{\frac{1}{2}}x-1\right)^{-1}-\left(2^{\frac{1}{2}}x-1\right)^{-1}\right)\times 2+v_{2}+1=\left(2x^{2}-1\right)\left(2^{\frac{1}{2}}x+1\right)^{-1}\left(v_{2}+1\right)$$

Use the distributive property to multiply 2x^{2}\left(2^{\frac{1}{2}}x-1\right)^{-1}-\left(2^{\frac{1}{2}}x-1\right)^{-1} by 2.

$$4\left(2^{\frac{1}{2}}x-1\right)^{-1}x^{2}-2\left(2^{\frac{1}{2}}x-1\right)^{-1}+v_{2}+1=\left(2x^{2}-1\right)\left(2^{\frac{1}{2}}x+1\right)^{-1}\left(v_{2}+1\right)$$

Use the distributive property to multiply 2x^{2}-1 by \left(2^{\frac{1}{2}}x+1\right)^{-1}.

$$4\left(2^{\frac{1}{2}}x-1\right)^{-1}x^{2}-2\left(2^{\frac{1}{2}}x-1\right)^{-1}+v_{2}+1=\left(2x^{2}\left(2^{\frac{1}{2}}x+1\right)^{-1}-\left(2^{\frac{1}{2}}x+1\right)^{-1}\right)\left(v_{2}+1\right)$$

Use the distributive property to multiply 2x^{2}\left(2^{\frac{1}{2}}x+1\right)^{-1}-\left(2^{\frac{1}{2}}x+1\right)^{-1} by v_{2}+1.

$$4\left(2^{\frac{1}{2}}x-1\right)^{-1}x^{2}-2\left(2^{\frac{1}{2}}x-1\right)^{-1}+v_{2}+1=2x^{2}\left(2^{\frac{1}{2}}x+1\right)^{-1}v_{2}+2x^{2}\left(2^{\frac{1}{2}}x+1\right)^{-1}-\left(2^{\frac{1}{2}}x+1\right)^{-1}v_{2}-\left(2^{\frac{1}{2}}x+1\right)^{-1}$$

Subtract 2x^{2}\left(2^{\frac{1}{2}}x+1\right)^{-1}v_{2} from both sides.

$$4\left(2^{\frac{1}{2}}x-1\right)^{-1}x^{2}-2\left(2^{\frac{1}{2}}x-1\right)^{-1}+v_{2}+1-2x^{2}\left(2^{\frac{1}{2}}x+1\right)^{-1}v_{2}=2x^{2}\left(2^{\frac{1}{2}}x+1\right)^{-1}-\left(2^{\frac{1}{2}}x+1\right)^{-1}v_{2}-\left(2^{\frac{1}{2}}x+1\right)^{-1}$$

Add \left(2^{\frac{1}{2}}x+1\right)^{-1}v_{2} to both sides.

$$4\left(2^{\frac{1}{2}}x-1\right)^{-1}x^{2}-2\left(2^{\frac{1}{2}}x-1\right)^{-1}+v_{2}+1-2x^{2}\left(2^{\frac{1}{2}}x+1\right)^{-1}v_{2}+\left(2^{\frac{1}{2}}x+1\right)^{-1}v_{2}=2x^{2}\left(2^{\frac{1}{2}}x+1\right)^{-1}-\left(2^{\frac{1}{2}}x+1\right)^{-1}$$

Reorder the terms.

$$4\times \frac{1}{\sqrt{2}x-1}x^{2}+v_{2}+1-2\times \frac{1}{\sqrt{2}x-1}-2\times \frac{1}{\sqrt{2}x+1}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Rationalize the denominator of \frac{1}{\sqrt{2}x-1} by multiplying numerator and denominator by \sqrt{2}x+1.

$$4\times \frac{\sqrt{2}x+1}{\left(\sqrt{2}x-1\right)\left(\sqrt{2}x+1\right)}x^{2}+v_{2}+1-2\times \frac{1}{\sqrt{2}x-1}-2\times \frac{1}{\sqrt{2}x+1}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Consider \left(\sqrt{2}x-1\right)\left(\sqrt{2}x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$4\times \frac{\sqrt{2}x+1}{\left(\sqrt{2}x\right)^{2}-1^{2}}x^{2}+v_{2}+1-2\times \frac{1}{\sqrt{2}x-1}-2\times \frac{1}{\sqrt{2}x+1}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Expand \left(\sqrt{2}x\right)^{2}.

$$4\times \frac{\sqrt{2}x+1}{\left(\sqrt{2}\right)^{2}x^{2}-1^{2}}x^{2}+v_{2}+1-2\times \frac{1}{\sqrt{2}x-1}-2\times \frac{1}{\sqrt{2}x+1}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

The square of \sqrt{2} is 2.

$$4\times \frac{\sqrt{2}x+1}{2x^{2}-1^{2}}x^{2}+v_{2}+1-2\times \frac{1}{\sqrt{2}x-1}-2\times \frac{1}{\sqrt{2}x+1}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Calculate 1 to the power of 2 and get 1.

$$4\times \frac{\sqrt{2}x+1}{2x^{2}-1}x^{2}+v_{2}+1-2\times \frac{1}{\sqrt{2}x-1}-2\times \frac{1}{\sqrt{2}x+1}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Express 4\times \frac{\sqrt{2}x+1}{2x^{2}-1} as a single fraction.

$$\frac{4\left(\sqrt{2}x+1\right)}{2x^{2}-1}x^{2}+v_{2}+1-2\times \frac{1}{\sqrt{2}x-1}-2\times \frac{1}{\sqrt{2}x+1}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Express \frac{4\left(\sqrt{2}x+1\right)}{2x^{2}-1}x^{2} as a single fraction.

$$\frac{4\left(\sqrt{2}x+1\right)x^{2}}{2x^{2}-1}+v_{2}+1-2\times \frac{1}{\sqrt{2}x-1}-2\times \frac{1}{\sqrt{2}x+1}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Rationalize the denominator of \frac{1}{\sqrt{2}x-1} by multiplying numerator and denominator by \sqrt{2}x+1.

$$\frac{4\left(\sqrt{2}x+1\right)x^{2}}{2x^{2}-1}+v_{2}+1-2\times \frac{\sqrt{2}x+1}{\left(\sqrt{2}x-1\right)\left(\sqrt{2}x+1\right)}-2\times \frac{1}{\sqrt{2}x+1}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Consider \left(\sqrt{2}x-1\right)\left(\sqrt{2}x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{4\left(\sqrt{2}x+1\right)x^{2}}{2x^{2}-1}+v_{2}+1-2\times \frac{\sqrt{2}x+1}{\left(\sqrt{2}x\right)^{2}-1^{2}}-2\times \frac{1}{\sqrt{2}x+1}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Expand \left(\sqrt{2}x\right)^{2}.

$$\frac{4\left(\sqrt{2}x+1\right)x^{2}}{2x^{2}-1}+v_{2}+1-2\times \frac{\sqrt{2}x+1}{\left(\sqrt{2}\right)^{2}x^{2}-1^{2}}-2\times \frac{1}{\sqrt{2}x+1}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

The square of \sqrt{2} is 2.

$$\frac{4\left(\sqrt{2}x+1\right)x^{2}}{2x^{2}-1}+v_{2}+1-2\times \frac{\sqrt{2}x+1}{2x^{2}-1^{2}}-2\times \frac{1}{\sqrt{2}x+1}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Calculate 1 to the power of 2 and get 1.

$$\frac{4\left(\sqrt{2}x+1\right)x^{2}}{2x^{2}-1}+v_{2}+1-2\times \frac{\sqrt{2}x+1}{2x^{2}-1}-2\times \frac{1}{\sqrt{2}x+1}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Express -2\times \frac{\sqrt{2}x+1}{2x^{2}-1} as a single fraction.

$$\frac{4\left(\sqrt{2}x+1\right)x^{2}}{2x^{2}-1}+v_{2}+1+\frac{-2\left(\sqrt{2}x+1\right)}{2x^{2}-1}-2\times \frac{1}{\sqrt{2}x+1}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply v_{2}+1 times \frac{2x^{2}-1}{2x^{2}-1}.

$$\frac{4\left(\sqrt{2}x+1\right)x^{2}}{2x^{2}-1}+\frac{\left(v_{2}+1\right)\left(2x^{2}-1\right)}{2x^{2}-1}+\frac{-2\left(\sqrt{2}x+1\right)}{2x^{2}-1}-2\times \frac{1}{\sqrt{2}x+1}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Since \frac{4\left(\sqrt{2}x+1\right)x^{2}}{2x^{2}-1} and \frac{\left(v_{2}+1\right)\left(2x^{2}-1\right)}{2x^{2}-1} have the same denominator, add them by adding their numerators.

$$\frac{4\left(\sqrt{2}x+1\right)x^{2}+\left(v_{2}+1\right)\left(2x^{2}-1\right)}{2x^{2}-1}+\frac{-2\left(\sqrt{2}x+1\right)}{2x^{2}-1}-2\times \frac{1}{\sqrt{2}x+1}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Do the multiplications in 4\left(\sqrt{2}x+1\right)x^{2}+\left(v_{2}+1\right)\left(2x^{2}-1\right).

$$\frac{4\sqrt{2}x^{3}+4x^{2}+2v_{2}x^{2}-v_{2}+2x^{2}-1}{2x^{2}-1}+\frac{-2\left(\sqrt{2}x+1\right)}{2x^{2}-1}-2\times \frac{1}{\sqrt{2}x+1}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Combine like terms in 4\sqrt{2}x^{3}+4x^{2}+2v_{2}x^{2}-v_{2}+2x^{2}-1.

$$\frac{4\sqrt{2}x^{3}+6x^{2}-v_{2}+2v_{2}x^{2}-1}{2x^{2}-1}+\frac{-2\left(\sqrt{2}x+1\right)}{2x^{2}-1}-2\times \frac{1}{\sqrt{2}x+1}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Since \frac{4\sqrt{2}x^{3}+6x^{2}-v_{2}+2v_{2}x^{2}-1}{2x^{2}-1} and \frac{-2\left(\sqrt{2}x+1\right)}{2x^{2}-1} have the same denominator, add them by adding their numerators.

$$\frac{4\sqrt{2}x^{3}+6x^{2}-v_{2}+2v_{2}x^{2}-1-2\left(\sqrt{2}x+1\right)}{2x^{2}-1}-2\times \frac{1}{\sqrt{2}x+1}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Do the multiplications in 4\sqrt{2}x^{3}+6x^{2}-v_{2}+2v_{2}x^{2}-1-2\left(\sqrt{2}x+1\right).

$$\frac{4\sqrt{2}x^{3}+6x^{2}-v_{2}+2v_{2}x^{2}-1-2\sqrt{2}x-2}{2x^{2}-1}-2\times \frac{1}{\sqrt{2}x+1}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Combine like terms in 4\sqrt{2}x^{3}+6x^{2}-v_{2}+2v_{2}x^{2}-1-2\sqrt{2}x-2.

$$\frac{4\sqrt{2}x^{3}+6x^{2}-2\sqrt{2}x+2v_{2}x^{2}-v_{2}-3}{2x^{2}-1}-2\times \frac{1}{\sqrt{2}x+1}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Rationalize the denominator of \frac{1}{\sqrt{2}x+1} by multiplying numerator and denominator by \sqrt{2}x-1.

$$\frac{4\sqrt{2}x^{3}+6x^{2}-2\sqrt{2}x+2v_{2}x^{2}-v_{2}-3}{2x^{2}-1}-2\times \frac{\sqrt{2}x-1}{\left(\sqrt{2}x+1\right)\left(\sqrt{2}x-1\right)}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Consider \left(\sqrt{2}x+1\right)\left(\sqrt{2}x-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{4\sqrt{2}x^{3}+6x^{2}-2\sqrt{2}x+2v_{2}x^{2}-v_{2}-3}{2x^{2}-1}-2\times \frac{\sqrt{2}x-1}{\left(\sqrt{2}x\right)^{2}-1^{2}}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Expand \left(\sqrt{2}x\right)^{2}.

$$\frac{4\sqrt{2}x^{3}+6x^{2}-2\sqrt{2}x+2v_{2}x^{2}-v_{2}-3}{2x^{2}-1}-2\times \frac{\sqrt{2}x-1}{\left(\sqrt{2}\right)^{2}x^{2}-1^{2}}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

The square of \sqrt{2} is 2.

$$\frac{4\sqrt{2}x^{3}+6x^{2}-2\sqrt{2}x+2v_{2}x^{2}-v_{2}-3}{2x^{2}-1}-2\times \frac{\sqrt{2}x-1}{2x^{2}-1^{2}}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Calculate 1 to the power of 2 and get 1.

$$\frac{4\sqrt{2}x^{3}+6x^{2}-2\sqrt{2}x+2v_{2}x^{2}-v_{2}-3}{2x^{2}-1}-2\times \frac{\sqrt{2}x-1}{2x^{2}-1}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Express -2\times \frac{\sqrt{2}x-1}{2x^{2}-1} as a single fraction.

$$\frac{4\sqrt{2}x^{3}+6x^{2}-2\sqrt{2}x+2v_{2}x^{2}-v_{2}-3}{2x^{2}-1}+\frac{-2\left(\sqrt{2}x-1\right)}{2x^{2}-1}v_{2}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Express \frac{-2\left(\sqrt{2}x-1\right)}{2x^{2}-1}v_{2} as a single fraction.

$$\frac{4\sqrt{2}x^{3}+6x^{2}-2\sqrt{2}x+2v_{2}x^{2}-v_{2}-3}{2x^{2}-1}+\frac{-2\left(\sqrt{2}x-1\right)v_{2}}{2x^{2}-1}x^{2}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Express \frac{-2\left(\sqrt{2}x-1\right)v_{2}}{2x^{2}-1}x^{2} as a single fraction.

$$\frac{4\sqrt{2}x^{3}+6x^{2}-2\sqrt{2}x+2v_{2}x^{2}-v_{2}-3}{2x^{2}-1}+\frac{-2\left(\sqrt{2}x-1\right)v_{2}x^{2}}{2x^{2}-1}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Since \frac{4\sqrt{2}x^{3}+6x^{2}-2\sqrt{2}x+2v_{2}x^{2}-v_{2}-3}{2x^{2}-1} and \frac{-2\left(\sqrt{2}x-1\right)v_{2}x^{2}}{2x^{2}-1} have the same denominator, add them by adding their numerators.

$$\frac{4\sqrt{2}x^{3}+6x^{2}-2\sqrt{2}x+2v_{2}x^{2}-v_{2}-3-2\left(\sqrt{2}x-1\right)v_{2}x^{2}}{2x^{2}-1}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Do the multiplications in 4\sqrt{2}x^{3}+6x^{2}-2\sqrt{2}x+2v_{2}x^{2}-v_{2}-3-2\left(\sqrt{2}x-1\right)v_{2}x^{2}.

$$\frac{4\sqrt{2}x^{3}+6x^{2}-2\sqrt{2}x+2v_{2}x^{2}-v_{2}-3-2\sqrt{2}v_{2}x^{3}+2x^{2}v_{2}}{2x^{2}-1}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Combine like terms in 4\sqrt{2}x^{3}+6x^{2}-2\sqrt{2}x+2v_{2}x^{2}-v_{2}-3-2\sqrt{2}v_{2}x^{3}+2x^{2}v_{2}.

$$\frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}-v_{2}+4v_{2}x^{2}-2\sqrt{2}x-3}{2x^{2}-1}+\frac{1}{\sqrt{2}x+1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Rationalize the denominator of \frac{1}{\sqrt{2}x+1} by multiplying numerator and denominator by \sqrt{2}x-1.

$$\frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}-v_{2}+4v_{2}x^{2}-2\sqrt{2}x-3}{2x^{2}-1}+\frac{\sqrt{2}x-1}{\left(\sqrt{2}x+1\right)\left(\sqrt{2}x-1\right)}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Consider \left(\sqrt{2}x+1\right)\left(\sqrt{2}x-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}-v_{2}+4v_{2}x^{2}-2\sqrt{2}x-3}{2x^{2}-1}+\frac{\sqrt{2}x-1}{\left(\sqrt{2}x\right)^{2}-1^{2}}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Expand \left(\sqrt{2}x\right)^{2}.

$$\frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}-v_{2}+4v_{2}x^{2}-2\sqrt{2}x-3}{2x^{2}-1}+\frac{\sqrt{2}x-1}{\left(\sqrt{2}\right)^{2}x^{2}-1^{2}}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

The square of \sqrt{2} is 2.

$$\frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}-v_{2}+4v_{2}x^{2}-2\sqrt{2}x-3}{2x^{2}-1}+\frac{\sqrt{2}x-1}{2x^{2}-1^{2}}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Calculate 1 to the power of 2 and get 1.

$$\frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}-v_{2}+4v_{2}x^{2}-2\sqrt{2}x-3}{2x^{2}-1}+\frac{\sqrt{2}x-1}{2x^{2}-1}v_{2}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Express \frac{\sqrt{2}x-1}{2x^{2}-1}v_{2} as a single fraction.

$$\frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}-v_{2}+4v_{2}x^{2}-2\sqrt{2}x-3}{2x^{2}-1}+\frac{\left(\sqrt{2}x-1\right)v_{2}}{2x^{2}-1}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Since \frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}-v_{2}+4v_{2}x^{2}-2\sqrt{2}x-3}{2x^{2}-1} and \frac{\left(\sqrt{2}x-1\right)v_{2}}{2x^{2}-1} have the same denominator, add them by adding their numerators.

$$\frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}-v_{2}+4v_{2}x^{2}-2\sqrt{2}x-3+\left(\sqrt{2}x-1\right)v_{2}}{2x^{2}-1}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Do the multiplications in 4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}-v_{2}+4v_{2}x^{2}-2\sqrt{2}x-3+\left(\sqrt{2}x-1\right)v_{2}.

$$\frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}-v_{2}+4v_{2}x^{2}-2\sqrt{2}x-3+\sqrt{2}xv_{2}-v_{2}}{2x^{2}-1}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Combine like terms in 4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}-v_{2}+4v_{2}x^{2}-2\sqrt{2}x-3+\sqrt{2}xv_{2}-v_{2}.

$$\frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}+4v_{2}x^{2}-2v_{2}-2\sqrt{2}x-3+\sqrt{2}xv_{2}}{2x^{2}-1}=2\times \frac{1}{\sqrt{2}x+1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Rationalize the denominator of \frac{1}{\sqrt{2}x+1} by multiplying numerator and denominator by \sqrt{2}x-1.

$$\frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}+4v_{2}x^{2}-2v_{2}-2\sqrt{2}x-3+\sqrt{2}xv_{2}}{2x^{2}-1}=2\times \frac{\sqrt{2}x-1}{\left(\sqrt{2}x+1\right)\left(\sqrt{2}x-1\right)}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Consider \left(\sqrt{2}x+1\right)\left(\sqrt{2}x-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}+4v_{2}x^{2}-2v_{2}-2\sqrt{2}x-3+\sqrt{2}xv_{2}}{2x^{2}-1}=2\times \frac{\sqrt{2}x-1}{\left(\sqrt{2}x\right)^{2}-1^{2}}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Expand \left(\sqrt{2}x\right)^{2}.

$$\frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}+4v_{2}x^{2}-2v_{2}-2\sqrt{2}x-3+\sqrt{2}xv_{2}}{2x^{2}-1}=2\times \frac{\sqrt{2}x-1}{\left(\sqrt{2}\right)^{2}x^{2}-1^{2}}x^{2}-\frac{1}{\sqrt{2}x+1}$$

The square of \sqrt{2} is 2.

$$\frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}+4v_{2}x^{2}-2v_{2}-2\sqrt{2}x-3+\sqrt{2}xv_{2}}{2x^{2}-1}=2\times \frac{\sqrt{2}x-1}{2x^{2}-1^{2}}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Calculate 1 to the power of 2 and get 1.

$$\frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}+4v_{2}x^{2}-2v_{2}-2\sqrt{2}x-3+\sqrt{2}xv_{2}}{2x^{2}-1}=2\times \frac{\sqrt{2}x-1}{2x^{2}-1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Express 2\times \frac{\sqrt{2}x-1}{2x^{2}-1} as a single fraction.

$$\frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}+4v_{2}x^{2}-2v_{2}-2\sqrt{2}x-3+\sqrt{2}xv_{2}}{2x^{2}-1}=\frac{2\left(\sqrt{2}x-1\right)}{2x^{2}-1}x^{2}-\frac{1}{\sqrt{2}x+1}$$

Express \frac{2\left(\sqrt{2}x-1\right)}{2x^{2}-1}x^{2} as a single fraction.

$$\frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}+4v_{2}x^{2}-2v_{2}-2\sqrt{2}x-3+\sqrt{2}xv_{2}}{2x^{2}-1}=\frac{2\left(\sqrt{2}x-1\right)x^{2}}{2x^{2}-1}-\frac{1}{\sqrt{2}x+1}$$

Rationalize the denominator of \frac{1}{\sqrt{2}x+1} by multiplying numerator and denominator by \sqrt{2}x-1.

$$\frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}+4v_{2}x^{2}-2v_{2}-2\sqrt{2}x-3+\sqrt{2}xv_{2}}{2x^{2}-1}=\frac{2\left(\sqrt{2}x-1\right)x^{2}}{2x^{2}-1}-\frac{\sqrt{2}x-1}{\left(\sqrt{2}x+1\right)\left(\sqrt{2}x-1\right)}$$

Consider \left(\sqrt{2}x+1\right)\left(\sqrt{2}x-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}+4v_{2}x^{2}-2v_{2}-2\sqrt{2}x-3+\sqrt{2}xv_{2}}{2x^{2}-1}=\frac{2\left(\sqrt{2}x-1\right)x^{2}}{2x^{2}-1}-\frac{\sqrt{2}x-1}{\left(\sqrt{2}x\right)^{2}-1^{2}}$$

Expand \left(\sqrt{2}x\right)^{2}.

$$\frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}+4v_{2}x^{2}-2v_{2}-2\sqrt{2}x-3+\sqrt{2}xv_{2}}{2x^{2}-1}=\frac{2\left(\sqrt{2}x-1\right)x^{2}}{2x^{2}-1}-\frac{\sqrt{2}x-1}{\left(\sqrt{2}\right)^{2}x^{2}-1^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}+4v_{2}x^{2}-2v_{2}-2\sqrt{2}x-3+\sqrt{2}xv_{2}}{2x^{2}-1}=\frac{2\left(\sqrt{2}x-1\right)x^{2}}{2x^{2}-1}-\frac{\sqrt{2}x-1}{2x^{2}-1^{2}}$$

Calculate 1 to the power of 2 and get 1.

$$\frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}+4v_{2}x^{2}-2v_{2}-2\sqrt{2}x-3+\sqrt{2}xv_{2}}{2x^{2}-1}=\frac{2\left(\sqrt{2}x-1\right)x^{2}}{2x^{2}-1}-\frac{\sqrt{2}x-1}{2x^{2}-1}$$

Since \frac{2\left(\sqrt{2}x-1\right)x^{2}}{2x^{2}-1} and \frac{\sqrt{2}x-1}{2x^{2}-1} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}+4v_{2}x^{2}-2v_{2}-2\sqrt{2}x-3+\sqrt{2}xv_{2}}{2x^{2}-1}=\frac{2\left(\sqrt{2}x-1\right)x^{2}-\left(\sqrt{2}x-1\right)}{2x^{2}-1}$$

Do the multiplications in 2\left(\sqrt{2}x-1\right)x^{2}-\left(\sqrt{2}x-1\right).

$$\frac{4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}+4v_{2}x^{2}-2v_{2}-2\sqrt{2}x-3+\sqrt{2}xv_{2}}{2x^{2}-1}=\frac{2\sqrt{2}x^{3}-2x^{2}-\sqrt{2}x+1}{2x^{2}-1}$$

Multiply both sides of the equation by 2x^{2}-1.

$$4\sqrt{2}x^{3}-2\sqrt{2}v_{2}x^{3}+6x^{2}+4v_{2}x^{2}-2v_{2}-2\sqrt{2}x-3+\sqrt{2}xv_{2}=2\sqrt{2}x^{3}-2x^{2}-\sqrt{2}x+1$$

Subtract 4\sqrt{2}x^{3} from both sides.

$$-2\sqrt{2}v_{2}x^{3}+6x^{2}+4v_{2}x^{2}-2v_{2}-2\sqrt{2}x-3+\sqrt{2}xv_{2}=2\sqrt{2}x^{3}-2x^{2}-\sqrt{2}x+1-4\sqrt{2}x^{3}$$

Combine 2\sqrt{2}x^{3} and -4\sqrt{2}x^{3} to get -2\sqrt{2}x^{3}.

$$-2\sqrt{2}v_{2}x^{3}+6x^{2}+4v_{2}x^{2}-2v_{2}-2\sqrt{2}x-3+\sqrt{2}xv_{2}=-2\sqrt{2}x^{3}-2x^{2}-\sqrt{2}x+1$$

Subtract 6x^{2} from both sides.

$$-2\sqrt{2}v_{2}x^{3}+4v_{2}x^{2}-2v_{2}-2\sqrt{2}x-3+\sqrt{2}xv_{2}=-2\sqrt{2}x^{3}-2x^{2}-\sqrt{2}x+1-6x^{2}$$

Combine -2x^{2} and -6x^{2} to get -8x^{2}.

$$-2\sqrt{2}v_{2}x^{3}+4v_{2}x^{2}-2v_{2}-2\sqrt{2}x-3+\sqrt{2}xv_{2}=-2\sqrt{2}x^{3}-8x^{2}-\sqrt{2}x+1$$

Add 2\sqrt{2}x to both sides.

$$-2\sqrt{2}v_{2}x^{3}+4v_{2}x^{2}-2v_{2}-3+\sqrt{2}xv_{2}=-2\sqrt{2}x^{3}-8x^{2}-\sqrt{2}x+1+2\sqrt{2}x$$

Combine -\sqrt{2}x and 2\sqrt{2}x to get \sqrt{2}x.

$$-2\sqrt{2}v_{2}x^{3}+4v_{2}x^{2}-2v_{2}-3+\sqrt{2}xv_{2}=-2\sqrt{2}x^{3}-8x^{2}+\sqrt{2}x+1$$

Add 3 to both sides.

$$-2\sqrt{2}v_{2}x^{3}+4v_{2}x^{2}-2v_{2}+\sqrt{2}xv_{2}=-2\sqrt{2}x^{3}-8x^{2}+\sqrt{2}x+1+3$$

Add 1 and 3 to get 4.

$$-2\sqrt{2}v_{2}x^{3}+4v_{2}x^{2}-2v_{2}+\sqrt{2}xv_{2}=-2\sqrt{2}x^{3}-8x^{2}+\sqrt{2}x+4$$

Combine all terms containing v_{2}.

$$\left(-2\sqrt{2}x^{3}+4x^{2}-2+\sqrt{2}x\right)v_{2}=-2\sqrt{2}x^{3}-8x^{2}+\sqrt{2}x+4$$

The equation is in standard form.

$$\left(-2\sqrt{2}x^{3}+4x^{2}+\sqrt{2}x-2\right)v_{2}=-2\sqrt{2}x^{3}-8x^{2}+\sqrt{2}x+4$$

Divide both sides by -2\sqrt{2}x^{3}+4x^{2}-2+\sqrt{2}x.

$$\frac{\left(-2\sqrt{2}x^{3}+4x^{2}+\sqrt{2}x-2\right)v_{2}}{-2\sqrt{2}x^{3}+4x^{2}+\sqrt{2}x-2}=\frac{-2\sqrt{2}x^{3}-8x^{2}+\sqrt{2}x+4}{-2\sqrt{2}x^{3}+4x^{2}+\sqrt{2}x-2}$$

Dividing by -2\sqrt{2}x^{3}+4x^{2}-2+\sqrt{2}x undoes the multiplication by -2\sqrt{2}x^{3}+4x^{2}-2+\sqrt{2}x.

$$v_{2}=\frac{-2\sqrt{2}x^{3}-8x^{2}+\sqrt{2}x+4}{-2\sqrt{2}x^{3}+4x^{2}+\sqrt{2}x-2}$$

Divide -2\sqrt{2}x^{3}-8x^{2}+\sqrt{2}x+4 by -2\sqrt{2}x^{3}+4x^{2}-2+\sqrt{2}x.

$$v_{2}=-\frac{\sqrt{2}\left(\sqrt{2}x+4\right)}{2\left(-x+\sqrt{2}\right)}$$

Solve for x

$$x=-\frac{\sqrt{2}\left(v_{2}+2\right)}{1-v_{2}}$$

$$v_{2}\neq -5\text{ and }|v_{2}|\neq 1$$