Solve x^2-1=frac{x^4-{x}^frac{4{2}}}{{x}^(2*2)-frac{4{2}}}

$${ x }^{ 2 } -1= \frac{ { x }^{ 4 } - { x }^{ \frac{ 4 }{ 2 } } }{ { x }^{ (2 \times 2)- \frac{ 4 }{ 2 } } }$$

Solve for x

$$x\neq 0$$

Short Solution Steps

Divide 4 by 2 to get 2.

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

Multiply 2 and 2 to get 4.

$$x^{2}-1=\frac{x^{4}-x^{2}}{x^{4-\frac{4}{2}}}$$

Divide 4 by 2 to get 2.

$$x^{2}-1=\frac{x^{4}-x^{2}}{x^{4-2}}$$

Subtract 2 from 4 to get 2.

$$x^{2}-1=\frac{x^{4}-x^{2}}{x^{2}}$$

Subtract \frac{x^{4}-x^{2}}{x^{2}} from both sides.

$$x^{2}-1-\frac{x^{4}-x^{2}}{x^{2}}=0$$

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

$$\frac{\left(x^{2}-1\right)x^{2}}{x^{2}}-\frac{x^{4}-x^{2}}{x^{2}}=0$$

Since \frac{\left(x^{2}-1\right)x^{2}}{x^{2}} and \frac{x^{4}-x^{2}}{x^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(x^{2}-1\right)x^{2}-\left(x^{4}-x^{2}\right)}{x^{2}}=0$$

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

$$\frac{x^{4}-x^{2}-x^{4}+x^{2}}{x^{2}}=0$$

Combine like terms in x^{4}-x^{2}-x^{4}+x^{2}.

$$\frac{0}{x^{2}}=0$$

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}.

$$0=0$$

This is true for any x.

$$x\in \mathrm{R}$$

Variable x cannot be equal to 0.

$$x\in \mathrm{R}\setminus 0$$