Solve 2+sqrt{2}+1/sqrt{2+2}+1/sqrt{2-2}=7

$$2+ \sqrt{ 2 } + \frac{ 1 }{ \sqrt{ 2 } +2 } + \frac{ 1 }{ \sqrt{ 2 } -2 } = 7$$

Verify

Short Solution Steps

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

$$2+\sqrt{2}+\frac{\sqrt{2}-2}{\left(\sqrt{2}+2\right)\left(\sqrt{2}-2\right)}+\frac{1}{\sqrt{2}-2}=7$$

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

$$2+\sqrt{2}+\frac{\sqrt{2}-2}{\left(\sqrt{2}\right)^{2}-2^{2}}+\frac{1}{\sqrt{2}-2}=7$$

Square \sqrt{2}. Square 2.

$$2+\sqrt{2}+\frac{\sqrt{2}-2}{2-4}+\frac{1}{\sqrt{2}-2}=7$$

Subtract 4 from 2 to get -2.

$$2+\sqrt{2}+\frac{\sqrt{2}-2}{-2}+\frac{1}{\sqrt{2}-2}=7$$

Multiply both numerator and denominator by -1.

$$2+\sqrt{2}+\frac{-\sqrt{2}+2}{2}+\frac{1}{\sqrt{2}-2}=7$$

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

$$2+\sqrt{2}+\frac{-\sqrt{2}+2}{2}+\frac{\sqrt{2}+2}{\left(\sqrt{2}-2\right)\left(\sqrt{2}+2\right)}=7$$

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

$$2+\sqrt{2}+\frac{-\sqrt{2}+2}{2}+\frac{\sqrt{2}+2}{\left(\sqrt{2}\right)^{2}-2^{2}}=7$$

Square \sqrt{2}. Square 2.

$$2+\sqrt{2}+\frac{-\sqrt{2}+2}{2}+\frac{\sqrt{2}+2}{2-4}=7$$

Subtract 4 from 2 to get -2.

$$2+\sqrt{2}+\frac{-\sqrt{2}+2}{2}+\frac{\sqrt{2}+2}{-2}=7$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2+\sqrt{2} times \frac{2}{2}.

$$\frac{2\left(2+\sqrt{2}\right)}{2}+\frac{-\sqrt{2}+2}{2}+\frac{\sqrt{2}+2}{-2}=7$$

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

$$\frac{2\left(2+\sqrt{2}\right)-\sqrt{2}+2}{2}+\frac{\sqrt{2}+2}{-2}=7$$

Do the multiplications in 2\left(2+\sqrt{2}\right)-\sqrt{2}+2.

$$\frac{4+2\sqrt{2}-\sqrt{2}+2}{2}+\frac{\sqrt{2}+2}{-2}=7$$

Do the calculations in 4+2\sqrt{2}-\sqrt{2}+2.

$$\frac{6+\sqrt{2}}{2}+\frac{\sqrt{2}+2}{-2}=7$$

Subtract 7 from both sides.

$$\frac{6+\sqrt{2}}{2}+\frac{\sqrt{2}+2}{-2}-7=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 7 times \frac{2}{2}.

$$\frac{6+\sqrt{2}}{2}+\frac{\sqrt{2}+2}{-2}-\frac{7\times 2}{2}=0$$

Since \frac{6+\sqrt{2}}{2} and \frac{7\times 2}{2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{6+\sqrt{2}-7\times 2}{2}+\frac{\sqrt{2}+2}{-2}=0$$

Do the multiplications in 6+\sqrt{2}-7\times 2.

$$\frac{6+\sqrt{2}-14}{2}+\frac{\sqrt{2}+2}{-2}=0$$

Do the calculations in 6+\sqrt{2}-14.

$$\frac{-8+\sqrt{2}}{2}+\frac{\sqrt{2}+2}{-2}=0$$

Multiply both sides of the equation by 2, the least common multiple of 2,-2.

$$-8+\sqrt{2}-\left(\sqrt{2}+2\right)=0$$

To find the opposite of \sqrt{2}+2, find the opposite of each term.

$$-8+\sqrt{2}-\sqrt{2}-2=0$$

Combine \sqrt{2} and -\sqrt{2} to get 0.

$$-8-2=0$$

Subtract 2 from -8 to get -10.

$$-10=0$$

Compare -10 and 0.

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