Solve 1-frac{5(-sqrt{2}+1)-5}{{left(-sqrt{2}+1right)}^2+1}

$$1- \frac{ 5(- \sqrt{ 2 } +1)-5 }{ { \left(- \sqrt{ 2 } +1 \right) }^{ 2 } +1 }$$

Evaluate

$$\frac{5\sqrt{2}+7}{2}\approx 7.035533906$$

Short Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-\sqrt{2}+1\right)^{2}.

$$1-\frac{5\left(-\sqrt{2}+1\right)-5}{\left(-\sqrt{2}\right)^{2}+2\left(-\sqrt{2}\right)+1+1}$$

Calculate -\sqrt{2} to the power of 2 and get \left(\sqrt{2}\right)^{2}.

$$1-\frac{5\left(-\sqrt{2}+1\right)-5}{\left(\sqrt{2}\right)^{2}+2\left(-\sqrt{2}\right)+1+1}$$

Add 1 and 1 to get 2.

$$1-\frac{5\left(-\sqrt{2}+1\right)-5}{\left(\sqrt{2}\right)^{2}+2\left(-\sqrt{2}\right)+2}$$

The square of \sqrt{2} is 2.

$$1-\frac{5\left(-\sqrt{2}+1\right)-5}{2+2\left(-\sqrt{2}\right)+2}$$

Add 2 and 2 to get 4.

$$1-\frac{5\left(-\sqrt{2}+1\right)-5}{4+2\left(-\sqrt{2}\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{4+2\left(-\sqrt{2}\right)}{4+2\left(-\sqrt{2}\right)}.

$$\frac{4+2\left(-\sqrt{2}\right)}{4+2\left(-\sqrt{2}\right)}-\frac{5\left(-\sqrt{2}+1\right)-5}{4+2\left(-\sqrt{2}\right)}$$

Since \frac{4+2\left(-\sqrt{2}\right)}{4+2\left(-\sqrt{2}\right)} and \frac{5\left(-\sqrt{2}+1\right)-5}{4+2\left(-\sqrt{2}\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4+2\left(-\sqrt{2}\right)-\left(5\left(-\sqrt{2}+1\right)-5\right)}{4+2\left(-\sqrt{2}\right)}$$

Do the multiplications in 4+2\left(-\sqrt{2}\right)-\left(5\left(-\sqrt{2}+1\right)-5\right).

$$\frac{4-2\sqrt{2}+5\sqrt{2}-5+5}{4+2\left(-\sqrt{2}\right)}$$

Do the calculations in 4-2\sqrt{2}+5\sqrt{2}-5+5.

$$\frac{4+3\sqrt{2}}{4+2\left(-\sqrt{2}\right)}$$

Multiply 2 and -1 to get -2.

$$\frac{4+3\sqrt{2}}{4-2\sqrt{2}}$$

Rationalize the denominator of \frac{4+3\sqrt{2}}{4-2\sqrt{2}} by multiplying numerator and denominator by 4+2\sqrt{2}.

$$\frac{\left(4+3\sqrt{2}\right)\left(4+2\sqrt{2}\right)}{\left(4-2\sqrt{2}\right)\left(4+2\sqrt{2}\right)}$$

Consider \left(4-2\sqrt{2}\right)\left(4+2\sqrt{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}.

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

Calculate 4 to the power of 2 and get 16.

$$\frac{\left(4+3\sqrt{2}\right)\left(4+2\sqrt{2}\right)}{16-\left(-2\sqrt{2}\right)^{2}}$$

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

$$\frac{\left(4+3\sqrt{2}\right)\left(4+2\sqrt{2}\right)}{16-\left(-2\right)^{2}\left(\sqrt{2}\right)^{2}}$$

Calculate -2 to the power of 2 and get 4.

$$\frac{\left(4+3\sqrt{2}\right)\left(4+2\sqrt{2}\right)}{16-4\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{\left(4+3\sqrt{2}\right)\left(4+2\sqrt{2}\right)}{16-4\times 2}$$

Multiply 4 and 2 to get 8.

$$\frac{\left(4+3\sqrt{2}\right)\left(4+2\sqrt{2}\right)}{16-8}$$

Subtract 8 from 16 to get 8.

$$\frac{\left(4+3\sqrt{2}\right)\left(4+2\sqrt{2}\right)}{8}$$

Use the distributive property to multiply 4+3\sqrt{2} by 4+2\sqrt{2} and combine like terms.

$$\frac{16+20\sqrt{2}+6\left(\sqrt{2}\right)^{2}}{8}$$

The square of \sqrt{2} is 2.

$$\frac{16+20\sqrt{2}+6\times 2}{8}$$

Multiply 6 and 2 to get 12.

$$\frac{16+20\sqrt{2}+12}{8}$$

Add 16 and 12 to get 28.

$$\frac{28+20\sqrt{2}}{8}$$