Solve frac{5sqrt{6}-2sqrt{3}}{5sqrt{6}+2sqrt{3}}=

$$\frac { 5 \sqrt { 6 } - 2 \sqrt { 3 } } { 5 \sqrt { 6 } + 2 \sqrt { 3 } } =$$

Evaluate

$$\frac{27-10\sqrt{2}}{23}\approx 0.559037582$$

Solution Steps

Rationalize the denominator of \frac{5\sqrt{6}-2\sqrt{3}}{5\sqrt{6}+2\sqrt{3}} by multiplying numerator and denominator by 5\sqrt{6}-2\sqrt{3}.

$$\frac{\left(5\sqrt{6}-2\sqrt{3}\right)\left(5\sqrt{6}-2\sqrt{3}\right)}{\left(5\sqrt{6}+2\sqrt{3}\right)\left(5\sqrt{6}-2\sqrt{3}\right)}$$

Consider \left(5\sqrt{6}+2\sqrt{3}\right)\left(5\sqrt{6}-2\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(5\sqrt{6}-2\sqrt{3}\right)\left(5\sqrt{6}-2\sqrt{3}\right)}{\left(5\sqrt{6}\right)^{2}-\left(2\sqrt{3}\right)^{2}}$$

Multiply 5\sqrt{6}-2\sqrt{3} and 5\sqrt{6}-2\sqrt{3} to get \left(5\sqrt{6}-2\sqrt{3}\right)^{2}.

$$\frac{\left(5\sqrt{6}-2\sqrt{3}\right)^{2}}{\left(5\sqrt{6}\right)^{2}-\left(2\sqrt{3}\right)^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5\sqrt{6}-2\sqrt{3}\right)^{2}.

$$\frac{25\left(\sqrt{6}\right)^{2}-20\sqrt{6}\sqrt{3}+4\left(\sqrt{3}\right)^{2}}{\left(5\sqrt{6}\right)^{2}-\left(2\sqrt{3}\right)^{2}}$$

The square of \sqrt{6} is 6.

$$\frac{25\times 6-20\sqrt{6}\sqrt{3}+4\left(\sqrt{3}\right)^{2}}{\left(5\sqrt{6}\right)^{2}-\left(2\sqrt{3}\right)^{2}}$$

Multiply 25 and 6 to get 150.

$$\frac{150-20\sqrt{6}\sqrt{3}+4\left(\sqrt{3}\right)^{2}}{\left(5\sqrt{6}\right)^{2}-\left(2\sqrt{3}\right)^{2}}$$

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$\frac{150-20\sqrt{3}\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}}{\left(5\sqrt{6}\right)^{2}-\left(2\sqrt{3}\right)^{2}}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$\frac{150-20\times 3\sqrt{2}+4\left(\sqrt{3}\right)^{2}}{\left(5\sqrt{6}\right)^{2}-\left(2\sqrt{3}\right)^{2}}$$

Multiply -20 and 3 to get -60.

$$\frac{150-60\sqrt{2}+4\left(\sqrt{3}\right)^{2}}{\left(5\sqrt{6}\right)^{2}-\left(2\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{150-60\sqrt{2}+4\times 3}{\left(5\sqrt{6}\right)^{2}-\left(2\sqrt{3}\right)^{2}}$$

Multiply 4 and 3 to get 12.

$$\frac{150-60\sqrt{2}+12}{\left(5\sqrt{6}\right)^{2}-\left(2\sqrt{3}\right)^{2}}$$

Add 150 and 12 to get 162.

$$\frac{162-60\sqrt{2}}{\left(5\sqrt{6}\right)^{2}-\left(2\sqrt{3}\right)^{2}}$$

Expand \left(5\sqrt{6}\right)^{2}.

$$\frac{162-60\sqrt{2}}{5^{2}\left(\sqrt{6}\right)^{2}-\left(2\sqrt{3}\right)^{2}}$$

Calculate 5 to the power of 2 and get 25.

$$\frac{162-60\sqrt{2}}{25\left(\sqrt{6}\right)^{2}-\left(2\sqrt{3}\right)^{2}}$$

The square of \sqrt{6} is 6.

$$\frac{162-60\sqrt{2}}{25\times 6-\left(2\sqrt{3}\right)^{2}}$$

Multiply 25 and 6 to get 150.

$$\frac{162-60\sqrt{2}}{150-\left(2\sqrt{3}\right)^{2}}$$

Expand \left(2\sqrt{3}\right)^{2}.

$$\frac{162-60\sqrt{2}}{150-2^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{162-60\sqrt{2}}{150-4\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{162-60\sqrt{2}}{150-4\times 3}$$

Multiply 4 and 3 to get 12.

$$\frac{162-60\sqrt{2}}{150-12}$$

Subtract 12 from 150 to get 138.

$$\frac{162-60\sqrt{2}}{138}$$