Solve 1/7sqrt{3+5sqrt{2}}

$$\frac { 1 } { 7 \sqrt { 3 } + 5 \sqrt { 2 } }$$

Evaluate

$$\frac{7\sqrt{3}-5\sqrt{2}}{97}\approx 0.052095751$$

Solution Steps

Rationalize the denominator of \frac{1}{7\sqrt{3}+5\sqrt{2}} by multiplying numerator and denominator by 7\sqrt{3}-5\sqrt{2}.

$$\frac{7\sqrt{3}-5\sqrt{2}}{\left(7\sqrt{3}+5\sqrt{2}\right)\left(7\sqrt{3}-5\sqrt{2}\right)}$$

Consider \left(7\sqrt{3}+5\sqrt{2}\right)\left(7\sqrt{3}-5\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}.

$$\frac{7\sqrt{3}-5\sqrt{2}}{\left(7\sqrt{3}\right)^{2}-\left(5\sqrt{2}\right)^{2}}$$

Expand \left(7\sqrt{3}\right)^{2}.

$$\frac{7\sqrt{3}-5\sqrt{2}}{7^{2}\left(\sqrt{3}\right)^{2}-\left(5\sqrt{2}\right)^{2}}$$

Calculate 7 to the power of 2 and get 49.

$$\frac{7\sqrt{3}-5\sqrt{2}}{49\left(\sqrt{3}\right)^{2}-\left(5\sqrt{2}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{7\sqrt{3}-5\sqrt{2}}{49\times 3-\left(5\sqrt{2}\right)^{2}}$$

Multiply 49 and 3 to get 147.

$$\frac{7\sqrt{3}-5\sqrt{2}}{147-\left(5\sqrt{2}\right)^{2}}$$

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

$$\frac{7\sqrt{3}-5\sqrt{2}}{147-5^{2}\left(\sqrt{2}\right)^{2}}$$

Calculate 5 to the power of 2 and get 25.

$$\frac{7\sqrt{3}-5\sqrt{2}}{147-25\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{7\sqrt{3}-5\sqrt{2}}{147-25\times 2}$$

Multiply 25 and 2 to get 50.

$$\frac{7\sqrt{3}-5\sqrt{2}}{147-50}$$

Subtract 50 from 147 to get 97.

$$\frac{7\sqrt{3}-5\sqrt{2}}{97}$$