Question

(7√3/√10+√3)-(2√5/√6+√5)-(3√2/√15+3√2). Simplify

Answer 1

Given:

The expression -

$\bf \longmapsto \: \dfrac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}} -\dfrac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}} - \dfrac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$

What To Find:

We have to -

• Simplify the expression.

Solution:

$\sf \longmapsto \: \dfrac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}} -\dfrac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}} - \dfrac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$

• Rationalising the denominator of -

$\sf \longmapsto \: \dfrac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}$

Here the conjugate of the denominator is -

$\sf \longmapsto \: \sqrt{10} - \sqrt{3}$

Multiply it with the expression,

$\sf \longmapsto \: \dfrac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}} \times \dfrac{\sqrt{10}-\sqrt{3}}{\sqrt{10}-\sqrt{3}}$

Take them as common,

$\sf \longmapsto \: \dfrac{7\sqrt{3} \times \sqrt{10}-\sqrt{3}}{\sqrt{10}+\sqrt{3} \times \sqrt{10}-\sqrt{3}}$

Solve the numerator,

$\sf \longmapsto \: \dfrac{7\sqrt{30}-21}{\sqrt{10}+\sqrt{3} \times \sqrt{10}-\sqrt{3}}$

Solve the denominator using the identity - (a + b) (a - b) = a² - b²,

$\sf \longmapsto \: \dfrac{7\sqrt{30}-21}{(\sqrt{10})^2-(\sqrt{3})^2}$

Find the squares,

$\sf \longmapsto \: \dfrac{7\sqrt{30}-21}{10-3}$

Subtract 3 from 10,

$\sf \longmapsto \: \dfrac{7\sqrt{30}-21}{7}$

Take 7 as a common factor,

$\sf \longmapsto \: \dfrac{7(\sqrt{30}-3)}{7}$

Cancel 7,

$\sf \longmapsto \: \sqrt{30}-3$

• Rationalising the denominator of -

$\sf \longmapsto \: \dfrac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}$

Here the conjugate of the denominator is -

$\sf \longmapsto \: \sqrt{6} - \sqrt{5}$

Multiply it with the expression,

$\sf \longmapsto \: \dfrac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}} \times \dfrac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}}$

Take them as common,

$\sf \longmapsto \: \dfrac{2\sqrt{5} \times \sqrt{6}-\sqrt{5}}{\sqrt{6}+\sqrt{5} \times \sqrt{6}-\sqrt{5}}$

Solve the numerator,

$\sf \longmapsto \: \dfrac{2\sqrt{30}-10}{\sqrt{6}+\sqrt{5} \times \sqrt{6}-\sqrt{5}}$

Solve the denominator using the identity - (a + b) (a - b) = a² - b²,

$\sf \longmapsto \: \dfrac{2\sqrt{30}-10}{(\sqrt{6})^2-(\sqrt{5})^2}$

Find the squares,

$\sf \longmapsto \: \dfrac{2\sqrt{30}-10}{6 - 5}$

Subtract 5 from 6,

$\sf \longmapsto \: \dfrac{2\sqrt{30}-10}{1}$

Can be written as,

$\sf \longmapsto \: 2\sqrt{30}-10$

• Rationalising the denominator of -

$\sf \longmapsto \: \dfrac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$

Here the conjugate of the denominator is -

$\sf \longmapsto \: \sqrt{15} - 3\sqrt{2}$

Multiply it with the expression,

$\sf \longmapsto \: \dfrac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}} \times \dfrac{\sqrt{15}-3\sqrt{2}}{\sqrt{15}-3\sqrt{2}}$

Take them as common,

$\sf \longmapsto \: \dfrac{3\sqrt{2} \times \sqrt{15}-3\sqrt{2}}{\sqrt{15}+3\sqrt{2} \times \sqrt{15}-3\sqrt{2}}$

Solve the numerator,

$\sf \longmapsto \: \dfrac{3\sqrt{30} - 18}{\sqrt{15}+3\sqrt{2} \times \sqrt{15}-3\sqrt{2}}$

Solve the denominator using the identity - (a + b) (a - b) = a² - b²,

$\sf \longmapsto \: \dfrac{3\sqrt{30} - 18}{(\sqrt{15})^2-(3\sqrt{2})^2}$

Find the squares,

$\sf \longmapsto \: \dfrac{3\sqrt{30} - 18}{15-18}$

Subtract 18 from 15,

$\sf \longmapsto \: \dfrac{3\sqrt{30} - 18}{-3}$

Also written as,

$\sf \longmapsto \: -\dfrac{3\sqrt{30} - 18}{3}$

Take 3 as a common factor,

$\sf \longmapsto \: -\dfrac{3(\sqrt{30} -6)}{3}$

Cancel 3,

$\sf \longmapsto \: -(\sqrt{30} -6)$

Remove the brackets,

$\sf \longmapsto \: 6 - \sqrt{30}$

• Simplifying the expression -

$\sf \longmapsto \: \dfrac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}} -\dfrac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}} - \dfrac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$

After rationalising each term,

$\sf \longmapsto \: (\sqrt{30}-3) - (2\sqrt{30}-10) - (6-\sqrt{30})$

Remove the brackets,

$\sf \longmapsto \: \sqrt{30}-3 - 2\sqrt{30}+10 - 6+\sqrt{30}$

Rearrange the terms,

$\sf \longmapsto \: \underline{\sqrt{30} +\sqrt{30}- 2\sqrt{30}} \:\: \underline{- 3 + 10 - 6}$

Solve the first term,

$\sf \longmapsto \: 2\sqrt{30}- 2\sqrt{30} \:\: \underline{- 3 + 10 - 6}$

Further, solve the first term,

$\sf \longmapsto \: \underline{- 3 + 10 - 6}$

Solve the second term,

$\sf \longmapsto \: 7 - 6$

Further, solve the second term,

$\sf \longmapsto \: 1$

Final Answer:

∴ Thus, the answer is 1 after simplifying the expression.