Question

Prove that: 15/√10+√20+√40−√5−√80

= √10 + √5
please tell it's urgent​

Answer 1

$\large\underline{\sf{Solution-}}$

Given

$\rm :\longmapsto\:\dfrac{15}{ \sqrt{10} + \sqrt{20} + \sqrt{40} - \sqrt{5} - \sqrt{80} }$

Let we first simply the square root terms in to their simplest form.

$\rm :\longmapsto\: \sqrt{20} = \sqrt{2 \times 2 \times 5} = 2 \sqrt{5}$

$\rm :\longmapsto\: \sqrt{40} = \sqrt{2 \times 2 \times 2 \times 5} = 2\sqrt{10}$

$\rm :\longmapsto\: \sqrt{80} = \sqrt{4 \times 4 \times 5} = 4 \sqrt{5}$

So, now given expression is

$\rm :\longmapsto\:\dfrac{15}{ \sqrt{10} + \sqrt{20} + \sqrt{40} - \sqrt{5} - \sqrt{80} }$

reduced to

$\rm \: = \:\:\dfrac{15}{ \sqrt{10} + 2\sqrt{5} + 2\sqrt{10} - \sqrt{5} - 4\sqrt{5} }$

$\rm \: = \:\dfrac{15}{3 \sqrt{10} - 3 \sqrt{5} }$

$\rm \: = \:\dfrac{15}{3 (\sqrt{10} - \sqrt{5}) }$

$\rm \: = \:\dfrac{5}{\sqrt{10} - \sqrt{5}}$

On rationalizing the denominator, we get

$\rm \: = \:\dfrac{5}{\sqrt{10} - \sqrt{5}} \times \dfrac{ \sqrt{10} + \sqrt{5} }{ \sqrt{10} + \sqrt{5} }$

$\rm \: = \:\dfrac{5( \sqrt{10} + \sqrt{5})}{ {( \sqrt{10} )}^{2} - {( \sqrt{5} )}^{2} }$

$\red{\bigg \{ \because \:(x + y)(x - y) = {x}^{2} - {y}^{2} \bigg \}}$

$\rm \: = \:\dfrac{5( \sqrt{10} + \sqrt{5})}{ 10 - 5}$

$\rm \: = \:\dfrac{5( \sqrt{10} + \sqrt{5})}{5}$

$\rm \: = \: \sqrt{10} + \sqrt{5}$

Hence,

$\bf :\longmapsto\:\dfrac{15}{ \sqrt{10} + \sqrt{20} + \sqrt{40} - \sqrt{5} - \sqrt{80} }$

$\bf \: = \: \sqrt{10} + \sqrt{5}$

More Identities to know:

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)