Question

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Answer 1

ANSWER:

Given:

• x = (√3-√2)/(√3+√2)
• y = (√3+√2)/(√3-√2)

To Find:

• x^2 + y^2 + xy

Solution:

We are given that,

$\implies x=\dfrac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}$

On rationalising,

$\implies x=\dfrac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}\times\dfrac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2}$

So,

$\implies x=\dfrac{(\sqrt3-\sqrt2)^2}{(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)}$

We know that,

$\hookrightarrow (a-b)(a+b)=a^2-b^2$

And,

$\hookrightarrow (a-b)^2=a^2-2ab+b^2$

So,

$\implies x=\dfrac{(\sqrt3)^2-2(\sqrt3)(\sqrt2)+(\sqrt2)^2}{(\sqrt3)^2-(\sqrt2)^2}$

$\implies x=\dfrac{3-2\sqrt6+2}{3-2}$

$\implies x=\dfrac{5-2\sqrt6}{1}$

Hence,

$\implies x=5-2\sqrt6$

We are also given,

$\implies y=\dfrac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}$

On rationalising,

$\implies y=\dfrac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\times\dfrac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}$

So,

$\implies y=\dfrac{(\sqrt3+\sqrt2)^2}{(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)}$

We know that,

$\hookrightarrow (a-b)(a+b)=a^2-b^2$

And,

$\hookrightarrow (a+b)^2=a^2+2ab+b^2$

So,

$\implies y=\dfrac{(\sqrt3)^2+2(\sqrt3)(\sqrt2)+(\sqrt2)^2}{(\sqrt3)^2-(\sqrt2)^2}$

$\implies y=\dfrac{3+2\sqrt6+2}{3-2}$

$\implies y=\dfrac{5+2\sqrt6}{1}$

Hence,

$\implies y=5+2\sqrt6$

Now, we need to find,

$\implies x^2+y^2+xy$

So substituting values of x and y,

$\implies (5-2\sqrt6)^2+(5+2\sqrt6)^2+(5-2\sqrt6)(5+2\sqrt6)$

So,

$\implies [(5)^2-2(5)(2\sqrt6)+(2\sqrt6)^2]+[(5)^2+2(5)(2\sqrt6)+(2\sqrt6)^2]+[(5)^2-(2\sqrt6)^2]$

$\implies 25-20\sqrt6+24+25+20\sqrt6+24+(25-24)$

Cancelling 20√6,

$\implies 25+24+25+24+(25-24)$

$\implies 98+1$

Hence,

$\implies\bf 99$

Therefore, the answer is 99.

Answer 2

Step-by-step explanation:

the answer is 99

hope it helps ^^

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