Question

plz help me......... ​

Answer 1

Given:-

$\longrightarrow\mathrm{x=\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}}$

$\longrightarrow\mathrm{y=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}}$

To Find:-

Value of  $\sf{\dfrac{x^{2}+xy+y^{2}}{x^{2}-xy+y^{2}}}$

Solution:-

We know that,

$\longrightarrow\bf{(a+b)^{2}=a^{2}+2ab+b^{2}}$

$\longrightarrow\bf{(a-b)^{2}=a^{2}-2ab+b^{2}}$

$\longrightarrow\bf{(a+b)(a-b)=a^{2}-b^{2}}$

Now,

$\mathrm{x+y=\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}}$

$\mathrm{x+y=\dfrac{(\sqrt{5}-\sqrt{3})^{2}+(\sqrt{5}+\sqrt{3})^{2}}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}}$

$\mathrm{x+y=\dfrac{5+3-2\sqrt{5}\sqrt{3}+5+3+2\sqrt{5}\sqrt{3}}{5-3}}$

$\mathrm{x+y=\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}}$

$\mathrm{x+y=\dfrac{16}{2}}$

$\mathrm{x+y=8}$

Now again,

$\mathrm{x-y=\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}-\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}}$

$\mathrm{x-y=\dfrac{(\sqrt{5}-\sqrt{3})^{2}-(\sqrt{5}+\sqrt{3})^{2}}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}}$

$\mathrm{x-y=\dfrac{5+3-2\sqrt{5}\sqrt{3}-(5+3+2\sqrt{5}\sqrt{3})}{5-3}}$

$\mathrm{x-y=\dfrac{5+3-2\sqrt{5}\sqrt{3}-5-3-2\sqrt{5}\sqrt{3}}{2}}$

$\mathrm{x-y=\dfrac{8-2\sqrt{15}-8-2\sqrt{15}}{2}}$

$\mathrm{x-y=\dfrac{-4\sqrt{15}}{2}}$

$\mathrm{x-y=-2\sqrt{15}}$

Also,

$\mathrm{xy=\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\times\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}}$

$\mathrm{xy=1}$

So,

$\rightarrow\sf{\dfrac{x^{2}+xy+y^{2}}{x^{2}-xy+y^{2}}}$

On adding and subtracting xy in numerator and denominator, we get

$\rightarrow\sf{\dfrac{x^{2}+xy+y^{2}+xy-xy}{x^{2}-xy+y^{2}+xy-xy}}$

$\rightarrow\sf{\dfrac{x^{2}+2xy+y^{2}-xy}{x^{2}-2xy+y^{2}+xy}}$

$\rightarrow\sf{\dfrac{(x+y)^{2}-xy}{(x-y)^{2}+xy}}$

On feeding values in above equation, we get

$\rightarrow\sf{\dfrac{(8)^{2}-1}{(-2\sqrt{15})^{2}+1}}$

$\rightarrow\sf{\dfrac{64-1}{60+1}}$

$\rightarrow\sf{\dfrac{63}{61}}$

Hence, the required value is  $\bf{\dfrac{63}{61}}$.

Answer 2

Answer:

Given:-

$\longrightarrow\mathrm{x=\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}}$

$\longrightarrow\mathrm{y=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}}$

To Find:-

Value of  $\sf{\dfrac{x^{2}+xy+y^{2}}{x^{2}-xy+y^{2}}}$

Solution:-

We know that,

$\longrightarrow\bf{(a+b)^{2}=a^{2}+2ab+b^{2}}$

$\longrightarrow\bf{(a-b)^{2}=a^{2}-2ab+b^{2}}$

$\longrightarrow\bf{(a+b)(a-b)=a^{2}-b^{2}}$

Now,

$\mathrm{x+y=\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}}$

$\mathrm{x+y=\dfrac{(\sqrt{5}-\sqrt{3})^{2}+(\sqrt{5}+\sqrt{3})^{2}}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}}$

$\mathrm{x+y=\dfrac{5+3-2\sqrt{5}\sqrt{3}+5+3+2\sqrt{5}\sqrt{3}}{5-3}}$

$\mathrm{x+y=\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}}$

$\mathrm{x+y=\dfrac{16}{2}}$

$\mathrm{x+y=8}$

Now again,

$\mathrm{x-y=\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}-\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}}$

$\mathrm{x-y=\dfrac{(\sqrt{5}-\sqrt{3})^{2}-(\sqrt{5}+\sqrt{3})^{2}}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}}$

$\mathrm{x-y=\dfrac{5+3-2\sqrt{5}\sqrt{3}-(5+3+2\sqrt{5}\sqrt{3})}{5-3}}$

$\mathrm{x-y=\dfrac{5+3-2\sqrt{5}\sqrt{3}-5-3-2\sqrt{5}\sqrt{3}}{2}}$

$\mathrm{x-y=\dfrac{8-2\sqrt{15}-8-2\sqrt{15}}{2}}$

$\mathrm{x-y=\dfrac{-4\sqrt{15}}{2}}$

$\mathrm{x-y=-2\sqrt{15}}$

Also,

$\mathrm{xy=\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\times\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}}$

$\mathrm{xy=1}$

So,

$\rightarrow\sf{\dfrac{x^{2}+xy+y^{2}}{x^{2}-xy+y^{2}}}$

On adding and subtracting xy in numerator and denominator, we get

$\rightarrow\sf{\dfrac{x^{2}+xy+y^{2}+xy-xy}{x^{2}-xy+y^{2}+xy-xy}}$

$\rightarrow\sf{\dfrac{x^{2}+2xy+y^{2}-xy}{x^{2}-2xy+y^{2}+xy}}$

$\rightarrow\sf{\dfrac{(x+y)^{2}-xy}{(x-y)^{2}+xy}}$

On feeding values in above equation, we get

$\rightarrow\sf{\dfrac{(8)^{2}-1}{(-2\sqrt{15})^{2}+1}}$

$\rightarrow\sf{\dfrac{64-1}{60+1}}$

$\rightarrow\sf{\dfrac{63}{61}}$

Hence, the required value is  $\bf{\dfrac{63}{61}}$.