[tex]\large\underline{\sf{Solution-}}[/tex]
(i)↓
[tex] \sf x=\dfrac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}[/tex]
On rationalising,
[tex]\sf\longmapsto x=\dfrac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}\times\dfrac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2}[/tex]
So,
[tex]\sf\longmapsto x=\dfrac{(\sqrt3-\sqrt2)^2}{(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)}[/tex]
We know that,
- [tex]\sf (a-b)(a+b)=a^2-b^2[/tex]
And,
- [tex]\sf (a-b)^2=a^2-2ab+b^2[/tex]
So,
[tex]\sf\longmapsto x=\dfrac{(\sqrt3)^2-2(\sqrt3)(\sqrt2)+(\sqrt2)^2}{(\sqrt3)^2-(\sqrt2)^2}[/tex]
[tex]\sf\longmapsto x=\dfrac{3-2\sqrt6+2}{3-2}[/tex]
[tex]\sf\longmapsto x=\dfrac{5-2\sqrt6}{1}[/tex]
Hence,
[tex]\sf\longmapsto x=5-2\sqrt6[/tex]
(ii)↓
[tex]\sf y=\dfrac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}[/tex]
On rationalising,
[tex]\sf\longmapsto y=\dfrac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\times\dfrac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}[/tex]
So,
[tex]\sf\longmapsto y=\dfrac{(\sqrt3+\sqrt2)^2}{(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)}[/tex]
We know that,
- [tex]\sf (a-b)(a+b)=a^2-b^2[/tex]
And,
- [tex]\sf (a+b)^2=a^2+2ab+b^2[/tex]
So,
[tex]\sf\longmapsto y=\dfrac{(\sqrt3)^2+2(\sqrt3)(\sqrt2)+(\sqrt2)^2}{(\sqrt3)^2-(\sqrt2)^2}[/tex]
[tex]\sf\longmapsto y=\dfrac{3+2\sqrt6+2}{3-2}[/tex]
[tex]\sf\longmapsto y=\dfrac{5+2\sqrt6}{1}[/tex]
Hence,
[tex]\sf\longmapsto y=5+2\sqrt6[/tex]
Now, finding the value of,↓↓
[tex]\sf x^2+y^2+xy[/tex]
Now, substituting the values of x and y in equation,
[tex]\sf\longmapsto (5-2\sqrt6)^2+(5+2\sqrt6)^2+(5-2\sqrt6)(5+2\sqrt6)[/tex]
So,
[tex]\sf\longmapsto [(5)^2-2(5)(2\sqrt6)+(2\sqrt6)^2]+[(5)^2+2(5)(2\sqrt6)+(2\sqrt6)^2]+[(5)^2-(2\sqrt6)^2][/tex]
[tex]\sf\longmapsto 25-20\sqrt6+24+25+20\sqrt6+24+(25-24)[/tex]
Cancelling 20√6,
[tex]\sf\longmapsto 25+24+25+24+(25-24)[/tex]
[tex]\sf\longmapsto 98+1[/tex]
[tex]\longmapsto\bf 99[/tex]
[tex]\textsf{Therefore, the value of x²+y²+xy is 99.}\\[/tex]
