Question

solve it fast...please​

Answer 1

Substitute the value of a and b in a² + b² - 4ab

$\sf {\left({\dfrac{{\sqrt{2}+1}}{\sqrt{2}-1}} \right)}^2 + \sf {\left({\dfrac{{\sqrt{2}-1}}{\sqrt{2}+1}} \right)}^2 - 4 \sf {\left({\dfrac{{\sqrt{2}+1}}{\sqrt{2}-1}} \right)} \sf {\left({\dfrac{{\sqrt{2}-1}}{\sqrt{2}+1}} \right)}$

$\sf{= {\left({\dfrac{{\sqrt{2}+1}}{\sqrt{2}-1}} \right)}^2 + \sf {\left({\dfrac{{\sqrt{2}-1}}{\sqrt{2}+1}} \right)}^2 - 4 × {\dfrac{{\sqrt{2}+1}}{\sqrt{2}-1}} × {\dfrac{{\sqrt{2}-1}}{\sqrt{2}+1}}}$

$\sf{\boxed {{\left({\dfrac{{\sqrt{2}+1}}{\sqrt{2}-1}} \right)}^2 = {\dfrac{3+2{\sqrt{2}}}{3-2{\sqrt{2}}}}}}$

$\sf{\boxed{{\left({\dfrac{{\sqrt{2}-1}}{\sqrt{2}+1}} \right)}^2 = {\dfrac{3-2{\sqrt{2}}}{3+2{\sqrt{2}}}}}}$

$\sf{\boxed{{4 × {\dfrac{{\sqrt{2}+1}}{\sqrt{2}-1}} × {\dfrac{{\sqrt{2}-1}}{\sqrt{2}+1 } = 4}}}}$

$\sf={\dfrac{3+2{\sqrt{2}}}{3-2{\sqrt{2}}}} + \sf{\dfrac{3-2{\sqrt{2}}}{3+2{\sqrt{2}}}-4}$

${\boxed{ {\sf{combine~the~fractions~ \sf={\dfrac{3+2{\sqrt{2}}}{3-2{\sqrt{2}}}} + \sf{\dfrac{3-2{\sqrt{2}}}{3+2{\sqrt{2}}}}}} = 34}}$

$\sf{=34-4}$

$\sf{=30}$