Question

已知x=√2=1分之1,y=√3+1分之2,求x²+2x-2y-y²分之x²y+y-xy&

Answer 1

x=1/(√2-1)=√2+1

y=2/(√3+1)=2(√3-1)/[(√3+1)(√3-1)]=√3-1

(x^2y+y-xy^2-x)/(x^2+2x-2y-y^2)

=[xy(x-y)+(y-x)]/[(x^2-y^2)+2(x-y)]

=[(x-y)(xy-1)]/[(x-y)(x+y+2)]

=(xy-1)/(x+y+2)

=[(√3-1)(√2+1)-1]/(√3+√2+2)

=(√6+√3-√2-2)/(√3+√2+2)