Question

1/ Cho x, y, z khác 0 và xy + yz + zx = 0. 

Tính S= (y+z)/x + (z+x)/y + (x+y)/z

2/ Cho x= y+1. C/m (x + y)(x² + y²)(x⁴ + y⁴)= (x⁸ - y⁸)

 

Answer 1

2) \(x=y+1\Rightarrow x-y=1\)

\(\Rightarrow\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)=x^8-y^8\)

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)=x^8-y^8\)

\(\Leftrightarrow\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)=x^8-y^8\)

\(\Leftrightarrow\left(x^4-y^4\right)\left(x^4+y^4\right)=x^8-y^8\)

\(\Leftrightarrow x^8-y^8=x^8-y^8\)(đúng)

Vậy \(\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)=x^8-y^8\)(đpcm)