Question

[ღHàn Thiên Băng ღ]

Cho x,y,z thỏa mãn: x + y + z = 7; x² + y² + z² = 23; xyz = 3

Tính giá trị \(A=\frac{1}{xy+z-6}+\frac{1}{yz+x-6}+\frac{1}{zx+y-6}\)

Answer 1

[alibaba nguyễn]

Ta có:

\(xy+yz+zx=\frac{\left(x+y+z\right)^2-x^2-y^2-z^2}{2}=\frac{7^2-23}{2}=13\)

Ta lại có:

\(xy+z-6=xy+z+1-x-y-z=\left(x-1\right)\left(y-1\right)\)

\(\Rightarrow A=\frac{1}{\left(x-1\right)\left(y-1\right)}+\frac{1}{\left(y-1\right)\left(z-1\right)}+\frac{1}{\left(z-1\right)\left(x-1\right)}\)

\(=\frac{x+y+z-3}{xyz-xy-yz-zx+x+y+z-1}=-1\)