Question

cho: x^2+y^2+z^2=xy+yz+zx tinh A= (1+x/y) .(1+y/z) .(1+z/x)

Answer 1

\(x^2+y^2+z^2=xy+yz+zx\)

\(\Leftrightarrow\)\(x^2+y^2+z^2-xy-yz-zx=0\)

\(\Leftrightarrow\)\(2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)

\(\Leftrightarrow\)\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}}\)\(\Leftrightarrow\)\(x=y=z\)

\(A=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)