Question

cho x,y,z khác 0 và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)

tính P=(x²⁵+y²⁵)(y³+z³)(z²⁰⁰⁶-x²⁰⁰⁶)

Answer 1

Ta có : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)

\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right):\left(\frac{1}{x+y+z}\right)=1\)

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

\(\Leftrightarrow3xyz+yz\left(y+z\right)+xz\left(x+z\right)+xy\left(x+y\right)=xyz\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\) hay B = 0