Question

Cho \(x,y,z\in R\) thỏa mãn \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\). Hãy tính giá trị của biểu thức \(M=\dfrac{3}{4}+\left(x^8-y^8\right)\left(y^9+z^9\right)\left(z^{10}-x^{10}\right)\).

Answer 1

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\\ \Leftrightarrow\dfrac{x+y}{xy}+\left(\dfrac{1}{z}-\dfrac{1}{x+y+z}\right)=0\\ \Leftrightarrow\dfrac{x+y}{xy}+\dfrac{x+y}{z\left(x+y+z\right)}=0\\ \Leftrightarrow\left(x+y\right)\left(\dfrac{1}{xy}+\dfrac{1}{xz+yz+z^2}\right)=0\\ \)

Nếu x+y=0 => x=-y

Nếu

\(\dfrac{1}{xy}+\dfrac{1}{xz+yz+z^2}=0\\ \Rightarrow xz+yz+z^2+xy=0\\ \Rightarrow\left(x+z\right)\left(y+z\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-z\\y=-z\end{matrix}\right.\)

Tự thế vào :v