Câu hỏi của Vân Nguyễn Thị

Question

Cho 3 số hữu tỉ x, y, z thỏa mãn với xyz(3x + y + z)(3y + z + x)(3z + x + y) $\neq$ 0 thỏa mãn điều kiện $\dfrac{x}{y+z+3x}=\dfrac{y}{z+x+3y}=\dfrac{z}{x+y+3z}$. Tính giá trị biểu thức:

A = $\left(2+\dfrac{y+z}{x}\right)\left(2+\dfrac{z+x}{y}\right)\left(2+\dfrac{x+y}{z}\right)$

Trên con đường thành côn... · Accepted Answer

Xét $x+y+z=0$$\Leftrightarrow\left\{{}\begin{matrix}y+z=-x\z+x=-y\x+y=-z\end{matrix}\right.$$\Rightarrow A=\left(2-1\right)\left(2-1\right)\left(2-1\right)=1$Xét $x+y+z\ne0$ thì ta có:$\dfrac{x}{y+z+3x}=\dfrac{y}{z+x+3y}=\dfrac{z}{x+y+3z}=\dfrac{x+y+z}{5x+5y+5z}=\dfrac{x+y+z}{5\left(x+y+z\right)}=\dfrac{1}{5}$$\Rightarrow\left\{{}\begin{matrix}5x=y+z+3x\5y=z+x+3y\5z=x+y+3z\end{matrix}\right.$$\Leftrightarrow\left\{{}\begin{matrix}2x=y+z\2y=z+x\2z=x+y\end{matrix}\right.$$\Rightarrow A=\left(2+2\right)\left(2+2\right)\left(2+2\right)=64$Vậy $\left[{}\begin{matrix}A=1\A=64\end{matrix}\right.$Câu trả lời: Xét $x+y+z=0$$\Leftrightarrow\left\{{}\begin{matrix}y+z=-x\z+x=-y\x+y=-z\end{matrix}\right.$$\Rightarrow A=\left(2-1\right)\left(2-1\right)\left(2-1\right)=1$Xét $x+y+z\ne0$ thì ta có:$\dfrac{x}{y+z+3x}=\dfrac{y}{z+x+3y}=\dfrac{z}{x+y+3z}=\dfrac{x+y+z}{5x+5y+5z}=\dfrac{x+y+z}{5\left(x+y+z\right)}=\dfrac{1}{5}$$\Rightarrow\left\{{}\begin{matrix}5x=y+z+3x\5y=z+x+3y\5z=x+y+3z\end{matrix}\right.$$\Leftrightarrow\left\{{}\begin{matrix}2x=y+z\2y=z+x\2z=x+y\end{matrix}\right.$$\Rightarrow A=\left(2+2\right)\left(2+2\right)\left(2+2\right)=64$Vậy $\left[{}\begin{matrix}A=1\A=64\end{matrix}\right.$