Câu hỏi của Nguyễn Khắc Quang

Question

Cho các số x,y,z thỏa mãn: $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$ và $\frac{2}{xy}-\frac{1}{z^2}=4$Tính giá trị của biểu thức:$C=\left(x+2y+z\right)^{2021}$

Greninja · Accepted Answer

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$$\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=4$$\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}=\frac{2}{xy}-\frac{1}{z^2}$$\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}-\frac{2}{xy}+\frac{1}{z^2}=0$$\Leftrightarrow\left(\frac{1}{x^2}+\frac{2}{xz}+\frac{1}{z^2}\right)+\left(\frac{1}{y^2}+\frac{2}{yz}+\frac{1}{z^2}\right)=0$$\Leftrightarrow\left(\frac{1}{x}+\frac{1}{z}\right)^2+\left(\frac{1}{y}+\frac{1}{z}\right)^2=0$$\Leftrightarrow\hept{\begin{cases}\frac{1}{x}+\frac{1}{z}=0\\frac{1}{y}+\frac{1}{z}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{1}{x}=\frac{1}{-z}\\frac{1}{y}=\frac{1}{-z}\end{cases}\Leftrightarrow}\frac{1}{x}=\frac{1}{y}=\frac{1}{-z}$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$$\Leftrightarrow\frac{1}{-z}+\frac{1}{-z}+\frac{1}{z}=2$$\Leftrightarrow z=\frac{-1}{2}$$x=y=\frac{1}{2}$$\Rightarrow C=\left(x+2y+z\right)^{2021}=\left(\frac{1}{2}+2.\frac{1}{2}-\frac{1}{2}\right)^{2021}=1^{2021}=1$Câu trả lời: $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$$\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=4$$\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}=\frac{2}{xy}-\frac{1}{z^2}$$\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}-\frac{2}{xy}+\frac{1}{z^2}=0$$\Leftrightarrow\left(\frac{1}{x^2}+\frac{2}{xz}+\frac{1}{z^2}\right)+\left(\frac{1}{y^2}+\frac{2}{yz}+\frac{1}{z^2}\right)=0$$\Leftrightarrow\left(\frac{1}{x}+\frac{1}{z}\right)^2+\left(\frac{1}{y}+\frac{1}{z}\right)^2=0$$\Leftrightarrow\hept{\begin{cases}\frac{1}{x}+\frac{1}{z}=0\\frac{1}{y}+\frac{1}{z}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{1}{x}=\frac{1}{-z}\\frac{1}{y}=\frac{1}{-z}\end{cases}\Leftrightarrow}\frac{1}{x}=\frac{1}{y}=\frac{1}{-z}$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$$\Leftrightarrow\frac{1}{-z}+\frac{1}{-z}+\frac{1}{z}=2$$\Leftrightarrow z=\frac{-1}{2}$$x=y=\frac{1}{2}$$\Rightarrow C=\left(x+2y+z\right)^{2021}=\left(\frac{1}{2}+2.\frac{1}{2}-\frac{1}{2}\right)^{2021}=1^{2021}=1$

Ngô Chi Lan · Answer

Ta có:$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\Rightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=4$$\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{2}{xy}-\frac{1}{z^2}$$\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}-\frac{2}{xy}+\frac{1}{z^2}=0$$\Leftrightarrow\left(\frac{1}{x^2}+\frac{2}{xz}+\frac{1}{z^2}\right)+\left(\frac{1}{y^2}+\frac{2}{yz}+\frac{1}{z^2}\right)=0$$\Leftrightarrow\left(\frac{1}{x}+\frac{1}{z}\right)^2+\left(\frac{1}{y}+\frac{1}{z}\right)^2=0$$\Leftrightarrow\hept{\begin{cases}\left(\frac{1}{x}+\frac{1}{z}\right)^2=0\\left(\frac{1}{y}+\frac{1}{z}\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{1}{x}=-\frac{1}{z}\\frac{1}{y}=-\frac{1}{z}\end{cases}}}$$\Leftrightarrow x=y=-z$Thay vào $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$ta được :$x=y=\frac{1}{2};z=-\frac{1}{2}$$\Rightarrow P=\left(\frac{1}{2}+2.\frac{1}{2}-\frac{1}{2}\right)^{2021}=1^{2020}=1$Câu trả lời: Ta có:$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\Rightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=4$$\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{2}{xy}-\frac{1}{z^2}$$\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}-\frac{2}{xy}+\frac{1}{z^2}=0$$\Leftrightarrow\left(\frac{1}{x^2}+\frac{2}{xz}+\frac{1}{z^2}\right)+\left(\frac{1}{y^2}+\frac{2}{yz}+\frac{1}{z^2}\right)=0$$\Leftrightarrow\left(\frac{1}{x}+\frac{1}{z}\right)^2+\left(\frac{1}{y}+\frac{1}{z}\right)^2=0$$\Leftrightarrow\hept{\begin{cases}\left(\frac{1}{x}+\frac{1}{z}\right)^2=0\\left(\frac{1}{y}+\frac{1}{z}\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{1}{x}=-\frac{1}{z}\\frac{1}{y}=-\frac{1}{z}\end{cases}}}$$\Leftrightarrow x=y=-z$Thay vào $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$ta được :$x=y=\frac{1}{2};z=-\frac{1}{2}$$\Rightarrow P=\left(\frac{1}{2}+2.\frac{1}{2}-\frac{1}{2}\right)^{2021}=1^{2020}=1$

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