Câu hỏi của lalalalala12345

Question

Cho $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$và x,y,x khác 0CM: $\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}$

Phùng Minh Quân · Accepted Answer

Ta có : $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$$\Leftrightarrow$$\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^3=0^3$$\Leftrightarrow$$\left(\frac{1}{x}\right)^3+\left(\frac{1}{y}\right)^3+\left(\frac{1}{z}\right)^3+3\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{z}+\frac{1}{x}\right)=0$$\Leftrightarrow$$\frac{1^3}{x^3}+\frac{1^3}{y^3}+\frac{1^3}{z^3}=-3\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{z}+\frac{1}{x}\right)$Lại có : $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$$\Rightarrow$$\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}=\frac{-1}{z}\\frac{1}{y}+\frac{1}{z}=\frac{-1}{x}\\frac{1}{z}+\frac{1}{x}=\frac{-1}{y}\end{cases}}$$\Leftrightarrow$$\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\left(-3\right).\frac{-1}{z}.\frac{-1}{x}.\frac{-1}{y}$$\Leftrightarrow$$\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}$ ( đpcm ) Vậy nếu $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$ thì $\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}$Chúc bạn học tốt ~ Câu trả lời: Ta có : $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$$\Leftrightarrow$$\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^3=0^3$$\Leftrightarrow$$\left(\frac{1}{x}\right)^3+\left(\frac{1}{y}\right)^3+\left(\frac{1}{z}\right)^3+3\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{z}+\frac{1}{x}\right)=0$$\Leftrightarrow$$\frac{1^3}{x^3}+\frac{1^3}{y^3}+\frac{1^3}{z^3}=-3\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{z}+\frac{1}{x}\right)$Lại có : $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$$\Rightarrow$$\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}=\frac{-1}{z}\\frac{1}{y}+\frac{1}{z}=\frac{-1}{x}\\frac{1}{z}+\frac{1}{x}=\frac{-1}{y}\end{cases}}$$\Leftrightarrow$$\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\left(-3\right).\frac{-1}{z}.\frac{-1}{x}.\frac{-1}{y}$$\Leftrightarrow$$\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}$ ( đpcm ) Vậy nếu $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$ thì $\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}$Chúc bạn học tốt ~

Đàm Thị Minh Hương · Answer

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{-1}{z}$$\Rightarrow\left(\frac{1}{x}+\frac{1}{y}\right)^3=\left(-\frac{1}{z}\right)^3\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{3}{x^2y}+\frac{3}{xy^2}=-\frac{1}{z^3}$$\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{-3}{x^2y}-\frac{3}{xy^2}=\frac{-3}{xy}.\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{-3}{xy}.-\frac{1}{z}=\frac{3}{xyz}$Câu trả lời: $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{-1}{z}$$\Rightarrow\left(\frac{1}{x}+\frac{1}{y}\right)^3=\left(-\frac{1}{z}\right)^3\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{3}{x^2y}+\frac{3}{xy^2}=-\frac{1}{z^3}$$\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{-3}{x^2y}-\frac{3}{xy^2}=\frac{-3}{xy}.\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{-3}{xy}.-\frac{1}{z}=\frac{3}{xyz}$

Đinh quang hiệp · Answer

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{1}{x}+\frac{1}{y}=-\frac{1}{z};\frac{1}{x}+\frac{1}{z}=-\frac{1}{y};\frac{1}{y}+\frac{1}{z}=-\frac{1}{x}$$2\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\left(\frac{1}{x^3}+\frac{1}{z^3}\right)+\left(\frac{1}{y^3}+\frac{1}{z^3}\right)$$=\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x^2}-\frac{1}{xy}+\frac{1}{y^2}\right)+\left(\frac{1}{x}+\frac{1}{z}\right)\left(\frac{1}{x^2}-\frac{1}{xz}+\frac{1}{z^2}\right)+\left(\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{y^2}-\frac{1}{yz}+\frac{1}{z^2}\right)$$=-\frac{1}{z}\left(\frac{1}{x^2}-\frac{1}{xy}+\frac{1}{y^2}\right)-\frac{1}{y}\left(\frac{1}{x^2}-\frac{1}{xz}+\frac{1}{z^2}\right)-\frac{1}{x}\left(\frac{1}{y^2}-\frac{1}{yz}+\frac{1}{z^2}\right)$$=-\frac{1}{x^2z}+\frac{1}{xyz}-\frac{1}{y^2z}-\frac{1}{x^2y}+\frac{1}{xyz}-\frac{1}{yz^2}-\frac{1}{xy^2}+\frac{1}{xyz}-\frac{1}{xz^2}$$=\left(-\frac{1}{x^2z}-\frac{1}{x^2y}\right)+\left(-\frac{1}{xy^2}-\frac{1}{y^2z}\right)+\left(-\frac{1}{xz^2}-\frac{1}{yz^2}\right)+\frac{3}{xyz}$$=-\frac{1}{x^2}\left(\frac{1}{z}+\frac{1}{y}\right)-\frac{1}{y^2}\left(\frac{1}{x}+\frac{1}{z}\right)-\frac{1}{z^2}\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{3}{xyz}$$=-\frac{1}{x^2}\cdot-\frac{1}{x}+-\frac{1}{y^2}\cdot-\frac{1}{y}+-\frac{1}{z^2}\cdot-\frac{1}{z}+\frac{3}{xyz}=\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}+\frac{3}{xyz}$$\Rightarrow2\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}+\frac{3}{xyz}\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}$(đpcm)Câu trả lời: $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{1}{x}+\frac{1}{y}=-\frac{1}{z};\frac{1}{x}+\frac{1}{z}=-\frac{1}{y};\frac{1}{y}+\frac{1}{z}=-\frac{1}{x}$$2\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\left(\frac{1}{x^3}+\frac{1}{z^3}\right)+\left(\frac{1}{y^3}+\frac{1}{z^3}\right)$$=\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x^2}-\frac{1}{xy}+\frac{1}{y^2}\right)+\left(\frac{1}{x}+\frac{1}{z}\right)\left(\frac{1}{x^2}-\frac{1}{xz}+\frac{1}{z^2}\right)+\left(\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{y^2}-\frac{1}{yz}+\frac{1}{z^2}\right)$$=-\frac{1}{z}\left(\frac{1}{x^2}-\frac{1}{xy}+\frac{1}{y^2}\right)-\frac{1}{y}\left(\frac{1}{x^2}-\frac{1}{xz}+\frac{1}{z^2}\right)-\frac{1}{x}\left(\frac{1}{y^2}-\frac{1}{yz}+\frac{1}{z^2}\right)$$=-\frac{1}{x^2z}+\frac{1}{xyz}-\frac{1}{y^2z}-\frac{1}{x^2y}+\frac{1}{xyz}-\frac{1}{yz^2}-\frac{1}{xy^2}+\frac{1}{xyz}-\frac{1}{xz^2}$$=\left(-\frac{1}{x^2z}-\frac{1}{x^2y}\right)+\left(-\frac{1}{xy^2}-\frac{1}{y^2z}\right)+\left(-\frac{1}{xz^2}-\frac{1}{yz^2}\right)+\frac{3}{xyz}$$=-\frac{1}{x^2}\left(\frac{1}{z}+\frac{1}{y}\right)-\frac{1}{y^2}\left(\frac{1}{x}+\frac{1}{z}\right)-\frac{1}{z^2}\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{3}{xyz}$$=-\frac{1}{x^2}\cdot-\frac{1}{x}+-\frac{1}{y^2}\cdot-\frac{1}{y}+-\frac{1}{z^2}\cdot-\frac{1}{z}+\frac{3}{xyz}=\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}+\frac{3}{xyz}$$\Rightarrow2\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}+\frac{3}{xyz}\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}$(đpcm)

Khoá học trên OLM (olm.vn)

Khoá học trên OLM (olm.vn)