Câu hỏi của Hiếu Minh

Question

Cmr:

Nếu $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\:$ và x+y+z=xyz thì $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=2$

Nguyễn Lê Phước Thịnh · Accepted Answer

Ta có: $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$

⇔$\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=4$

⇔$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\left(\frac{1}{xy}+\frac{1}{zy}+\frac{1}{xz}\right)=4$

⇔$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2\left(x+y+z\right)}{xyz}=4$

⇔$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2xyz}{xyz}=4$

⇔$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2=4$

⇔$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=2$(đpcm)Câu trả lời: Ta có: $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$

⇔$\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=4$

⇔$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\left(\frac{1}{xy}+\frac{1}{zy}+\frac{1}{xz}\right)=4$

⇔$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2\left(x+y+z\right)}{xyz}=4$

⇔$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2xyz}{xyz}=4$

⇔$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2=4$

⇔$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=2$(đpcm)

Nguyễn Lê Phước Thịnh · Answer

Ta có: $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$

$\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=2$

$\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)=2$

$\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2\left(x+y+z\right)}{xyz}=2$

$\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2xyz}{xyz}=2$

$\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2=2$

$\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=0$

=> Đề có vấn đề rồi bạnCâu trả lời: Ta có: $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$