Question

giup mình vơi bài này

cho x khác 0,y khác 0,zkhác 0

\(\frac{1}{x}-\frac{1}{y}-\frac{1}{z}=1\)và x=y+z

chứng minh rằng:\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)=1

Answer 1

theo bài ra ta có : \(\left(\frac{1}{x}-\frac{1}{y}-\frac{1}{z}\right)^2=1^2=1\)

Ta thấy

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

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

\(=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-2\left(\frac{0}{xyz}\right)=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\) vì x = y+z nê y+z-x = 0

Vậy \(\left(\frac{1}{x}-\frac{1}{y}-\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=1ĐPCM\)