Các bạn giúp Shin với nha!!!
Cho \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0\) và \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2\) ( với \(a;b;c;x;y;z\ne0\) )
Khi đó giá trị của biểu thức \(A=\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}\) là:
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=2\Leftrightarrow\left(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}\right)^2=4\Leftrightarrow\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}+2\left(\dfrac{ab}{xy}+\dfrac{ac}{xz}+\dfrac{bc}{yz}\right)=\dfrac{4\Leftrightarrow a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}+2\dfrac{abc}{xyz}\left(\dfrac{z}{c}+\dfrac{y}{b}+\dfrac{x}{a}\right)=\dfrac{4\Leftrightarrow a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}+2\dfrac{abc}{xyz}.0=\dfrac{4\Leftrightarrow a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}=4\)
Ta có: \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0\)
\(\Rightarrow\frac{xbc+yac+zab}{abc}=0\\ \Rightarrow xbc+yac+zab=0\\ \)
Và:\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2\)
\(\Rightarrow\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)^2=2^2\\ \Rightarrow\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}+2\left(\frac{ab}{xy}+\frac{bc}{yz}+\frac{ac}{xy}\right)=4\\ \Rightarrow\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}+2\frac{abz+bcx+acy}{xyz}=4\\ \Rightarrow\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}=4\)
(Vì xab + yac + zab = 0)
