Question

 

Cho 

\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0\)     ;       \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2\)

Tính giá trị của biểu thức

\(A=\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}\)   

Answer 1

\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0\)  =>\(\frac{xbc+yac+zab}{abc}=0\)=>\(xbc+yac+zab=0\)(1)

\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2\)=>\(\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)^2=4\)<=>\(\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}+2\left(\frac{ab}{xy}+\frac{bc}{yz}+\frac{ac}{xz}\right)=4\)

<=>\(\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}+2\left(\frac{abz+bcx+acy}{xyz}\right)=4\)mà abz+bcx+acy=0 ( từ 1) nên \(\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}=4\)

Nhớ . hehe ^_^ 

                               

=>\(xbc+yac+zab=0\)