Ta có: \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0\)
=>\(\frac{x}{a}+\frac{y}{b}=0-\frac{z}{c}\)
=>\(\frac{bx+ay}{ab}=-\frac{z}{c}\)
=>\(\frac{1}{\frac{bx+ay}{ab}}=\frac{1}{-\frac{z}{c}}\)
=>\(\frac{ab}{bx+ay}=-\frac{c}{z}\)
=>ab.z=-c.(bx+ay)
=>abz=-(bcx+acy)
Lại có: \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2\)
=>\(\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)^2=2^2\)
=>\(\frac{a}{x}.\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{x}\right)+\frac{b}{y}.\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)+\frac{c}{z}.\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)=4\)
=>\(\frac{a^2}{x^2}+\frac{ab}{xy}+\frac{ac}{xz}+\frac{ab}{xy}+\frac{b^2}{y^2}+\frac{bc}{yz}+\frac{ac}{xz}+\frac{bc}{yz}+\frac{c^2}{z^2}=4\)
=>\(\left(\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}\right)+2.\left(\frac{ab}{xy}+\frac{ac}{xz}+\frac{bc}{yz}\right)=4\)
=>\(A+2.\left(\frac{abz}{xyz}+\frac{acy}{xyz}+\frac{bcx}{xyz}\right)=4\)
=>\(A+2.\frac{abz+acy+bcx}{xyz}=4\)
Vì abz=-(acy+bcx)
=>\(A+2.\frac{-\left(acy+bcx\right)+\left(acy+bcx\right)}{xyz}=4\)
=>\(A+2.\frac{0}{xyz}=4\)
=>\(A+2.0=4\)
=>A+0=4
=>A=4
Vậy A=4