Question

cho \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\)=0 (1) và \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\)=2 (2)

a)Tính giá trị biểu thức A=\(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)

b)Tính P=\(\frac{ab}{a^2+b^2-c^2}+\frac{bc}{b^2+a^2-c^2}+\frac{ca}{c^2+a^2-b^2}\)

Answer 1

a) Từ (1) => bcx + acy + abz = 0

Từ (2) => \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{ab}{xy}+\frac{ac}{xz}+\frac{bc}{yz}\right)=0\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=4-2\left(\frac{abz+acy+bcx}{xyz}\right)=4\)

b) Từ a+b+c = 0 => a+b= -c => \(a^2+b^2-c^2=-2ab\)

Tương tự : \(b^2+c^2-a^2=-2bc;c^2+a^2-b^2=-2ac\)

\(\Rightarrow B=\frac{ab}{-2ab}+\frac{bc}{-2bc}+\frac{ac}{-2ac}=-\frac{3}{2}\)