Question

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

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

b) Biết a+b+c = 0

Tính: B\(=\dfrac{ab}{a^2+b^2-c^2}+\dfrac{bc}{b^2+c^2-a^2}+\dfrac{ac}{c^2+a^2-b^2}\)

Answer 1

a,Sửa lại đề nha bạn:Tính A = \(\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}\)

\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=\dfrac{bcx+acy+abz}{abc}=0\)

\(\Rightarrow bcz+acy+abz=0\)

(2) \(\Rightarrow\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}+2\left(\dfrac{ab}{xy}+\dfrac{ac}{xz}+\dfrac{bc}{xz}\right)=4\)\(\Rightarrow A=\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}=4-2.\left(\dfrac{abz+acy+bcz}{xyz}\right)=4\)b, \(a+b+c=0\Rightarrow a+b=-c\)

\(\Rightarrow a^2+2ab+b^2=c^2\Rightarrow a^2+b^2-c^2=-2ab\)Tương tự: \(b^2+c^2-a^2=-2bc\)

\(c^2+a^2-b^2=-2bc\)

Vậy \(B=\dfrac{ab}{-2ab}+\dfrac{bc}{-2bc}+\dfrac{ac}{-2ac}=\dfrac{-3}{2}\)