Question

Cho a,b,c khác 0 ,a+ b +c=0

Chứng minh :\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)Tính giá trị biểu thức P=\(\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}\)

Answer 1

\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{ac}+\frac{2}{bc}\)

\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{a+b+c}{abc}\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\) (đpcm)

Do \(a+b+c=0\Rightarrow a+b=-c\)

\(a^3+b^3+c^3=a^3+b^3+3ab\left(a+b\right)+c^3-3ab\left(a+b\right)\)

\(=\left(a+b\right)^3+c^3-3ab\left(-c\right)\)

\(=\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2\right)+3abc=3abc\)

Vậy \(a^3+b^3+c^3=3abc\)

\(\Rightarrow P=\frac{a^3}{abc}+\frac{b^3}{abc}+\frac{c^3}{abc}=\frac{a^3+b^3+c^3}{abc}=\frac{3abc}{abc}=3\)