Question

[Treallagx]

Giúp mình với mn ơi

Answer 1

[HT.Phong (9A5)]

Bài 1: Ta có:

\(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\)

\(A=\dfrac{a^3}{abc}+\dfrac{b^3}{abc}+\dfrac{c^3}{abc}\)

\(A=\dfrac{a^3+b^3+c^3}{abc}\) (2) 

Mà: \(a+b+c=0\)

\(\Rightarrow\left(a+b+c\right)^3=0\)

\(\Rightarrow a^3+b^3+c^3+3a^2b+3ab^2+3b^2c+3bc^2+3a^2c+3ac^2+6abc=0\)

\(\Rightarrow a^3+b^3+c^3+\left(3a^2b+3ab^2+3abc\right)+\left(3b^2c+3bc^2+3abc\right)+\left(3a^2c+3ac^2+3abc\right)-3abc=0\)

\(\Rightarrow a^3+b^3+c^3+3ab\left(a+b+c\right)+3ac\left(a+b+c\right)+3bc\left(a+b+c\right)-3abc=0\)

\(\Rightarrow a^3+b^3+c^3+\left(a+b+c\right)\left(3ab+3ac+3bc\right)-3abc=0\) (1)

Thay \(a+b+c=0\) (1) ta có:
\(a^3+b^3+c^3+0\cdot\left(3ab+3ac+3bc\right)-3abc=0\)

\(\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow a^3+b^3+c^3=3abc\)

Thay vào (2) ta có:

\(\dfrac{3abc}{abc}=3\)

ậy

Answer 2

[Nguyễn Lê Phước Thịnh]

1:

a+b=c=0

=>a+b=-c; a+c=-b; b+c=-a

\(A=\dfrac{a^3+b^3+c^3}{abc}\)

\(=\dfrac{\left(a+b\right)^3-3ab\left(a+b\right)+c^3}{abc}=\dfrac{\left(-c\right)^3+3bac+c^3}{abc}\)

=3abc/abc=3