Question

[Blue Frost]

Các bạn ơi cho mình hỏi:

Cho a,c,b khác 0 và phân biệt t/m: a^3 +b^3 +c^3=3abc

Tính M= ab^2/a^2+b^2-c^2 + bc^2/b^2+c^2-a^2 + ca^2/ c^2+a^2-b^2

CẢM ƠN CÁC BẠN NHIỀU!!!

Answer 1

[Đinh quang hiệp]

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

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

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

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

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

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\Rightarrow\hept{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-ac-bc=0\end{cases}}\)

\(\left(a-b\right)^2>=0\Rightarrow a^2-2ab+b^2>=0\Rightarrow a^2+b^2>=2ab\)

tương tự \(a^2+c^2>=2ac;b^2+c^2>=2bc\)

\(\Rightarrow a^2+b^2+a^2+c^2+b^2+c^2>=2ab+2ac+2bc\Rightarrow2\left(a^2+b^2+c^2\right)>=2\left(ab+ac+bc\right)\)

\(\Rightarrow a^2+b^2+c^2.=ab+ac+bc\)dấu = xảy ra khi a=b=c

mà nếu \(a^2+b^2+c^2-ab-ac-bc=0\Rightarrow a^2+b^2+c^2=ab+ac+bc\Rightarrow a=b=c\)

th1:a+b+c=0

\(\Rightarrow a+b=-c;a+c=-b;b+c=-a\)

\(M=\frac{ab^2}{a^2+b^2-c^2}+\frac{bc^2}{b^2+c^2-a^2}+\frac{ca^2}{c^2+a^2-b^2}=\frac{ab^2}{a^2+b^2-\left(-c\right)^2}+\frac{bc^2}{b^2+c^2-\left(-a\right)^2}+\frac{ca^2}{c^2+a^2-\left(-b\right)^2}\)

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

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

\(=\frac{ab^2}{-2ab}+\frac{bc^2}{-2bc}+\frac{ca^2}{-2ac}=\frac{b}{-2}+\frac{c}{-2}+\frac{a}{-2}=\frac{a+b+c}{-2}=\frac{0}{-2}=0\)

th2:a=b=c tự lm nhá