Ta có :

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

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

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

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

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

\(=-ab\left(a+b+c\right)=0\)

Ta có: \(a+b+c=0\)

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

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

\(M=a^3+b^3+ca^2+cb^2-abc\)

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

\(M=a^2.\left(-b\right)+b^2.\left(-a\right)\)

\(M=-a^2b-b^2a\)

\(M=-ab.\left(a+b\right)\)

\(M=-ab.\left(-c\right)\)

\(M=abc\)

Tham khảo nhé~

Ta có a+b+c=0 <=> a+b=-c

M= (a+b)(a²-ab+b²)+a²c+b²c-abc

=-ca²+abc-cb²+a²c+b²c-abc

=0

Chúc học tốt!!!!

Ta có A=\(\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)

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

\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2=a^2-ab+b^2+3ab\left(1-2ab\right)+6a^2b^2\)

=\(\left(a+b\right)^2-3ab+3ab-6a^2b^2+6a^2b^2=1\)

2) Ta có \(A=\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc-ab-bc-ca+a+b+c-1=0\)

bài 3 : Ta có \(A=\left(x-y\right)\left(x^2+xy+y^2\right)-36xy=12\left(x^2+xy+y^2\right)-36xy=12\left(x^2-2xy+y^2\right)\)

\(=12\left(x-y\right)^2=12.12^2=1728\)

\(a+b+c=0\Leftrightarrow a+b=-c\) thay vào : 

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

\(=-c^3-3ab.\left(-c\right)+c\left[c^2-2ab\right]-abc\)

\(=-c^3+3abc+c^3-2abc-abc=0\)

hello

ta có : a+b+c=0=>a+b=-c ; b+c=-a ; a+c=-b 

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

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

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

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

M=\(\frac{2ab}{-2ab}+\frac{2bc}{-2bc}+\frac{2ca}{-2ca}\)

M=-1-1-1=-3

Vậy với a+b+c=0 thì M=-3

Bài 1:

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

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

\(\Leftrightarrow ab+bc+ac=0\Leftrightarrow bc=-ab-ac\)

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

CMTT: \(\left\{{}\begin{matrix}\dfrac{b^2}{b^2+2ca}=\dfrac{b^2}{\left(b-c\right)\left(b-a\right)}\\\dfrac{c^2}{c^2+2ab}=\dfrac{c^2}{\left(b-c\right)\left(a-c\right)}\end{matrix}\right.\)

\(M=\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=\dfrac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)

Bài 2:

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

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

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

\(\Rightarrow A=\dfrac{0}{\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3}=0\)

\(a+b+c=0\Leftrightarrow a=-b-c\Leftrightarrow a^2=b^2+c^2+2bc\Leftrightarrow a^2-b^2-c^2=2bc\)

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

Ta có : \(T=\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-a^2-b^2}=\frac{a^2}{2bc}+\frac{b^2}{2ca}+\frac{c^2}{2ab}\)

\(=\frac{1}{2abc}\left(a^3+b^3+c^3\right)\)(1)

Ta sẽ chứng minh nếu a + b + c = 0 thì \(a^3+b^3+c^3=3abc\)

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

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

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

= 0

=> \(a^3+b^3+c^3=3abc\) thay vào (1) được : 

\(T=\frac{a^3+b^3+c^3}{2abc}=\frac{3abc}{2abc}=\frac{3}{2}\)

\(a.\) Với  \(a+b+c=0\)  thì  \(\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{\left(-c\right).\left(-a\right).\left(-b\right)}{abc}=\frac{-abc}{abc}=-1\)

\(b.\)   Công thức tổng quát:  \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

Ta có:

\(\frac{1}{x\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)

\(\frac{1}{\left(x+1\right)\left(x+2\right)}=\frac{1}{x+1}-\frac{1}{x+2}\)

\(\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{1}{x+2}-\frac{1}{x+3}\)

\(\frac{1}{\left(x+3\right)\left(x+4\right)}=\frac{1}{x+3}-\frac{1}{x-4}\)

\(\frac{1}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+4}-\frac{1}{x+5}\)

Do đó, suy ra được:  \(A=\frac{1}{x}-\frac{1}{x+5}=\frac{x+5-x}{x\left(x+5\right)}=\frac{5}{x\left(x+5\right)}\)

Sửa đề cho nó đẹp

\(\frac{\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3}{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}\)

\(=\frac{3\left(a-b\right)\left(a-c\right)\left(c-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=-3\)

em ms hok lớp 1