M=a(a+b)(a+c)

N=b(b+c)(b+a)

P=c(c+a)(c+b)

Có a+b+c=0

=> a+b=-c

b+c=-a

a+c=-b

=> M=a(-c)(-b)

=abc

N=b(-a)(-c)

=bac

P=c(-b)(-a)

=cba

=> M=N=P(đpcm)

\(a+b+c=0\)

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

THAY \(a+b=-c;a+c=-b;b+c=-a\)VÀO M;N;P TA CÓ:
\(M=a.\left(-c\right).\left(-b\right)=a.b.c\)(1)

\(N=b.\left(-a\right).\left(-c\right)=a.b.c\)(2)

\(P=c.\left(-b\right).\left(-a\right)=a.b.c\)(3)
Từ (1) ; (2) ; (3) Ta có 

\(M=N=P\left(=a.b.c\right)\)(đpcm)

\(a+b+c=0\Rightarrow\hept{\begin{cases}a+b=-c\\a+c=-b\\b+c=-a\end{cases}}\)

Lần lượt thay vào M, N, P ta có :

\(\Rightarrow\hept{\begin{cases}M=a\cdot\left(-c\right)\cdot\left(-b\right)=a\cdot b\cdot c\\N=b\cdot\left(-a\right)\cdot\left(-c\right)=a\cdot b\cdot c\\P=c\cdot\left(-b\right)\cdot\left(-a\right)=a\cdot b\cdot c\end{cases}}\)

\(\Rightarrow M=N=P\left(đpcm\right)\)

a: M,N,P đều nằm trên trục tung

b; Hoành độ bằng 0

a,b,c phân biệt \(\Rightarrow a\ne b\ne c\)

\(a^2\left(b+c\right)=b^2\left(c+a\right)=2012\)

\(\Rightarrow a^2b-ab^2+a^2c-b^2c=0\)

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

\(\Rightarrow\left(a-b\right)\left(ab+bc+ca\right)=0\)

\(\Rightarrow ab+bc+ca=0\) vì \(a\ne b\)

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

\(\Rightarrow\dfrac{a^2}{a+c}=\dfrac{b^2}{b+c}=\dfrac{a^2-b^2}{a-b}=a+b\)

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

\(\Rightarrow2012=a^2\left(b+c\right)=\left(a+b\right)\left(b+c\right)\left(c+a\right)=\left(a+b\right)\left(ab+bc+ca+c^2\right)=c^2\left(a+b\right)\)

Vậy....................

bn cs tự tin vs câu trả lời of mk ko?

nếu bn lam Đ thì cho mk thank nha!

Bài 1:

a, Ta có:

\(\left(a+b+c\right)^2-\left(ab+bc+ca\right)=0\Leftrightarrow a^2+b^2+c^2+ab+bc+ca=0\)\(\Leftrightarrow2a^2+2b^2+2c^2+2ab+2bc+2ca=0\)

\(\Leftrightarrow\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2=0\Leftrightarrow a+b=b+c=c+a=0\)

\(\Leftrightarrow a=b=c=0\)

Vậy điều kiện để phân thức M được xác định là a, b, c không đồng thời = 0

b, Ta có:

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

Đặt: \(a^2+b^2+c^2=x,ab+bc+ca=y\)

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

Ta cũng có:

\(M=\dfrac{x\left(x+2y\right)+y^2}{x+2y-y}=\dfrac{x^2+2xy+y^2}{x+y}=\dfrac{\left(x+y\right)^2}{x+y}=x+y\)

\(=a^2+b^2+c^2+ab+bc+ca\)

2 ) b )

\(a+b+c+d=0\)

\(\Leftrightarrow a+b=-\left(c+d\right)\)

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

\(\Leftrightarrow a^3+b^3+3a^2b+3b^2a=-c^3-3c^2d-3d^2c-d^3\)

\(\Leftrightarrow a^3+b^3+3a^2b+3b^2a+c^3+3c^2d+3d^2c+d^3=0\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=-3a^2b-3b^2a-3c^2d-3d^2c\)

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

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

\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(ab-cd\right)\left(c+d\right)\) \(\left(đpcm\right)\)