P = (a/ab+a-2012)+(b/b+bc+1)-(2012c/ac-2012c-2012)

theo đề,có abc=-2012 hoăc 2012=-abc,thay vào,có

P=(a/ab+a+abc)+(b/b+bc+1)+(abc²/ac+abc²+abc)

P=(ac/ac+abc+abc²)+(abc/ac+abc+abc²)+(abc²/ac+abc+abc²)  (cái này là bước quy đồng mẫu số)

P=1

Thay abc=2012 vào ta có:

\(A=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{abc^2}{ca+abc^2+abc}\)

\(=\frac{a}{a\left(b+1+bc\right)}+\frac{b}{bc+b+1}+\frac{ca\left(bc\right)}{ca\left(1+bc+b\right)}\)

\(=\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{bc}{1+bc+b}\)

\(=\frac{1+b+bc}{1+b+bc}=1\)

Câu hỏi của Hà Văn Minh Hiếu - Toán lớp 8 - Học toán với OnlineMath

Ta có : \(a+b+c=6\)

\(\Rightarrow\left(a+b+c\right)^2=36\)

\(\Rightarrow a^2+b^2+c^2+2.\left(ab+bc+ca\right)=36\)

\(\Rightarrow a^2+b^2+c^2=36-2.12=12\)

Do đó : \(a^2+b^2+c^2=ab+bc+ca\left(=12\right)\)

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

Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

Khi đó biểu thức :

\(\left(a-b\right)^{2012}+\left(b-c\right)^{2013}+\left(c-a\right)^{2014}=0+0+0=0\)

|\(x\)| = 1 ⇒ (|\(x\)|)² = 1 ⇒ \(x^2\) = 1

Thay \(x^2\) = 1 vào biểu thức: M = (\(x^{2^{ }}\) + a)(\(x^2\) + b)(\(x^2\) + c) ta có:

M = (1 + a)(1 + b)(1 + c)

M = (1 + b + a + ab)(1 + c)

M = 1 + b + a + ab + c + bc + ac + abc

M = 1 + ( a + b + c) + (ab + bc + ac) + abc

M = 1 + 2 + (-5) +  3

M = (1+2+3) - 5

M = 1

Ta có: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)

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

\(=2\left(a^2+b^2+c^2+2ab+2ac+2bc\right)-6ab-6bc-6ac\)

\(=2\left(a+b+c\right)^2-6\left(ab+bc+ac\right)\)

\(=2.6^2-6.12=0\)

Mà : \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(a-c\right)^2\ge0\)

nên \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

Do đó: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

<=> \(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(a-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\Leftrightarrow a=b=c\)

Vậy \(\left(a-b\right)^{2012}+\left(b-c\right)^{2013}+\left(c-a\right)^{2014}=0\)