2012 . | x - 2011| + (x-2011)² = 2013 . | 2011 - x|

|x-2011|.|x-2011| + 2012 . | x - 2011| - 2013 . | 2011- x| =0

|x - 2011|.| x - 2011| + 2012 .| x - 2011| - 2013 | x - 2011| = 0

| x- 2011| .| x -2011|  - | x - 2011| = 0

| x - 2011|. { | x - 2011| - 1} = 0

\(\left[{}\begin{matrix}\left|x-2011\right|=0\\\left|x-2011\right|-1=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=2011\\x=2012\\x=2010\end{matrix}\right.\)

Kết luận x \(\in\) { 2010; 2011; 2012}

\(\frac{2011.2013-2011.2012}{2012.2011+2011.2013}\)

\(=\frac{2011.\left(2013-2012\right)}{2011.\left(2012+2013\right)}\)

\(=\frac{2011.1}{2011.4025}\)

\(=\frac{1}{4025}\)

bài này là bài Tính nha mọi người giải rõ ra giúp mik nha

\(\frac{2011x2013-2012x2011}{2012x2011+2011x2013}\)

\(=\frac{2011x\left(2013-2012\right)}{2011x\left(2012+2013\right)}\)

\(=\frac{2011x1}{2011x4025}\)

\(=\frac{1}{4025}\)

=>2012|x-2011|-|x-2011|+(x-2011)^2+2013>0

=>2011|x-2011|+(x-2011)^2+2013>0(luôn đúng)

bạn ơi,đáp án bằng 2024 đó.

đáp án 100% là 2024

Kết quả là 2024 nha bạn

Đặt \(\hept{\begin{cases}a=x+2011\\b=y+2011\\c=z+2011\end{cases}}\) Ta có Hệ:

\(\hept{\begin{cases}\sqrt{a}+\sqrt{b+1}+\sqrt{c+2}\left(A\right)=\sqrt{b}+\sqrt{c+1}+\sqrt{a+2}\left(B\right)\\\sqrt{b}+\sqrt{c+1}+\sqrt{a+2}\left(B\right)=\sqrt{c}+\sqrt{a+1}+\sqrt{b+2}\left(C\right)\end{cases}}\)

Vai trò \(x,y,z\) bình đẳng

Giả sử \(c=Max\left(a;b;c\right)\) vì \(A=C\) ta có:

\(\sqrt{a}+\sqrt{b+1}+\sqrt{c+2}=\sqrt{c}+\sqrt{a+1}+\sqrt{b+2}\)

\(\Leftrightarrow\left(\sqrt{a+1}-\sqrt{a}\right)+\left(\sqrt{b+2}-\sqrt{b+1}\right)\)

\(=\sqrt{c+2}-\sqrt{c}=\left(\sqrt{c+2}-\sqrt{c+1}\right)+\left(\sqrt{c+1}-\sqrt{c}\right)\)

\(\Leftrightarrow\frac{1}{\sqrt{a+1}+\sqrt{a}}+\frac{1}{\sqrt{b+2}+\sqrt{b+1}}\)

\(=\frac{1}{\sqrt{c+2}+\sqrt{c+1}}+\frac{1}{\sqrt{c+1}+\sqrt{c}}\left(1\right)\)

Mặt khác \(\hept{\begin{cases}c\ge a\Rightarrow\frac{1}{\sqrt{a+1}+\sqrt{a}}\le\frac{1}{\sqrt{c+1}+\sqrt{c}}\\c\ge b\Rightarrow\frac{1}{\sqrt{b+2}+\sqrt{b+1}}\le\frac{1}{\sqrt{c+2}+\sqrt{c+1}}\end{cases}}\)

Suy ra \(\left(1\right)\) xảy ra khi \(a=b=c\Leftrightarrow x=y=z\) (Đpcm)

=>\(-\left|x-2011\right|+\left(x-2011\right)^2=0\)

\(\Leftrightarrow\left|x-2011\right|\left(\left|x-2011\right|-1\right)=0\)

\(\Leftrightarrow x\in\left\{2011;2012;2010\right\}\)