2012 x 2013 - 1013 x 2013 + 2013

= 2012 x 2013 - 1013 x 2013 + 2013 x 1

= 2013 x ( 2012 - 1013 + 1 )

= 2013 x         1000

=       2013000

a) 2012x2013-1013x2013+2013

=2013(2012-1013+1)

=2013x1000

=2013000

\(\frac{2011\times2012+2013\times21+1991}{2012\times2013-2012\times2012}\)

\(=\frac{2011\times2012+2013\times\left(21+1991\right)}{2012\times2013-2012\times2012}\)

\(=\frac{2011\times2012+2013\times2012}{2012\times2013-2012\times2012}=\frac{2011}{2012}\)

<=> \(\frac{\left(x+2014\right)}{2011}+1+\frac{\left(x+2013\right)}{2012}+1=\frac{\left(x+2012\right)}{2013}+1+\frac{\left(x+2011\right)}{2014}+1\)

\(\Rightarrow\frac{\left(x+4025\right)}{2011}+\frac{\left(x+4025\right)}{2012}=\frac{\left(x+4025\right)}{2013}+\frac{\left(x+4025\right)}{2014}\)

=> \(\frac{\left(x+4025\right)}{2011}+\frac{\left(x+4025\right)}{2012}-\frac{\left(x+4025\right)}{2013}-\frac{\left(x+4025\right)}{2014}=0\)

=> \(\left(x+4025\right)\left\lbrack\left(\frac{1}{2011}+\frac{1}{2012}\right)-\left(\frac{1}{2013}+\frac{1}{2014}\right)\right\rbrack=0\)

vì \(\left(\frac{1}{2011}+\frac{1}{2012}\right)>\left(\frac{1}{2013}+\frac{1}{2014}\right)\)

=> \(\left\lbrack\left(\frac{1}{2011}+\frac{1}{2012}\right)-\left(\frac{1}{2013}+\frac{1}{2014}\right)\right\rbrack>0\) hay ≠0

=> \(x+4025=0\)

\(x=-4025\)

bạn ơi,đáp án bằng 2024 đó.

đáp án 100% là 2024

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)

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

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

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

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

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

1.

<=> B=\(3^{24}-\left\lbrack\left(3^3\right)^4+1\right\rbrack\left\lbrack\left(3^2\right)^6-1\right\rbrack\)

\(B=3^{24}-\left(3^{12}+1\right)\left(3^{12}-1\right)\)

\(B=3^{24}-3^{24}+1\)

\(B=1\)

2.

xét vế đầu tiên

\(2011\cdot2013+2012\cdot2014\)

\(=\left(2012-1\right)\left(2012+1\right)+\left(2013-1\right)\left(2013+1\right)\)

\(=2012^2-1+2013^2-1\)

\(=2012^2+2013^2-2\)

=> \(2011\cdot2013+2012\cdot2014=2012^2+2013^2-2\)