$\frac{\frac{2010}{2011}}{\frac{2012}{2013}}+\frac{\frac{2011}{2012}}{\frac{2013}{2014}}+\frac{\frac{2012}{2013}}{\frac{2014}{2015}}$

$\frac{\frac{2010}{2011}}{\frac{2012}{2013}}+\frac{\frac{2011}{2012}}{\frac{2013}{2014}}+\frac{\frac{2012}{2013}}{\frac{2014}{2015}}$

$\frac{\frac{2010+2011+2012}{2011+2012+2013}}{\frac{2012+2013+2014}{2013+2014+2015}}$

$\frac{\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}}{\frac{2012+2013+2014}{2013+2014+2015}}$

$\frac{\frac{2010+2011+2012}{2011+2012+2013}}{\frac{2012}{2013}+\frac{2013}{2014}+\frac{2014}{2015}}$

dễ ợt nhưng éo biết làm thông cảm nha

ban Dang Ha Trang an noi gi ki vay 

\(P=\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}=\frac{2010}{8144863716}+\frac{2011}{8144863716}+\frac{2012}{8144863716}\)

\(=\frac{6033}{8144863716}=\frac{1}{1350052}\)

\(Q=2010+2011+\frac{2012}{2011}+2012+2013\)

\(=2010+2011+2012+2013+\frac{2012}{2011}\)

\(=8046+\frac{2012}{2011}=\frac{8046}{1}+\frac{2012}{2011}\)

\(=\frac{16180506}{2011}+\frac{2012}{2011}=\frac{16182518}{2011}\)

ai giải dùm vs

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)

\(\frac{2010+2011+2012}{2011+2012+2013}=\frac{2010}{2011+2012+2013}+\frac{2011}{2011+2012+2013}+\frac{2012}{2011+2012+2013}\)

Vì \(\frac{2010}{2011+2012+2013}<\frac{2010}{2011};\frac{2011}{2011+2012+2013}<\frac{2011}{2012};\frac{2012}{2011+2012+2013}<\frac{2012}{2013}\)

nên phép dưới nhỏ hơn phép trên

Ta có : \(\frac{2011}{2012}=1-\frac{1}{2012}\)

           \(\frac{2012}{2013}=1-\frac{1}{2013}\)

            \(\frac{2013}{2011}=1+\frac{2}{2011}\)

Ta có : \(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}=\left(1-\frac{1}{2012}\right)+\left(1-\frac{1}{2013}\right)+\left(1+\frac{2}{2011}\right)\)

       =   \(\left(1+1+1\right)+\left(\frac{2}{2011}-\frac{1}{2012}-\frac{1}{2013}\right)\)

       =  \(3+\frac{2}{2011}-\left(\frac{1}{2012}+\frac{1}{2013}\right)\)

Ta có :  

\(\frac{1}{2012}+\frac{1}{2013}< \frac{1}{2012}+\frac{1}{2012}=\frac{2}{2012}\)

mà : \(\frac{2}{2012}< \frac{2}{2011}=>\frac{1}{2012}+\frac{1}{2013}< \frac{2}{2011}\)

=> \(\frac{2}{2011}-\left(\frac{1}{2012}+\frac{1}{2013}\right)>0\)

Vậy : \(3+\frac{2}{2011}-\left(\frac{1}{2012}+\frac{1}{2013}\right)>3\)

Vậy : \(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}>3\)

ủng hộ mik nhá các bạn ơiii ^_^"

bạn tham khảo:

2010/2011+2012+2013 > 2010+2011+2012/2011+2012+2013

2011/2011+2012+2013 > 2010+2011+2012/2011+2012+2013

2012/2011+2012+2013 > 2010+2011+2012/2011+2012+2013

=> 2010/2011+2011/2012+2012/2013 > 2010+2011+2012/2011+2012+2013

2010/2011+2012+2013 > 2010+2011+2012/2011+2012+2013

2011/2011+2012+2013 > 2010+2011+2012/2011+2012+2013

2012/2011+2012+2013 > 2010+2011+2012/2011+2012+2013

=> 2010/2011+2011/2012+2012/2013 > 2010+2011+2012/2011+2012+2013

\(\overline{ }\)

Giả sử z là số lớn nhất trong 3 số 

Từ đề bài ta có:

\(\sqrt{x+2011}+\sqrt{y+2012}+\sqrt{z+2013}=\sqrt{z+2011}+\sqrt{x+2012}+\sqrt{y+2013}\)

\(\Leftrightarrow\sqrt{x+2012}-\sqrt{x+2011}+\sqrt{y+2013}-\sqrt{y+2012}=\sqrt{z+2012}-\sqrt{z+2011}+\sqrt{z+2013}-\sqrt{z+2012}\)

\(\Leftrightarrow\frac{1}{\sqrt{x+2012}+\sqrt{x+2011}}+\frac{1}{\sqrt{y+2013}+\sqrt{y+2012}}=\frac{1}{\sqrt{z+2012}+\sqrt{z+2011}}+\frac{1}{\sqrt{z+2013}+\sqrt{z+2012}}\)

Ta lại có:

\(\hept{\begin{cases}\frac{1}{\sqrt{x+2012}+\sqrt{x+2011}}\ge\frac{1}{\sqrt{z+2012}+\sqrt{z+2011}}\\\frac{1}{\sqrt{y+2013}+\sqrt{y+2012}}\ge\frac{1}{\sqrt{z+2013}+\sqrt{z+2012}}\end{cases}}\)

Dấu = xảy ra khi x = y = z

Tương tự cho trường hợp x lớn nhất với y lớn nhất.

fdy 'rshniytguo;yhuyt65edip;ioy86fo87ogtb eubuiltgr6sdwjhytguyh8 ban oi bai nay mac kho giai vao cut sit

2011/2012+2012/2013+2013/2011

=2011/2012+2012/2013+1+2/2011

(1/2011+2011/2012)+(2012/2013+1/2012)+1

Vì 1/2011<1/2012 nên 1/2011+2011/2012<1

Vì 1/2011<1/2013 nên 1/2011+2012/2013<1

Suy ra C>1+1+1=3

Vậy C>3