Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left[12\cdot15-x\right]\cdot\frac{1}{4}=120\cdot\frac{1}{4}\)
\(\Leftrightarrow\left[180-x\right]\cdot\frac{1}{4}=30\)
\(\Leftrightarrow180-x=30:\frac{1}{4}\)
\(\Leftrightarrow180-x=120\)
\(\Leftrightarrow x=60\)
$\frac{\frac{2010}{2011}}{\frac{2012}{2013}}+\frac{\frac{2011}{2012}}{\frac{2013}{2014}}+\frac{\frac{2012}{2013}}{\frac{2014}{2015}}$
$\frac{\frac{2010}{2011}}{\frac{2012}{2013}}+\frac{\frac{2011}{2012}}{\frac{2013}{2014}}+\frac{\frac{2012}{2013}}{\frac{2014}{2015}}$
$\frac{\frac{2010+2011+2012}{2011+2012+2013}}{\frac{2012+2013+2014}{2013+2014+2015}}$
$\frac{\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}}{\frac{2012+2013+2014}{2013+2014+2015}}$
$\frac{\frac{2010+2011+2012}{2011+2012+2013}}{\frac{2012}{2013}+\frac{2013}{2014}+\frac{2014}{2015}}$
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 ^_^"