Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\left(x-\frac{1}{2}\right)^4=\frac{1}{81}\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^4=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{3}\\x-\frac{1}{2}=\frac{-1}{3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{1}{6}\end{cases}}\)
Vậy ...
Với x\(\ge1\)\(x-1-\sqrt{x-1}=0< =>x-1=\sqrt{x-1}< =>\left(x-1\right)^2=x-1< =>\left(x-1\right)^2-\left(x-1\right)=0< =>\left(x-1\right)\left(x-1-1\right)=0< =>\left(x-1\right)\left(x-2\right)=0\)\(< =>\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=1\left(TM\right)\\x=2\left(TM\right)\end{matrix}\right.\)
Ta thấy
3^2017 > 3^2015
5^2015/3 > 1^2013/3
5 >1
Suy ra A = 3^2017 + 5^2015/3 + 5 > B= 3^2015 + 1^2013/3 + 1
Vậy A>B
\(\Rightarrow\frac{x+5}{2015}+1+\frac{x+4}{2016}+1+\frac{x+3}{2017}+1=\frac{x+2015}{5}+1+\frac{x+2016}{4}+1+\frac{x+2017}{3}+1\)
\(\Rightarrow\frac{x+2020}{2015}+\frac{x+2020}{2016}+\frac{x+2020}{2017}=\frac{x+2020}{5}+\frac{x+2020}{4}+\frac{x+2020}{3}\)
\(\Rightarrow\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
\(\Rightarrow x=-2020\)
\(\left|x+1\right|và\left|x+2\right|\ge0\)
\(\Rightarrow\orbr{\begin{cases}\left(x+1\right)+\left(x+2\right)=3\\\left(x+1\right)+\left(x+2\right)=-3\end{cases}}\)
\(\orbr{\begin{cases}2x+3=3\\2x+3=-3\end{cases}}\)
\(\orbr{\begin{cases}2x=0\\2x=-6\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)
\(\left|x+1\right|+\left|x+2\right|=3\)
Xét \(x+1\ge0;x+2\ge0\Leftrightarrow x\ge-1;x\ge-2\Rightarrow x\ge-1\) ta có : \(\hept{\begin{cases}\left|x+1\right|=x+1\\\left|x+2\right|=x+2\end{cases}}\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|=3\Leftrightarrow x+1+x+2=3\Leftrightarrow2x+3=3\Rightarrow x=0\)(TM)
Xét \(x+1\le0;x+2\ge0\Leftrightarrow-2\le x\le-1\) ta có : \(\hept{\begin{cases}\left|x+1\right|=-x-1\\\left|x+2\right|=x+2\end{cases}}\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|=3\Leftrightarrow-x-1+x+2=3\Leftrightarrow1=3\) (loại)
Xét \(x+1\le0;x+2\le0\Leftrightarrow x\le-1;x\le-2\Leftrightarrow x\le-2\) ta có : \(\hept{\begin{cases}\left|x+1\right|=-x-1\\\left|x+2\right|=-x-2\end{cases}}\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|=-x-1-x-2=-2x-3=3\Rightarrow x=-3\)(TM)
Vậy \(x=\left\{-3;0\right\}\)
