![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. |x-1| = x-1 khi:
\(x-1\ge0\:\Leftrightarrow\:x\:\ge1\)
Vậy x là 1 số thực bất kì lớn hơn hoặc bằng 1
b. \(\left|x-1\right|=1-x=-x+1=-\left(x-1\right)\)
|x-1| = -(x-1) khi: \(x-1\:\le\:0\Leftrightarrow\:x\:\le1\)
Vậy x là 1 số thực bất kì ko vượt quá 1.
c. x-1 < |x-1|
Mà |x-1| = x-1 hoặc |x-1| = 1-x
=> |x-1| = 1-x
=> |x-1| = -(x-1)
Do đó x - 1 âm => \(x\le1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{2}\left(\frac{4}{9}-x\right)-\frac{3}{2}\left(16-x\right)+\frac{1}{2}\left(5x+10\right)=0\)
\(\Leftrightarrow\frac{2}{9}-\frac{1}{2}x-24+\frac{3}{2}x+\frac{5}{2}x+5=0\)
\(\Leftrightarrow-\frac{169}{9}=\frac{7}{2}x\Leftrightarrow x=-\frac{338}{63}\)
Sai thì thông cảm cho mk nha
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left[\frac{4}{7}-\frac{1}{2}.x\right]^3=2^3\)
\(\Rightarrow\frac{4}{7}-\frac{1}{2}.x=2\)
\(\frac{1}{2}.x=-\frac{10}{7}\)
\(x=-\frac{20}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)Ta có:
\(\left(x-3,5\right)^2+\left(y-\dfrac{1}{10}\right)^4\le0\)
\(\Rightarrow x-3,5=y-\dfrac{1}{10}=0\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\y=\dfrac{1}{10}=0,1\end{matrix}\right.\)
b) Ta có:
\(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=\dfrac{-6}{7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{35}\\x=\dfrac{-13}{35}\end{matrix}\right.\)
b: ta có: \(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{-1}{7}\\5x=\dfrac{-13}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{35}\\x=\dfrac{-13}{35}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(9^{x-1}=\frac{1}{9}\)
\(9^{x-1}=9^{-1}\)
=> x - 1 = - 1
x = -1
x = 0
9x-1= \(\frac{1}{9}\)
9x-1= 9-1
=> x-1=-1
x= -1-1
x= 0
tk nhé
`(1)^2+x=3/4`
`1+x=3/4`
`x=3/4-1`
`x=-1/4`