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=>(2x-7)^2017[(2x-7)^2-1]=0
=>(2x-7)(2x-8)(2x-6)=0
hay \(x\in\left\{3;3.5;4\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left(2x-3\right)\left(2x-\dfrac{7}{2}\right)\left(2x-\dfrac{5}{2}\right)=0\)
\(\Leftrightarrow x\in\left\{\dfrac{3}{2};\dfrac{7}{4};\dfrac{5}{4}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
|2x+2017|=2019
=>\(|^{2x+2017=2019}_{2x+2017=-2019}\)
=>\(|^{2x=2019-2017}_{2x=-2019-2017}\)
=>\(|^{2x=2}_{2x=-4036}\)
=>\(|^{x=2:2}_{x=-4036:2}\)
=>\(|^{x=1}_{x=-2018}\)
Vậy x\(\in\left\{1;-2018\right\}\)
Chúc bn học tốt!
|2x + 2017|= 2019
\(\Rightarrow\)2x + 2017 = 2019 & 2x + 2017 = -2019
TH1: 2x + 2017 = 2019
2x = 2019 - 2017
2x = 2
x = 2 ÷ 2
x = 1
TH2: 2x + 2017 = -2019
2x = -2019 - 2017
2x = - 4036
x = -4036 ÷ 2
x = - 2018
Vây x \(\in\){ 1 ; -2018}
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>x-2017=0 và y-2018=0
=>x=2017; y=2018
b: =>3x-y=0 và y+2/3=0
=>y=-2/3 và 3x=-2/3
=>x=-2/9 và y=-2/3
c: =>3/4x-1/2=0 và 4/5y+6/25=0
=>x=2/3 và y=-3/10
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(3x-7\right)^{2019}=\left(3x-7\right)^{2017}\)
\(\Rightarrow\left(3x-7\right)^{2019}-\left(3x-7\right)^{2017}=0\)
\(\Rightarrow\left(3x-7\right)^{2017}\left[\left(3x-7\right)^2-1\right]=0\)
\(\Rightarrow\left(3x-7\right)^{2017}=0\text{ hoặc }\left[\left(3x-7\right)^2-1\right]=0\)
\(\Rightarrow3x-7=0\text{ hoặc }\left(3x-7\right)^2=1\)
\(\Rightarrow3x-7=0\text{ hoặc } \hept{\begin{cases}3x-7=1\\3x-7=-1\end{cases}}\)
\(\Rightarrow3x=7\text{ hoặc }3x=8\text{ hoặc }3x=6\)
\(\Rightarrow x=\frac{7}{3}\text{ hoặc }x=\frac{8}{3}\text{ hoặc }x=2\)
\(\left(3x-7\right)^{2019}=\left(3x-7\right)^{2017}\)
\(\left(3x-7\right)^{2019}-\left(3x-7\right)^{2017}=0\)
\(\left(3x-7\right)^{2017}\cdot\left[\left(3x-7\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(3x-7\right)^{2017}=0\\\left(3x-7\right)^2-1=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}3x-7=0\\\left(3x-7\right)^2=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=7\\3x-7=\pm1\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\3x=6\text{ hoặc }3x=8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=2\text{ hoặc }x=\frac{8}{3}\end{cases}}\)
\(\Rightarrow x\in\left\{\frac{7}{3};2;\frac{8}{3}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 2017-Ix-2017I=x
\(TH1:x>0\)
\(\Rightarrow2017-\left(x-2017\right)=x\)
\(\Rightarrow2017-x+2017\)
\(\Rightarrow4034-x=x\) ( loại )
\(TH2:x\le0\)
\(\Rightarrow2017-\left[-\left(x-2017\right)\right]=x\)
\(\Rightarrow2017-\left(-x+2017\right)=x\)
\(\Rightarrow2017+x-2017=x\)
\(\Rightarrow x+0=x\)
\(\Rightarrow x=x\) \(\left(x\in Z\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) để \(\left(x-7\right)^{x+2015}-\left(x-7\right)^{x+2016}=0\)
thì \(\left(x-7\right)^{x+2015}=\left(x-7\right)^{x+2016}\)
mà \(x+2015
bài a)
|2x+3|=x+2
2x+3=x+2 hoặc -(2x+3)=x+2
2x-x=2-3 -2x-3=x+2
1x=-1 -2x-x=2+3
x=-1 -3x =5
x=\(\frac{-5}{3}\)
x = 4
\(\Leftrightarrow\left(2x-7\right)^{2017}\left[\left(2x-7\right)^2-1\right]=0\)
=>(2x-7)(2x-6)(2x-8)=0
hay \(x\in\left\{3;\dfrac{7}{2};4\right\}\)