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câu đầu bạn dưới làm rồi nên mình k làm lại
(2x+9)2=0
=> 2x+9=0
=> 2x=-9
=> x=-9/2
(2x-1)3=8
=> 2x-1=2
=> 2x=3
=> x=3/2
(1-3x)2=16
=> 1-3x=4
=> 3x=-3
=> x=-1
(3x+1)+1=-26
=> 3x=-27
=> x=-9
(x+1)+(x+3)+(x+5)+...+(x+2017)=0
(x+x+x+...+x)+(1+3+5+...+2017)=0
=> 1009x+1018081=0
1009x=-1018081
=> x=-1009
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(|3x+1|=4\)
\(\Rightarrow\orbr{\begin{cases}3x+1=4\\3x+1=-4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=4-1\\3x=-4-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=3\\3x=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\div3\\x=-5\div3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1,6667\end{cases}}\)
Vậy x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+2\right)-2=0\)
\(\Rightarrow x+2-2=0\)
\(\Rightarrow x=0\)
\(\left(x+3\right)+1=7\)
\(\Rightarrow x+3+1=7\)
\(\Rightarrow x+4=7\)
\(\Rightarrow x=3\)
\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)
\(\Rightarrow3x=12\)
\(\Rightarrow x=4\)
\(\left(5x+4\right)-1=13\)
\(\Rightarrow5x+4-1=13\)
\(\Rightarrow5x+3=13\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\)
\(\left(4x-8\right)-3=5\)
\(\Rightarrow4x-8-3=5\)
\(\Rightarrow4x-11=5\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
\(8-\left(2x+4\right)=2\)
\(\Rightarrow8-2x-4=2\)
\(\Rightarrow4-2x=2\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
\(7+\left(5x+2\right)=14\)
\(\Rightarrow7+5x+2=14\)
\(\Rightarrow9+5x=14\)
\(\Rightarrow5x=5\)
\(\Rightarrow x=1\)
\(5-\left(3x-11\right)=1\)
\(\Rightarrow5-3x+11=1\)
\(\Rightarrow16-3x=1\)
\(\Rightarrow3x=15\)
\(\Rightarrow x=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 : Bài giải
\(-2\left(-3-4x\right)-3\left(3x-7\right)=31\)
\(6+8x-9x-21=31\)
\(-x-15=31\)
\(-x=31+15\)
\(-x=46\)
\(x=-46\)
b, \(10\left(x-7\right)=8\left(x-4\right)+x\)
\(10x-70=8x-32+x\)
\(10x-70=9x-32\)
\(10x-9x=-32+70\)
\(x=38\)
c, \(2\left|x-1\right|=3\cdot\left(5-1\right)\)
\(2\left|x-1\right|=15-3\)
\(2\left|x-1\right|=12\)
\(\left|x-1\right|=12\text{ : }2\)
\(\left|x-1\right|=6\)
\(\Rightarrow\orbr{\begin{cases}x-1=-6\\x-1=6\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5\\x=7\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-5\text{ ; }7\right\}\)
d, \(2\left(x+3\right)-2\left(x-3\right)=x\)
\(2\left(x+3-x-3\right)=x\)
\(2\cdot0=x\)
\(x=0\)
e, \(\left(x+3\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-3\text{ ; }4\right\}\)
a, \(-2\left(-3-4x\right)-3\left(3x-7\right)=31\)
\(6+8x-9x-21=31\)
\(-x-15=31\)
\(-x=31+15\)
\(-x=46\)
\(x=-46\)
b, \(10\left(x-7\right)=8\left(x-4\right)+x\)
\(10x-70=8x-32+x\)
\(10x-70=9x-32\)
\(10x-9x=-32+70\)
\(x=38\)
c, \(2\left|x-1\right|=3\cdot\left(5-1\right)\)
\(2\left|x-1\right|=15-3\)
\(2\left|x-1\right|=12\)
\(\left|x-1\right|=12\text{ : }2\)
\(\left|x-1\right|=6\)
\(\Rightarrow\orbr{\begin{cases}x-1=-6\\x-1=6\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5\\x=7\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-5\text{ ; }7\right\}\)
d, \(2\left(x+3\right)-2\left(x-3\right)=x\)
\(2\left(x+3-x-3\right)=x\)
\(2\cdot0=x\)
\(x=0\)
e, \(\left(x+3\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-3\text{ ; }4\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
1. $(x+2)-2=0$
$x+2=2$
$x=0$
2.
$(x+3)+1=7$
$x+3=7-1=6$
$x=6-3=3$
3.
$(3x-4)+4=12$
$3x-4+4=12$
$3x=12$
$x=12:3=4$
4.
$(5x+4)-1=13$
$5x+4=13+1=14$
$5x=14-4=10$
$x=10:5=2$
5.
$(4x-8)-3=5$
$4x-8=5+3=8$
$4x=8+8=16$
$x=16:4=4$
6.
$3+(x-5)=7$
$x-5=7-3=4$
$x=4+5=9$
7.
$8-(2x-4)=2$
$2x-4=8-2=6$
$2x=6+4=10$
$x=10:2=5$
8.
$7+(5x+2)=14$
$5x+2=14-7=7$
$5x=7-2=5$
$x=5:5=1$
9.
$5-(3x-11)=1$
$3x-11=5-1=4$
$3x=11+4=15$
$x=15:3=5$
10.
$16-(8x+2)=6$
$8x+2=16-6=10$
$8x=10-2=8$
$x=8:8=1$
![](https://rs.olm.vn/images/avt/0.png?1311)
a,\(\frac{1}{3}x=-2-\frac{2}{3}=\frac{-8}{3}\)
\(x=\frac{-8}{3}:\frac{1}{3}=\frac{-8}{3}.\frac{3}{1}=-8\)
\(b,\left[x+\frac{1}{1}\right]^2+\frac{5}{6}=\frac{7}{8}\)
\(\Rightarrow\left[x+1\right]^2=\frac{7}{8}-\frac{5}{6}\)
\(\Rightarrow\left[x+1\right]^2=\frac{7\cdot3}{24}-\frac{5\cdot4}{24}\)
\(\Rightarrow\left[x+1\right]^2=\frac{21}{24}-\frac{20}{24}\)
\(\Rightarrow\left[x+1\right]^2=\frac{1}{24}\)
\(\Rightarrow x\in\left\{\varnothing\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2^5.3^2+2^5.11-2^6.5-5\)
\(=2^5.9+2^5.11-2^5.2^1.5-5\)
\(=2^5.\left(9+11-2.5\right)-5\)
\(=32.\left(9+11-10\right)-5\)
\(=32.10-5\)
\(=320-5\)
\(=315\)
\(-2017-\left[\left(15-2017\right)+\left(-115\right)\right]\)
\(=-2017-\left[\left(-2002\right)+\left(-115\right)\right]\)
\(=-2017-\left(-2117\right)\)
\(=-2017+2117\)
\(=100\)
Vì 24 chia hết cho x, 120 chia hết cho x và 10<x<20 nên x ƯC(24,120)
Ta có : 24 =12.2 ; 120= 10.12
ƯCLN(24,120) = 12
Mà Ư(12) = { 1,2,3,4,6,12}
=>ƯC(24,120) = { 1,2,3,4,6,12}
Vì 10<x<20
=> x = 12
Vậy x = 12
3.|x-1| = 28:23 + 20170
3.|x-1| = 25+ 1
3.|x-1| = 32 + 1
3.|x-1| = 33
|x-1| = 33 : 3
|x-1| =11
=> x -1 =11
x = 11+1
x = 22
\(3^x+3^{x+1}=3^8.2+2.3^8.2017^0\)
\(3^x+3^x.3=3^8.2^2\)
\(3^x=3^8.2^2:3\)
\(3^x=3^7.2^2\)
3x+3x+1=38.2+2.38.201703x+3x+1=38.2+2.38.20170
3x+3x.3=38.223x+3x.3=38.22
3x=38.22:33x=38.22:3
3x=37.223x=37.22