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a)(2x+4).8-40=24
(2x+4).8=24+40
(2x+4).8=64
2x+4=64:8
2x+4=8
2x=8+4
2x=12
x=12:2
x=6
b)(2x+42)-8=64:2
(2x+16)-8=32
2x+16=32+8
2x+16=40
2x=40-16
2x=24
x=24:2
x=12
c)(3x+2).4-20=24
(3x+2).4=24+20
(3x+2).4=44
3x+2=44:4
3x+2=11
3x=11-2
3x=9
x=9:3
x=3
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2: 12-10x=25-30x
=>20x=13
=>x=13/20
3: \(3\left(2x+3\right)-2\left(4x-5\right)=10x+21\)
=>6x+9-8x+10=10x+21
=>10x+21=-2x+19
=>12x=-2
=>x=-1/6
4: \(\Leftrightarrow25x-15-6x+12=11-5x\)
=>19x-3=11-5x
=>24x=14
=>x=7/12
5: \(\Leftrightarrow8-12x-5+10x=4-6x\)
=>4-6x=-2x+3
=>-4x=-1
=>x=1/4
6: \(\Leftrightarrow32x-24-6+9x=13-40x\)
=>41x-30=13-40x
=>81x=43
=>x=43/81
7: \(\Leftrightarrow10x-5+20x=5x-11\)
=>30x-5=5x-11
=>25x=-6
=>x=-6/25
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Bài giải
a, \(-2\left(2x-8\right)+3\left(4-2x\right)=-72-5\left(3x-7\right)\)
\(-4x+8+12-6x=-72-15x+7\)
\(-10x+20=-65-15x\)
\(-10x+15x=-65-20\)
\(5x=-85\)
\(x=-85\text{ : }5\)
\(x=-17\)
b, \(3\left|2x^2-7\right|=33\)
\(\left|2x^2-7\right|=33\text{ : }3\)
\(\left|2x^2-7\right|=11\)
\(\Rightarrow\orbr{\begin{cases}2x^2-7=-11\\2x^2-7=11\end{cases}}\Rightarrow\orbr{\begin{cases}2x^2=-4\text{ ( loại ) }\\2x^2=18\end{cases}}\Rightarrow\text{ }x^2=9\text{ }\Rightarrow\text{ }x=\pm3\)
\(\Rightarrow\text{ }x=\pm3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>1/3:x=3/5-2/3=9/15-10/15=-1/15
=>x=-1/3:1/15=5
b: \(\Leftrightarrow x\cdot\dfrac{2}{3}-3=\dfrac{2}{5}\cdot\left(-10\right)=-4\)
=>x*2/3=-1
=>x=-3/2
c: =>2x+1=4 hoặc 2x+1=-4
=>x=3/2 hoặc x=-5/2
h: =>x-3=4
=>x=7
g: =>2x-1=3
=>2x=4
=>x=2
f: \(\Leftrightarrow x\cdot\left(\dfrac{3}{2}-\dfrac{7}{3}\right)=\dfrac{3}{2}-\dfrac{2}{3}\)
=>x*-5/6=5/6
=>x=-1
d: =>|2x-1|=3
=>2x-1=3 hoặc 2x-1=-3
=>x=-1 hoặc x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Rightarrow x=3\)
\(b,\Rightarrow2x-1=2\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\dfrac{3}{2}\)
\(c,\Rightarrow x-2=4\)
\(\Rightarrow x=6\)
\(d,\Rightarrow2x-3=3\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
`(2x-2)^3=8`
`(2x-2)^3=2^3`
`=>2x-2=2`
`x=2`
\(\left(2x-2\right)^3=8\)
\(\left(2x-2\right)^3=2^3\)
\(\Rightarrow2x-2=2\)
\(2x=2+2\)
\(2x=4\)
\(x=4:2\)
\(x=2\)
\(#WendyDang\)