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(3x-9).(x-7)=0 TH1;3x-9=0 TH2;x-7=0 3x=0+9 x=0+7 3x=9 x=7 x=3 Vậy x có 2 giá trị x=3 và x=7
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\(a.\)
\(\Leftrightarrow x-5+x-7=-12.\)
\(\Leftrightarrow x=-12+-12\)
\(\Leftrightarrow x=-24\)
\(b.\)
\(\Leftrightarrow3x+2x=-109+44\)
\(\Leftrightarrow5x=-65\)
\(\Rightarrow x=-13\)
\(c.\)
\(\Leftrightarrow2x-10-15+5x=-109\)
\(\Leftrightarrow2x-5x=-109+15+10\)
\(\Leftrightarrow3x=-84\)
\(\Rightarrow x=-28\)
\(d.\)
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a)5-(13-3x)=109-(32+109)
5-(13-3x)=-32
13-3x =37
3x =-24
3x =-6
a)5-(13-3x)=109-(32+109)
5-13+3x =109-32-109
-8+3x=-32
3x=-32-(-8)
3x=-24
x=-24:3
x=-8
b)|3x-6|+(x-22)=0
(3x-6)+(x-22)=0
3x-6+x-22=0
3x+x=0+6+22
4x=10
x=\(\dfrac{5}{2}\)
c)/7-2x/=-13-5.(-8)
|7-2x|=27
⇒7-2x=27 hoặc 7-2x=-27
x=-10 hoặc x=-17
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\(a,3x-31=-40\Rightarrow3x=-9\Rightarrow x=-3\)
\(b,-3x+37=\left(-4\right)^2\Rightarrow-3x=-21\Rightarrow x=7\)
\(c,\left|2x+7\right|=5\)
\(\Rightarrow\left\{{}\begin{matrix}2x+7=5\Rightarrow x=-1\\2x+7=-5\Rightarrow x=-6\end{matrix}\right.\)
\(d,-x+21=15+2x\Rightarrow3x=6\Rightarrow x=2\)
a) Ta có: 3x-31=-40
\(\Leftrightarrow3x=-9\)
hay x=-3
Vậy: x=-3
b) Ta có: \(-3x+37=\left(-4\right)^2\)
\(\Leftrightarrow-3x+37=16\)
\(\Leftrightarrow-3x=16-37=-21\)
hay x=7
Vậy: x=7
![](https://rs.olm.vn/images/avt/0.png?1311)
2.(x-7)-3.(5-x)=-109
=>2x-14-15+3x=-109
=>5x-29=-109
=>5x=-80
=>x=-16
1280:[109-(3x-7)]=40
109-(3x-7)=1280/40
109-(3x-7)=32
3x-7=109-32
3x-7=77
3x=77+7
3x=84
x=84/3