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![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x\left(x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b) \(\left(-7-x\right)\left(-x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)
c) \(\left(x+3\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)
d) \(\left(x-3\right)\left(x^2+12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)
\(\Rightarrow x=3\)
e) \(\left(x+1\right)\left(2-x\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)
\(\Rightarrow-1\le x\le2\)
f) \(\left(x-3\right)\left(x-5\right)\le0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow3\le x\le5\)
a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)
d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
c) \(\left(34-2x\right)\left(2x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6-0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
d) \(\left(2019-x\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.\)
e) \(57\left(9x-27\right)=0\)
\(\Rightarrow9x-27=0\)
\(\Rightarrow9\left(x-3\right)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
f) \(25+\left(15-x\right)=30\)
\(\Rightarrow25+15-x=30\)
\(\Rightarrow40-x=30\)
\(\Rightarrow x=40-30\)
\(\Rightarrow x=10\)
g) \(43-\left(24-x\right)=20\)
\(\Rightarrow43-24+x=20\)
\(\Rightarrow19+x=20\)
\(\Rightarrow x=20-19\)
\(\Rightarrow x=1\)
h) \(2\left(x-5\right)-17=25\)
\(\Rightarrow2\left(x-5\right)=17+25\)
\(\Rightarrow x-5=21\)
\(\Rightarrow x=21+5\)
\(\Rightarrow x=26\)
i) \(3\left(x+7\right)-15=27\)
\(\Rightarrow3\left(x+7\right)=27+15\)
\(\Rightarrow x+7=14\)
\(\Rightarrow x=14-7\)
\(\Rightarrow x=7\)
j) \(15+4\left(x-2\right)=95\)
\(\Rightarrow4\left(x-2\right)=95-15\)
\(\Rightarrow4\left(x-2\right)=80\)
\(\Rightarrow x-2=20\)
\(\Rightarrow x=20+2\)
\(\Rightarrow x=22\)
k) \(20-\left(x+14\right)=5\)
\(\Rightarrow x+14=20-5\)
\(\Rightarrow x+14=15\)
\(\Rightarrow x=15-14\)
\(\Rightarrow x=1\)
l) \(14+3\left(5-x\right)=27\)
\(\Rightarrow3\left(5-x\right)=27-14\)
\(\Rightarrow3\left(5-x\right)=13\)
\(\Rightarrow5-x=\dfrac{13}{3}\)
\(\Rightarrow x=5-\dfrac{13}{3}\)
\(\Rightarrow x=\dfrac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 124 + (118 - x) = 217
118 - x = 217 - 124
118 - x = 93
x = 118 - 93
x = 25
b) 156 - (x + 61) = 82
x + 61 = 156 - 82
x + 61 = 74
x = 74 - 61
x = 13
c) 219 - 7(x + 1) = 100
7(x + 1) = 219 - 100
7(x + 1) = 119
x + 1 = 119 : 7
x + 1 = 17
x = 17 - 1
x = 16
d) (3x - 6).3 = 3⁴
3x - 6 = 3⁴ : 3
3x - 6 = 3³
3x - 6 = 27
3x = 27 + 6
3x = 33
x = 33 : 3
x = 11
e) 231 - (x - 6) = 1339 : 13
231 - (x - 6) = 103
x - 6 = 231 - 103
x - 6 = 128
x = 128 + 6
x = 134
a) 124 + (118 - x) = 217
118 - x = 217 - 124
118 - x = 93
x = 118 - 93
x = 25
b) 156 - (x + 61) = 82
x + 61 = 156 - 82
x + 61 = 74
x = 74 - 61
x = 13
c) 219 - 7(x + 1) = 100
7(x + 1) = 219 - 100
7(x + 1) = 119
x + 1 = 119 : 7
x + 1 = 17
x = 17 - 1
x = 16
d) (3x - 6) . 3 = 34
(3x - 6) . 3 = 81
(3x - 6) = 81 : 3
(3x - 6) = 27
3x = 27 + 6
3x = 33
x = 33 : 3
x = 11
e) 231 - (x - 6) = 1339 : 13
231 - (x - 6) = 103
(x - 6) = 231 - 103
x - 6 = 128
x = 128 + 6
x = 134
![](https://rs.olm.vn/images/avt/0.png?1311)
\(25\div x=x\Leftrightarrow25=x.x\Leftrightarrow25=x^2\)
\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=5\end{cases}}\)
\(3358\div23=2x-6\)
\(\Leftrightarrow146=2x-6\)
\(\Leftrightarrow2x=152\Leftrightarrow x=76\)
\(2782\div\left[216-\left(x-3\right)\right]=13\)
\(\Leftrightarrow216-\left(x-3\right)=214\)
\(\Leftrightarrow x-3=2\)
\(\Leftrightarrow x=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2x-15=25
2x=25+15
2x=40
x=40:2
x=20
63:x=21
x=63:21
x=3
x-8=96
x=96+8
x=104
120-(x-3)=35
x-3=120-35
x-3=85
x=85+3
x=88
35-5(x-2)=15
5(x-2)=35-15
5(x-2)=20
x-2=20:5
x-2=4
x=4+2
x=6
(x-9)-20=19
x-9=19+20
x-9=39
x=39+9
x=48
2782:(216-x-3)=15
câu này mk thấy đề sai rồi bạn nha
130-30(x-2)=100
30(x-2)=130-100
30(x-2)=30
x-2=30:30
x-2=1
x=1+2
x=3
chúc bạn học tốt nha
ủng hộ mk bằng cach tk nha
a/ 2x -15 = 25
2x = 25 - 15
2x = 10
x = 10 : 2
x = 5
b/ 63 : x =21
x = 63 : 21
x = 3
c/ x - 8 = 96
x = 96 + 8
x =104
d/ 120 - (x - 3) = 65
x - 3 = 65
x - 3 = 120 - 65
x - 3 = 55
x = 55 + 3
x = 58
e/ 35 - 5 (x - 2) = 15
5 (x - 2) = 35 - 15
5 (x - 2) = 20
x - 2 = 20 : 5
x - 2 = 4
x = 4 + 2
x = 6
f/ (x - 9) - 20 = 19
x - 9 = 19 + 20
x - 9 = 39
x = 39 - 9
x = 30
g/ 2782 : (216 - x - 3) = 15 (bài này ra số thập phân bạn nên coi lại)
h/ 130 - 30 (x - 2) = 100
30 (x - 2) = 130 -100
30 (x - 2) = 30
x - 2 = 30 : 30
x - 2 = 1
x = 1 + 2
x =3
![](https://rs.olm.vn/images/avt/0.png?1311)
- a, Do \(\left[3x-15\right]^7=0\)
=> \(3x-15=0\)
=> \(3x=0+15\)
=> \(3x=15\)
=> \(x=15:3\)
=> \(x=5\)
\(\left(3x-5\right)^7=0\)
\(\Rightarrow3x-5=0\)
\(\Rightarrow3x=5\)
\(\Rightarrow x=\frac{5}{3}\)