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a) 3x – 15 = 25 – 5x
=> 3x + 5x = 25 + 15
=> 8x = 40
=> x = 5
b) 3x - 17 = 2x – 7
=> 3x - 2x = -7 + 17
=> x = 10
c) 2x – 17 = – (3x – 18)
=> 2x - 17 = -3x + 18
=> 2x + 3x = 18 + 17
=> 5x = 35
=> x = 7
d) 3x – 14 = 2(x – 9) + 1
=> 3x - 14 = 2x - 18 + 1
=> 3x - 2x = -18 + 1 + 14
=> x = -3
f) (x – 5)2 = 9
\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
a) Ta có: \(3x-15=25-5x\)
\(\Leftrightarrow3x-15-25+5x=0\)
\(\Leftrightarrow8x-40=0\)
\(\Leftrightarrow8x=40\)
hay x=5
Vậy: x=5
b) Ta có: \(3x-17=2x-7\)
\(\Leftrightarrow3x-17-2x+7=0\)
\(\Leftrightarrow x-10=0\)
hay x=10
Vậy: x=10
c) Ta có: \(2x-17=-\left(3x-18\right)\)
\(\Leftrightarrow2x-17=-3x+18\)
\(\Leftrightarrow2x-17+3x-18=0\)
\(\Leftrightarrow5x-35=0\)
\(\Leftrightarrow5x=35\)
hay x=7
Vậy: x=7
d) Ta có: \(3x-14=2\left(x-9\right)+1\)
\(\Leftrightarrow3x-14=2x-18+1\)
\(\Leftrightarrow3x-14-2x+18-1=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy: x=-3
f) Ta có: \(\left(x-5\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;8\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Ta có: \(\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy \(x=2\) hoặc \(x=-1\)
2) Ta có: \(\left(3-x\right)x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3-x=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=0\)
3) Ta có: \(2x-17=-\left(3x-18\right)\)
\(\Leftrightarrow2x-17=18-3x\)
\(\Leftrightarrow2x+3x=18+17\)
\(\Leftrightarrow5x=35\Leftrightarrow x=\dfrac{35}{5}=7\)
Vậy \(x=7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tìm số nguyên x, biết:
1) -16 + 23 + x = - 16
7+x=-16
x=-16-7
x=-23
2) 2x – 35 = 15
2x=15+35
2x=50
x=50:2
x=25
3) 3x + 17 = 12
3x=12-17
3x=-5
x=-5/3
4) (2x – 5) + 17 = 6
2x-5=6-17
2x-5=-11
2x=-11+5
2x=-6
x=-6:2
x=-3
5) 10 – 2(4 – 3x) = -4
2(4-3x)=10-(-4)
2(4-3x)=14
4-3x=14:2
4-3x=7
3x=4-7
3x=-3
x=-3:3
x=-1
6) - 12 + 3(-x + 7) = -18
3(-x+7)=-18-(-12)
3(x+7)=-6
x+7=-6:3
x+7=-2
x=-2-7
x=-9
![](https://rs.olm.vn/images/avt/0.png?1311)
( 2x - 5 ) + 17 = 6
2x - 5 = 6 - 17
2x - 5 = - 11
2x = - 11 + 5
2x = -6
x = - 6 : 2
x = -3
Vậy x = - 3
10 - 2 ( 4 - 3 x ) = - 4
2 ( 4 - 3 x ) = 10 - ( - 4 )
2 ( 4 - 3 x ) = 14
4 - 3x = 14 :2
4 - 3x = 7
3x = 4 - 7
3x = -3
x = -3 : 3
x = 1
Vậy x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1 ) \(\left(2x-5\right)+17=6\)
\(\Leftrightarrow2x-5=-11\)
\(\Leftrightarrow2x=-6\)
\(\Leftrightarrow x=-3\)
Vậy \(x=-3.\)
2 ) \(10-2\left(4-3x\right)=-4\)
\(\Leftrightarrow2+6x=-4\)
\(\Leftrightarrow6x=-6\)
\(\Leftrightarrow x=-1\)
Vậy \(x=-1.\)
3 ) \(-12+3\left(-x+7\right)=-18\)
\(\Leftrightarrow3x+9=-18\)
\(\Leftrightarrow3x=-27\)
\(\Leftrightarrow x=-9\)
Vậy \(x=-9.\)
4 ) \(24:\left(3x-2\right)=-3\)
\(\Leftrightarrow\left(3x-2\right)=-8\)
\(\Leftrightarrow3x=-6\)
\(\Leftrightarrow x=-2\)
Vậy \(x=-2.\)
5 ) \(-45:5.\left(-3-2x\right)=3\)
\(\Leftrightarrow-9\left(-3-2x\right)=3\)
\(\Leftrightarrow18x+27=3\)
\(\Leftrightarrow18x=-24\)
\(\Leftrightarrow x=-\frac{4}{3}\)
Vậy \(x=-\frac{4}{3}.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(\left(x+1\right)\left(y-2\right)=5\\ \Rightarrow\left(x+1\right),\left(y-2\right)\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
Ta có bảng:
x+1 | 1 | -1 | 5 | -5 |
y-2 | 5 | -5 | 1 | -1 |
x | 0 | -2 | 4 | -6 |
y | 7 | -3 | 3 | 1 |
Vậy \(\left(x;y\right)=\left(0;7\right),\left(-2;-3\right),\left(4;3\right),\left(-6;1\right)\)
b)
\(\left(x-5\right)\left(y+4\right)=-7\\ \Rightarrow\left(x-5\right),\left(y+4\right)\inƯ\left(-7\right)=\left\{1;-1;7;-7\right\}\)
Ta có bảng:
x-5 | 1 | -1 | 7 | -7 |
y+4 | -7 | 7 | -1 | 1 |
x | 6 | 4 | 12 | -2 |
y | -11 | 3 | -5 | -3 |
Vậy \(\left(x;y\right)=\left(6;-11\right),\left(4;3\right),\left(12;-5\right),\left(-2;-3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(12\left(x-3\right)=5\left(x-1\right)+4\)
\(\Rightarrow12x-36=5x-5+4\)
\(\Rightarrow12x-5x=-5+4+36\)
\(\Rightarrow7x=35\)
\(\Rightarrow x=5\)
\(-6\left(x-7\right)+2\left(x+10\right)=x-18\)
\(-6x+42+2x+20=x-18\)
\(-6x+2x-x=-18-20-42\)
\(-5x=-80\)
\(x=16\)
\(7\left(2x-1\right)-12\left(x-5\right)=3\)
\(\Rightarrow14x-7-12x+60=3\)
\(\Rightarrow14x-12x=3-60+7\)
\(\Rightarrow2x=-50\)
\(\Rightarrow x=-25\)
\(\left|2x-1\right|=17\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=17\\2x-1=-17\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=9\\x=-8\end{cases}}\)
\(\left|8-3x\right|=14\)
\(\Leftrightarrow\orbr{\begin{cases}8-3x=14\\8-3x=-14\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{22}{3}\end{cases}}\)
\(\left(2x-1\right)^2+17=18\\ \Leftrightarrow\left(2x-1\right)^2=1\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=1\\2x-1=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)
\(\left(2x-1\right)^2+17=18\)
\(\Leftrightarrow\left(2x-1\right)^2-1=0\)
\(\Leftrightarrow\left(2x-1-1\right)\left(2x-1+1\right)=0\)
\(\Leftrightarrow2x\left(2x-2\right)=0\)
\(\Rightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)