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17x + 3. ( -16x – 37) = 2x + 43 - 4x
<=>17x-48x-111=-2x+43
<=>-29x=154
<=> \(x=-\frac{154}{29}\)
-3. (2x + 5) -16 < -4. (3 – 2x)
\(\Leftrightarrow-6x-31< -12+8x.\)
\(\Leftrightarrow-14x< 19\Rightarrow x< -\frac{19}{14}\)
1. \(3-|2x+1|=-5\)
\(\Rightarrow|2x+1|=8\)
\(\Rightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{9}{2}\end{cases}}\)
Vậy \(x\in\left\{\frac{7}{2};-\frac{9}{2}\right\}\)
2.\(12+|3-x|=9\)
\(\Rightarrow|3-x|=-3\)
Mà \(|3-x|\ge0\forall x\)
\(\Rightarrow\)Vô lí
Vậy không có x
3.\(|x+9|=12+\left(-9\right)+2\)
\(\Rightarrow|x+9|=5\)
\(\Rightarrow\orbr{\begin{cases}x+9=5\\x+9=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-4\\x=-14\end{cases}}\)
Vậy \(x\in\left\{-4;-14\right\}\)
4.\(5x-16=40+x\)
\(\Rightarrow5x-x=40+16\)
\(\Rightarrow4x=56\)
\(\Rightarrow x=14\)
Vậy \(x=14\)
5.\(5x-7=-21-2x\)
\(\Rightarrow5x+2x=-21+7\)
\(\Rightarrow7x=-14\)
\(\Rightarrow x=-2\)
Vậy \(x=-2\)
6.\(\left(2x-1\right)\left(y-2\right)=12\)
Vì \(x,y\inℤ\)nên \(2x-1;y-2\inℤ\)
\(\Rightarrow2x-1;y-2\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
Ta có bảng : (em tự xét bảng nhé)
1.
a, \(x-14=3x+18\)
\(\Rightarrow x-3x=18+14\)
\(\Rightarrow-2x=32\Rightarrow x=\frac{32}{-2}=-16\)
b, \(\left(x+7\right).\left(x-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+7=0\\x-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-7\\x=9\end{cases}}}\)
c, \(\left|2x-5\right|-7=22\)
\(\Rightarrow\left|2x-5\right|=22+7\)
\(\Rightarrow\left|2x-5\right|=29\)
\(\Rightarrow\orbr{\begin{cases}2x+5=29\\2x-5=29\end{cases}}\Rightarrow\orbr{\begin{cases}2x=24\\2x=34\end{cases}\Rightarrow}\orbr{\begin{cases}x=12\\x=17\end{cases}}\)
d, \(\left(\left|2x\right|-5\right)-7=22\)
\(\Rightarrow\left(\left|2x\right|-5\right)=29\)
\(\Rightarrow\left|2x\right|=29+5\Rightarrow\left|2x\right|=34\Rightarrow x=\pm17\)
e, \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\)
Vì \(\left|x+3\right|\ge0;\left|x+9\right|\ge0;\left|x+5\right|\ge0;4x\ge0\)
Nên \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\ge0\)
\(\Rightarrow\left|x+3\right|>0\Rightarrow\left|x+3\right|=x+3\)
\(\left|x+9\right|>0\Rightarrow\left|x+9\right|=x+9\)
\(\left|x+5\right|>0\Rightarrow\left|x+5\right|=x+5\)
Ta có :
\(x+3+x+9+x+5=4x\)
\(\Rightarrow3x+\left(3+9+5\right)=4x\)
\(\Rightarrow4x-3x=17\)
\(\Rightarrow x=17\)
2. a , b sai đề bn
c, \(\left(5x+1\right).\left(y-1\right)=4\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)
\(\text{ }Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau :
| 5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
| y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
| x | 0 | -2/5 | 1/5 | -3/5 | 3/5 | -1 |
| y | -3 | 5 | -1 | 3 | 0 | 2 |
d, \(5xy-5x+y=5\)
\(\Rightarrow\left(5xy-5x\right)+y=5\)
\(\Rightarrow5x.\left(y-1\right)+y=5\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)=4\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)
\(Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau :
| 5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
| y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
| x | 0 | -2 | 1/5 | -3/5 | 3/5 | -1 |
| y | -3 | 5 | -1 | 3 | 0 | 2 |
Câu 1:
a: =>-2x-x+17=34+x-25
=>-3x+17=x+9
=>-4x=-8
hay x=2
b: =>17x+16x+27=2x+43
=>33x+27=2x+43
=>31x=16
hay x=16/31
c: =>-2x-3x+51=34+2x-50
=>-5x+51=2x-16
=>-7x=-67
hay x=67/7
e: 3x-32>-5x+1
=>8x>33
hay x>33/8
a; - \(\dfrac{1}{3}\).(15\(x-9\)) + \(\dfrac{2}{7}\).(- \(x-34\)) = 1 - \(\dfrac{3}{4}\).(-16\(x+4\))
- 5\(x\) + 3 - \(\dfrac{2}{7}\)\(x\) - \(\dfrac{68}{7}\) = 1 + 12\(x\) - 3
12\(x\) + 5\(x\) + \(\dfrac{2}{7}x\) = 3 - \(\dfrac{68}{7}\) - 1 + 3
17\(x\) + \(\dfrac{2}{7}x\) = (3 - 1 + 3) - \(\dfrac{68}{7}\)
\(\dfrac{121}{7}\)\(x\) = 5 - \(\dfrac{68}{7}\)
\(\dfrac{121}{7}\) \(x\) = - \(\dfrac{33}{7}\)
\(x\) = - \(\dfrac{33}{7}\): \(\dfrac{121}{7}\)
\(x\) = - \(\dfrac{3}{11}\)
Vậy \(x\) = - \(\dfrac{3}{11}\)