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theo dau bai ta co
23x + 23x +23x +.......+23x ( 16 so hang )
23x . 16 = 1024
23x = 1024 : 16
23x = 64
23x = 26
=> 3x = 6
x = 6 : 3
x = 2
2x + 11 = 3(x - 9)
=> 2x + 11 = 3x - 27
=> 2x - 3x = -11 - 27
=> -x = -38
=> x = 38
\(\Leftrightarrow\left(3-x\right)\left(3+x\right)2\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=2\end{matrix}\right.\)
`Answer:`
\(12x-144=0\)
\(\Leftrightarrow12x=144\)
\(\Leftrightarrow x=144:12\)
\(\Leftrightarrow x=12\)
\(5x-32:18=13\)
\(\Leftrightarrow5x-\frac{16}{9}=13\)
\(\Leftrightarrow5x=13+\frac{16}{9}\)
\(\Leftrightarrow5x=\frac{133}{9}\)
\(\Leftrightarrow x=\frac{133}{9}:5\)
\(\Leftrightarrow x=\frac{133}{45}\)
\(3x+6=15\)
\(\Leftrightarrow3x=9\)
\(\Leftrightarrow x=3\)
a ) 6 |x - 7| = 18 : (-3)
\(\Rightarrow6\left|x-7\right|=-6\)
\(\Rightarrow\left|x-7\right|=-1\) (1)
Mà \(\hept{\begin{cases}\left|x-7\right|\ge0\forall x\\-1< 0\end{cases}}\)
\(\Rightarrow\) | x - 7| = - 1 ( vô lí ) (2)
Từ (1) và (2) \(\Rightarrow\) \(x\in\varnothing\)
Vậy \(x\in\varnothing\)
Câu c tương tự nhé
b) -7 | x + 4| = 21 : (-3)
\(\Rightarrow-7\left|x+4\right|=-7\)
\(\Rightarrow\left|x+4\right|=1\)
\(\Rightarrow\orbr{\begin{cases}x+4=1\\x+4=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=-5\end{cases}}\)
Vậy \(x\in\left\{-3;-5\right\}\)
@@ Hc tốt @@
## Chiyuki Fujito
6 | x - 7 | = 18 : ( -3 )
6 | x - 7 | = ( -6 )
| x - 7 | = ( -6 ) : 6
| x - 7 | = ( -1 )
\(\Rightarrow\)x - 7 = ( -1 ) hoặc x - 7 = 1
x = ( -1 ) + 7 x = 1 + 7
x = 6 x = 8
\(\Rightarrow\) x = 6 ; x = 8
-7 | x + 4 | = 21 : ( -3 )
-7 | x + 4 | = ( -7 )
| x + 4 | = ( -7 ) : ( -7 )
| x + 4 | = 1
\(\Rightarrow\)x + 4 = 1 hoặc x + 4 = ( -1 )
x = 1 - 4 x = ( -1 ) - 4
x = -3 x = -5
\(\Rightarrow\) x = ( -3 ) ; x = ( -5 )
3 | x + 5 | = ( -9 )
| x + 5 | = ( -9 ) : 3
| x + 5 | = ( -3 )
\(\Rightarrow\)x + 5 = ( -3 ) hoặc x + 5 = 3
x = ( -3 ) - 5 x = 3 - 5
x = ( -8 ) x = ( -2 )
\(\Rightarrow\)x = ( -8 ) ; x = ( -2 )
@ Học tốt @
Nhớ k cho mình nha !!!!! Thank
a, \(\Rightarrow x-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
x-2 | 1 | -1 | 3 | -3 |
x | 3 | 1 | 5 | -1 |
b, \(3\left(x-2\right)+13⋮x-2\Rightarrow x-2\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)
x-2 | 1 | -1 | 13 | -13 |
x | 3 | 1 | 15 | -11 |
c, \(x\left(x+7\right)+2⋮x+7\Rightarrow x+7\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
x+7 | 1 | -1 | 2 | -2 |
x | -6 | -8 | -5 | -9 |
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=2\end{matrix}\right.\)
\(\Leftrightarrow3\left(x-2\right)\left(3-x\right)\left(3+x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=2\end{matrix}\right.\)