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\((-12)^2\cdot x=56-(-10)\cdot13x\)
\(\Rightarrow144\cdot x=56-(-130x)\)
\(\Rightarrow144x=56+130x\)
\(\Rightarrow144x-130x=56\)
\(\Rightarrow14x=56\)
\(\Rightarrow x=4\)
P/S : Hoq chắc :D
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a) 2|x+1|=10
\(\left|x+1\right|\) = 10:2
\(\left|x+1\right|\) = 5
\(\left[{}\begin{matrix}x+1=5\\x+1=-5\end{matrix}\right.\) \(\Rightarrow\) \(\left[{}\begin{matrix}x=5-1\\x=\left(-5\right)-1\end{matrix}\right.\) \(\Rightarrow\) \(\left[{}\begin{matrix}x=4\\x=-6\end{matrix}\right.\)
b) (−12)2x=56+10.13x
144x= 56+130x
144x-130x= 56
14x= 56
x= 56: 14
x= 4
a) 2\(|\)x + 1\(|\) = 10 \(\Rightarrow\) \(|\)x + 1\(|\) = 5
\(\Rightarrow\) x + 1 = 5 hay x = 4
hoặc x + 1 = \(-\)5 hay x = \(-\)6.
ĐS : x = 4, x = \(-\)6.
b) x = 4.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(x+\dfrac{1}{3}=\dfrac{2}{6}\)
\(\Leftrightarrow x+\dfrac{1}{3}=\dfrac{1}{3}\)
hay x=0
Vậy: x=0
b) Ta có: \(x-\dfrac{1}{4}=\dfrac{1}{-2}\)
\(\Leftrightarrow x-\dfrac{1}{4}=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-1}{2}+\dfrac{1}{4}=\dfrac{-2}{4}+\dfrac{1}{4}=\dfrac{-1}{4}\)
Vậy: \(x=-\dfrac{1}{4}\)
c) Ta có: \(\dfrac{-1}{6}=\dfrac{3}{2}x\)
\(\Leftrightarrow x=\dfrac{-1}{6}:\dfrac{3}{2}=\dfrac{-1}{6}\cdot\dfrac{2}{3}\)
hay \(x=\dfrac{-1}{9}\)
Vậy: \(x=\dfrac{-1}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(-45:5.\left(-3-2x\right)=3\)
\(-9.\left(-3-2x\right)=3\)
\(-3-2x=\left(3:-9\right)\)
\(-3-2x=\dfrac{-1}{3}\)
\(-2x=-3-\dfrac{1}{3}\)
-2x=\(\dfrac{-10}{3}\)
\(x=\dfrac{-10}{3}:-2\)
\(x=\dfrac{5}{3}\)
b)
3x - 28 = x + 36
<=> 3x - x = 36 + 28
<=> 2x = 64
<=> x = 32
Vậy x = 32
c)
(-12)2.x = 56 + 10.13.x
144.x = 56 + 130.x
144x – 130 x = 56
14x = 56
x = 56: 14
x = 4
Vậy x = 4
Đáp án là B
Ta có:
- 12 2 .x = 56 + 10.13x
144x = 56 + 130x
144x - 130x = 56
14x = 56
x = 4