a) Ta có: -16-15(x-4)=-14
\(\Leftrightarrow-16-15x+60+14=0\)
\(\Leftrightarrow-15x+58=0\)
\(\Leftrightarrow-15x=-58\)
hay \(x=\frac{58}{15}\)
Vậy: \(x=\frac{58}{15}\)
b) Ta có: 25+2(3-x)=23
\(\Leftrightarrow25+6-2x-23=0\)
\(\Leftrightarrow-2x+8=0\)
\(\Leftrightarrow-2x=-8\)
hay x=4
Vậy: x=4
c) Ta có: 3x+5=x-7
\(\Leftrightarrow3x+5-x+7=0\)
\(\Leftrightarrow2x+12=0\)
\(\Leftrightarrow2x=-12\)
hay x=-6
Vậy: x=-6
d) Ta có: \(3⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(3\right)\)
\(\Leftrightarrow x+1\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{0;-2;2;-4\right\}\)
Vậy: \(x\in\left\{0;-2;2;-4\right\}\)
a, -16 - 15(x - 4) = -14
-16 - 15x + 60 = -14
44 - 15x = -14
-15x = -14 - 44
-15x = -58
x = \(\frac{58}{15}\)
Vậy x = \(\frac{58}{15}\)
b, 25 + 2(3 - x) = 23
25 + 6 - 2x = 23
31 - 2x = 23
-2x = 23 - 31
-2x = -8
x = 4
Vậy x = 4
c, 3x + 5 = x - 7
3x - x = -7 - 5
2x = -12
x = -6
Vậy x = -6
d, 3 \(⋮\) x + 1
\(\Rightarrow\) x + 1 \(\in\) Ư(3)
Ta có Ư(3) = {1; -1; 3; -3}
Xét các TH:
\(\left[{}\begin{matrix}x+1=1\\x+1=-1\\x+1=3\\x+1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\\x=2\\x=-4\end{matrix}\right.\)
Vậy x \(\in\) {-1; -2; 2; -4}
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