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Bài 2:
a: Ta có: \(2^{x+1}\cdot3^y=12^x\)
\(\Leftrightarrow2^{x+1}\cdot3^y=2^{2x}\cdot3^x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=2x\\x=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
a) (x-35)-120=0
x-35= 120
➩x = 155
b)310-(118-x)=217
118 -x = 93
➩ x = 25
c)156-(x+61)=82
➩ x + 61 = 74
➩ x = 13
a: Ta có: \(x-35-120=0\)
\(\Leftrightarrow x-155=0\)
hay x=155
b: Ta có: \(310-\left(118-x\right)=217\)
\(\Leftrightarrow118-x=93\)
hay x=25
c: Ta có: \(156-\left(x+61\right)=82\)
\(\Leftrightarrow x+61=74\)
hay x=13
\(a,\Rightarrow x\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ b,\Rightarrow2\left(x+1\right)-1⋮x+1\\ \Rightarrow x+1\inƯ\left(1\right)=\left\{-1;1\right\}\\ \Rightarrow x\in\left\{-2;0\right\}\)
\(a,\left(8-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}8-x=0\\x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-5\end{matrix}\right.\\ b,2x\left(x+81\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=0\\x+81=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-81\end{matrix}\right.\)
a)\(\left(8-x\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}8-x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-5\end{matrix}\right.\)
b)\(2x\left(x+81\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x+81=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-81\end{matrix}\right.\)
a) (35+x)-81=54
35+x=54+81
35+x=135
x=135-35
x=100
vậy x=100
b)143+(120-x)=190
120-x=190-143
120-x=47
x=120-47
x=73
vậy x=73
c) 175-(x+32)=67
x+32=175-67
x+32=90
x=90-32
x=58
vậy x=58