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a/ \(x-4⋮x-1\)
Mà \(x-1⋮x-1\)
\(\Leftrightarrow5⋮x-1\)
\(\Leftrightarrow x-1\inƯ\left(5\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-1=5\\x-1=-1\\x-1=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\\x=0\\x=-4\end{matrix}\right.\)
Vậy ...
b/ \(2x+5⋮x-1\)
Mà \(x-1⋮x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+5⋮x-1\\2x-2⋮x-1\end{matrix}\right.\)
\(\Leftrightarrow7⋮x-1\)
\(\Leftrightarrow x-1\inƯ\left(7\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-1=7\\x-1=-1\\x-1=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=8\\x=0\\x=-6\end{matrix}\right.\)
Vậy ...
c/ \(x^2+3x+4⋮x+3\)
Mà \(x+3⋮x+3\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+3x+4⋮x+3\\x^2+3x⋮x+3\end{matrix}\right.\)
\(\Leftrightarrow4⋮x+3\)
\(\Leftrightarrow x+3\inƯ\left(4\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=1\\x+3=2\\x+3=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-1\\x=1\end{matrix}\right.\)
Vậy ..
a: =>3x+17=14
=>3x=-3
hay x=-1
b: =>|x+9|=-8(vô lý)
c: =>3x+2=17
=>3x=15
hay x=5
d: \(\Leftrightarrow\left(x-2\right)\left(x+2\right)\cdot3\cdot\left(2-x\right)=0\)
hay \(x\in\left\{2;-2\right\}\)
e: =>2x+4=0
hay x=-2
f: =>2|2x-1|=34
=>|2x-1|=17
=>2x-1=17 hoặc 2x-1=-17
=>2x=18 hoặc 2x=-16
=>x=9 hoặc x=-8
1e) Để \(\frac{2x-1}{x-3}\) nguyên thì \(2x-1⋮x-3\)
\(\Leftrightarrow2x-6+5⋮x-3\)
\(\Leftrightarrow2\left(x-3\right)+5⋮x-3\)
Do \(2\left(x-3\right)⋮x-3\) \(\Rightarrow5⋮x-3\)
\(\Rightarrow x-3\in\left\{-5;-1;1;5\right\}\)
\(\Leftrightarrow x\in\left\{-2;2;4;8\right\}\)
Vậy:...................
a, \(\left(2x-4\right).\left(x+3\right)=0\)
\(TH1:2x-4=0\)
\(2x=4+0\)
\(2x=4\)
\(x=4:2\)
\(x=2\)
\(TH2:x+3=0\)
\(x=0+3\)
\(x=3\)
Vậy \(x\in\left\{2;3\right\}\)
a: \(A=x^2+2+3x^2+2-2x^2-2=2x^2+2\)
b: \(B=2x-3-\left(3x-2\right)-\left(2x-4\right)\)
\(=2x-3-3x+2-2x+4=-3x+3\)
c: \(C=6-3x+3x+10=16\)
d: \(D=\left|3x-8\right|+\left|x+2\right|\)
\(=3x-8+x+2=4x-6\)