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\(\Leftrightarrow x-3\in\left\{1;-1;17\right\}\)
hay \(x\in\left\{4;2;20\right\}\)
\(\left(x-3\right)\inƯ\left(17\right)=\left\{1;17\right\}\)
\(=>\left[{}\begin{matrix}x=4\\x=20\end{matrix}\right.\)
\(x⋮24\\ x⋮40\\ x⋮168\\ \Rightarrow x⋮BCNN\left(24;40;168\right)\\ \Rightarrow x⋮840\\ \Rightarrow x\in B\left(840\right)=\left\{0;840;...\right\}\)
Do \(x< 250\Rightarrow x=0\)
a) \(\left(x-7\right)-\left(5-x\right)=12-\left(-8+x\right)\)
\(\Leftrightarrow x-7-5+x=12+8-x\)
\(\Leftrightarrow2x-12=20-x\)
\(\Leftrightarrow2x+x=12+20\)
\(\Leftrightarrow3x=32\)
\(\Leftrightarrow x=\frac{32}{3}\)
b) 124 + ( 13 - 16 ) = 162 - ( x + 162 )
<=> 124 - 3 = 162 - x - 162
<=> 121 = -x
<=> x = -121
|x+19|+|x+5|+|x+2020|=5x(*)
+)Ta có:|x+19|\(\ge\)0;|x+5|\(\ge\)0;|x+2020|\(\ge\)0
=>VT(*)=|x+19|+|x+5|+|x+2020|\(\ge\)0
Mà |x+19|+|x+5|+|x+2020|=5x
=>5x\(\ge\)0
=>x\(\ge\)0
+)Ta lại có:x\(\ge\)0=>x+19\(\ge\)19=>|x+19|=x+19
x\(\ge\)0=>x+5\(\ge\)5=>|x+5|=x+5
x\(\ge\)0=>x+2020\(\ge\)2020=>|x+2020|=x+2020
=>VT(*)=x+19+x+5+x+2020=5x
x+x+x+19+5+2020=5x
3x+2044 =5x
2044 =5x-3x
2044 =2x
=> 2x =2044
x =\(\frac{2044}{2}=1022\)\(\in\)Z
Vậy x=1022
Chúc bn học tốt
\(\left(19-x:2\right)\inƯ\left(13\right)=\left\{1;13\right\}\)
\(=>\left[{}\begin{matrix}19-x:2=1\\19-x:2=13\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=36\\x=12\end{matrix}\right.\)