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Vì x⋮6;x⋮24;x⋮40
→xϵ BC[6;24;40]
TA CÓ:
6=2.3
24=23.3
40=23.5
→BCNN[6;24;40]=23.3.5=60
BC[6;24;40]=B[60]={1;2;3;4;5;6;10;12;15;20;30;60}
hay x ϵ {1;2;3;4;5;6;10;12;15;20;30;60}
CÂU SAU TRÌNH BÀY NHƯ THẾ NHƯNG LÀ ƯỚC THÔI !
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b: \(x+1⋮x-2\)
\(\Leftrightarrow x-2+3⋮x-2\)
\(\Leftrightarrow x-2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{3;1;5;-1\right\}\)
c: \(\Leftrightarrow2x-1\in\left\{12;24;36;48\right\}\)
\(\Leftrightarrow x\in\left\{\dfrac{13}{2};\dfrac{25}{2};\dfrac{37}{2};\dfrac{49}{2}\right\}\)
d: \(\Leftrightarrow n+3+2⋮n+3\)
\(\Leftrightarrow n+3\in\left\{1;-1;2;-2\right\}\)
hay \(n\in\left\{-2;-4;-1;-5\right\}\)
e: \(2n-1⋮3n+6\)
\(\Leftrightarrow6n-3⋮3n+6\)
\(\Leftrightarrow6n+12-15⋮3n+6\)
\(\Leftrightarrow3n+6\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)
hay \(n\in\left\{-\dfrac{5}{3};-\dfrac{7}{3};-1;-3;-\dfrac{1}{3};-\dfrac{11}{3};3;-7\right\}\)
x-6:2-[48-24]:2:6-3=0
x-6:2-24:2:6-3=0
x-3-2-3=0
x-(-2)=0
x+2=0
x=0-2
x=-2
Vậy x=-2