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Ta có: \(\frac{8c+36}{c+7}=\frac{8c+56-20}{c+7}=\frac{8\left(c+7\right)}{c+7}-\frac{20}{c+7}=8-\frac{20}{c+7}\)
\(\Rightarrow\frac{8c+36}{c+7}\in Z\Leftrightarrow\frac{20}{c+7}\in Z\Leftrightarrow c+7\inƯ20\)
\(\Leftrightarrow c+7\in\left\{\pm1;\pm2;\pm4;\pm5;\pm10;\pm20\right\}\)
\(\Leftrightarrow c\in\left\{-27;-17;-12;-11;-9;-8;-6;-5;-3;-2;3;13\right\}\)
Vậy \(\Rightarrow\frac{8c+36}{c+7}\in Z\Leftrightarrow\frac{20}{c+7}\in Z\Leftrightarrow c+7\inƯ20\)
\(\Leftrightarrow c+7\in\left\{\pm1;\pm2;\pm4;\pm5;\pm10;\pm20\right\}\)
\(\Leftrightarrow c\in\left\{-27;-17;-12;-11;-9;-8;-6;-5;-3;-2;3;13\right\}\)
Vậy \(c\in\left\{-27;-17;-12;-11;-9;-8;-6;-5;-3;-2;3;13\right\}\) thì \(\frac{8c+36}{c+7}\) là số nguyên
b) Để M là số nguyên thì \(2n-7⋮n-5\)
\(\Leftrightarrow2n-10+3⋮n-5\)
mà \(2n-10⋮n-5\)
nên \(3⋮n-5\)
\(\Leftrightarrow n-5\inƯ\left(3\right)\)
\(\Leftrightarrow n-5\in\left\{1;-1;3;-3\right\}\)
hay \(n\in\left\{6;4;8;2\right\}\)
Vậy: \(n\in\left\{6;4;8;2\right\}\)
a) Ta có: \(\left|x-3\right|=2x+4\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x+4\left(x\ge3\right)\\x-3=-2x-4\left(x< 3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2x=4+3\\x+2x=-4+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=7\\3x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-7\left(loại\right)\\x=-\dfrac{1}{3}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(x=-\dfrac{1}{3}\)
\(\frac{3a+45}{a+9}\)là số nguyên
\(\Rightarrow3a+45⋮a+9\)
Ta có : \(3a+45⋮a+9\)
\(\Rightarrow3a+27+18⋮a+9\)
\(\Rightarrow3\left(a+9\right)+18⋮a+9\)
\(\Rightarrow18⋮a+9\)
\(\Rightarrow a+9\inƯ\left(18\right)=\left\{-18;-9;-6;-3;-2;-1;1;2;3;6;9;18\right\}\)
\(\Rightarrow a\in\left\{-27;-18;-15;-12;-11;-10;-8;-7;-6;-3;0;9\right\}\)
Học tốt!
4x-37 chia hết cho x-6
4x-24-13
=>13 chia hết cho x-6
x=7,19,5,-7
a) Số nguyên n phải: n-7 \(\inƯ\left(7\right)\)
b) Nếu n= -7 thì \(B=\frac{7}{-7}=-1\)
c) Muốn B nguyên thì n \(\in\left\{0;6;8;14\right\}\)
Ta đặt A\(=\dfrac{4c-4+8}{c-1}\) \(\Rightarrow A=\dfrac{4c-4+8}{c-1}=\dfrac{4\left(c-1\right)+8}{c-1}=4+\dfrac{8}{c-1}\)
Để A∈Z \(\Leftrightarrow\) \(4+\dfrac{8}{c-1}\in Z\) \(\Rightarrow\dfrac{8}{c-1}\in Z\) \(\Rightarrow8⋮\left(c-1\right)\) \(\Rightarrow c-1\in\left\{-8;-4;-2;-1;1;2;4;8\right\}\) \(\Rightarrow c\in\left\{-7;-3;-1;0;2;3;5;9\right\}\)
Ta đặt A=\(\dfrac{4n-2}{n-4}\)\(\Rightarrow A=\dfrac{4n-16+14}{n-4}=\dfrac{4\left(n-4\right)+14}{n-4}=4+\dfrac{14}{n-4}\)
Để A\(\in Z\) \(\Leftrightarrow4+\dfrac{14}{n-4}\in Z\) \(\Rightarrow\dfrac{14}{n-4}\in Z\) \(\Rightarrow14⋮\left(n-4\right)\Rightarrow n-4\in\left\{-14;-7;-2;-1;1;2;7;14\right\}\)
\(\Rightarrow n\in\left\{-10;-3;2;3;5;6;11;18\right\}\)
\(\text{Ta có:}\)
\(\text{Để}\)\(\frac{4b+42}{b+7}\)\(\text{nguyên thì}\)\(4b+42⋮b+7\)
\(\text{Lại có:}\)
\(\text{4b + 42 = 4b + 28 + 14 = 4( b+7 ) + 14}\)
\(\text{Vì}\)\(b+7⋮b+7\)\(\Rightarrow4\left(b+7\right)⋮b+7\)
\(\text{Do đó:}\)\(14⋮b+7\)
\(\Rightarrow b+7\inƯ\left(14\right)=\left\{1;2;7;14\right\}\)
\(\Rightarrow b\in\left\{-6;-5;0;7\right\}\)