\(A=\dfrac{6x+5}{2x-1}=\dfrac{3\left(2x-1\right)+8}{2x-1}=3+\dfrac{8}{2x-1}\)

\(\Rightarrow2x-1\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

(a) \(f\left(x\right)⋮g\left(x\right)\Rightarrow\dfrac{x^2-5x+9}{x-3}\in Z\)

Ta có: \(\dfrac{x^2-5x+9}{x-3}\left(x\ne3\right)=\dfrac{x\left(x-3\right)-2\left(x-3\right)+3}{x-3}=x-2+\dfrac{3}{x-3}\)nguyên khi và chỉ khi: \(\left(x-3\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\\x-3=3\\x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\\x=6\\x=0\end{matrix}\right.\) (thỏa mãn).

Vậy: \(x\in\left\{0;2;4;6\right\}\).

(b) \(f\left(x\right)⋮g\left(x\right)\Rightarrow\dfrac{2x^3-x^2+6x+2}{2x-1}\in Z\left(x\ne\dfrac{1}{2}\right)\)

Ta có: \(\dfrac{2x^3-x^2+6x+2}{2x-1}=\dfrac{x^2\left(2x-1\right)+3\left(2x-1\right)+5}{2x-1}=x^2+3+\dfrac{5}{2x-1}\)

nguyên khi và chỉ khi: \(\left(2x-1\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=1\\2x-1=-1\\2x-1=5\\2x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\\x=3\\x=-2\end{matrix}\right.\) (thỏa mãn).

Vậy: \(x\in\left\{-2;0;1;3\right\}\).

a: f(x) chia hết cho g(x)

=>x^2-3x-2x+6+3 chia hết cho x-3

=>3 chia hết cho x-3

=>x-3 thuộc {1;-1;3;-3}

=>x thuộc {4;2;6;0}

b: f(x) chia hết cho g(x)

=>2x^3-x^2+6x-3+5 chia hết cho 2x-1

=>5 chia hết cho 2x-1

=>2x-1 thuộc {1;-1;5;-5}

=>x thuộc {2;0;3;-2}

Chào bạn!

Tìm số nguyên x để A;B;C là số nguyên

ta có: \(A=\frac{3x-1}{x+2}=\frac{3x+6-7}{x+2}=\frac{3.\left(x+2\right)-7}{x+2}=3-\frac{7}{x+2}\)

Để A nguyên

=>7/(x+2) nguyên

=> 7 chia het cho x +2

=> x + 2 thuộc U(7)={1;-1;7;-7}

...

bn tu lm tiep nha

ta có: \(B=\frac{2x+5}{x-1}=\frac{2x-2+7}{x-1}=\frac{2.\left(x-1\right)+7}{x-1}\)

...

c) ta co: \(C=\frac{6x+5}{2x-3}=\frac{6x-9+14}{2x-3}=\frac{3.\left(2x-3\right)+14}{2x-3}=3+\frac{14}{2x-3}\)

...

\(\dfrac{6x}{2x-1}\in Z\Rightarrow\dfrac{6x-3+3}{2x-1}\in Z\Rightarrow\dfrac{3\left(2x-1\right)+3}{2x-1}\in Z\)

\(\Rightarrow3+\dfrac{3}{2x-1}\in Z\)

\(\Rightarrow\dfrac{3}{2x-1}\in Z\)

\(\Rightarrow2x-1=Ư\left(3\right)=\left\{-3;-1;1;3\right\}\)

\(\Rightarrow x=\left\{-1;0;1;2\right\}\)

Để \(A\) là số nguyên thì \(\left(6x-4\right)⋮\left(2x+1\right)\)

Ta có : 

 \(6x-4=6x+3-7=3\left(2x+1\right)-7\) chia hết cho \(2n+1\) \(\Rightarrow\) \(\left(-7\right)⋮\left(2x+1\right)\) \(\Rightarrow\) \(\left(2x+1\right)\inƯ\left(-7\right)\)

Mà \(Ư\left(-7\right)=\left\{1;-1;7;-7\right\}\)

Suy ra : 

Vậy \(x\in\left\{0;-1;3;-4\right\}\)

Năm mới zui zẻ nhá ^^