\(P=\left(\frac{9}{x^2-3x}+\frac{x-2}{x}-\frac{x}{x-3}\right).\frac{x}{3-3x}\)

a,\(ĐKXĐ:x\ne0;x\ne3;x\ne1\)

\(P=\left(\frac{9}{x^2-3x}+\frac{x-2}{x}-\frac{x}{x-3}\right).\frac{x}{3-3x}=\left(\frac{9}{x\left(x-3\right)}+\frac{x-2}{x}-\frac{x}{x-3}\right).\frac{x}{3\left(1-x\right)}\)

\(=\left(\frac{9+\left(x-2\right)\left(x-3\right)-x.x}{x\left(x-3\right)}\right).\frac{x}{3\left(1-x\right)}=\frac{9+x^2-5x+6-x^2}{x\left(x-3\right)}.\frac{x}{3\left(1-x\right)}\)

\(=\frac{-5x+15}{x\left(x-3\right)}.\frac{x}{3\left(1-x\right)}=\frac{-5\left(x-3\right)}{x\left(x-3\right)}.\frac{x}{3\left(1-x\right)}=-\frac{5}{3\left(1-x\right)}\)

b, \(x=\frac{1}{2}\)

\(\Rightarrow P=-\frac{5}{3\left(1-\frac{1}{2}\right)}=-\frac{5}{3.\frac{1}{2}}=-5:\frac{3}{2}=-\frac{10}{3}\)

c, Để \(P\in z\)thì \(3\left(1-x\right)\inƯ\left(5\right)=\left(-5;-1;1;5\right)\)

\(3\left(1-x\right)=-5\Rightarrow1-x=-\frac{5}{3}\Rightarrow x=\frac{8}{3}\)

\(3\left(1-x\right)=-1\Rightarrow1-x=-\frac{1}{3}\Rightarrow x=\frac{4}{3}\)

\(3\left(1-x\right)=1\Rightarrow1-x=\frac{1}{3}\Rightarrow x=\frac{2}{3}\)

\(3\left(1-x\right)=5\Rightarrow1-x=\frac{5}{3}\Rightarrow x=-\frac{2}{3}\)

a, 2x+1 chia hết cho x-1

=>2x-2+3 chia hết cho x-1

=>2(x-1)+3 chia hết cho x-1

=>3 chia hết cho x-1

=>x-1 E Ư(3)={1;-1;3;-3}

=>x E {2;0;4;-2}

b, 3x+2 chia hết cho 2x-1

=>2(3x+2)-3(2x-1) chia hết cho 2x-1

=>6x+4-6x-3 chia hết cho 2x-1

=>1 chia hết cho 2x-1

=>2x-1 E Ư(1)={1;-1}

=>x E {1;0}

cảm ơn bạn nhìu nha

1,x=3 hoặc x=-2

2,x=12

3,không có x nào thỏa mãn

Bài 1 : 

Ta có : 

\(\left|2x-1\right|=5\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}2x-1=5\\2x-1=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=6\\2x=-4\end{cases}}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{6}{2}\\x=\frac{-4}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}}\)

Vậy \(x=-2\) hoặc \(x=3\)

Bài 2 : 

Đặt \(A=\frac{3x+4}{x-1}\) ta có : 

\(A=\frac{3x+4}{x-1}=\frac{3x-3+7}{x-1}=\frac{3x-3}{x-1}+\frac{7}{x-1}=\frac{3\left(x-1\right)}{x-1}+\frac{7}{x-1}=3+\frac{7}{x-1}\)

Để A là số nguyên thì \(\frac{7}{x-1}\) phải nguyên \(\Rightarrow\)\(7⋮\left(x-1\right)\)\(\Rightarrow\)\(\left(x-1\right)\inƯ\left(7\right)\)

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

Suy ra : 

Vậy \(x\in\left\{-6;0;2;8\right\}\) thì \(A\inℤ\)

Chúc bạn học tốt ~ 

                                                           Bài giải

Ta có :

\(x^2+3x+x+3=x^2+0,5x+4x+2\)

\(x^2+4x+3=x^2+4,5x+2\)

\(\Rightarrow\text{ }\left(x^2+4,5x+2\right)-\left(x^2+4x+3\right)=0\)

\(x^2+4,5x+2-x^2-4x-3=0\)

\(0,5x-1=0\)

\(\frac{1}{2}x=1\)

\(x=1\text{ : }\frac{1}{2}\)

\(x=2\)