\(A=\frac{3x-1}{x-1}=\frac{3\left(x-1\right)+2}{x-1}=3+\frac{2}{x-1}\)

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

Để A,B đều là số nguyên thì \(x-1\in\left\{1;2;-1;-2\right\}\) và \(x+2\in\left\{1;5;-1;-5\right\}\)

Bạn tự làm nốt

a) bài 1

để \(x\in Z\)thì \(3x-1⋮x-1\)

mà \(x-1⋮x-1\)

\(\Rightarrow3\left(x-1\right)⋮x-1\)

\(\Rightarrow\left(3x-1\right)-\left[3x-3\right]⋮x-1\)

\(\Rightarrow2⋮x-1\)

\(\Rightarrow x-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

ta có bảng

vậy \(x\in\left\{2;0;3;-1\right\}\)

còn nữa mà bạn

câu c bài 2:

x chỉ nhận giá trị là 2

thật vậy

\(2^2=4\)

\(2.2=4\)

a) A xác định \(\Leftrightarrow\hept{\begin{cases}3x\ne0\\x+1\ne0\\2-4x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne\frac{1}{2}\end{cases}}}\)

\(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\)

\(A=\left[\frac{\left(x+2\right)\left(x+1\right)}{3x\left(x+1\right)}+\frac{2\cdot3x}{3x\left(x+1\right)}-\frac{3\cdot3x\left(x+1\right)}{3x\left(x+1\right)}\right]\cdot\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

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

\(A=\frac{\left(-8x^2+2\right)\left(x+1\right)}{3x\left(x+1\right)2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{2\left(1-4x^2\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{2\left(1-2x\right)\left(1-2x\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{1+2x}{3x}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{2x+1-3x-1+x^2}{3x}\)

\(A=\frac{x^2-x}{3x}\)

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

\(A=\frac{x-1}{3}\)

b) Thay x = 4 ta có :

\(A=\frac{4-1}{3}=\frac{3}{3}=1\)

c) Để A thuộc Z thì \(x-1⋮3\)

\(\Rightarrow x-1\in B\left(3\right)=\left\{0;3;6;...\right\}\)

\(\Rightarrow x\in\left\{1;4;7;...\right\}\)

Vậy.....

Cho Bt 

a,Tìm điều kiện xác định và rút gọn bt A

b,Tính giá trị bt A tại x=4

c,tìm x thuộc Z để a thuộc Z

a ) ĐKXĐ : \(\hept{\begin{cases}x+1\ne0\\x-1\ne0\\x^2-1\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne-1\\x\ne1\end{cases}}}\)

b ) \(P=\frac{2x+3}{x+1}-\frac{x+2}{x-1}+\frac{3x+5}{x^2-1}\)

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

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

\(=\frac{x^2-x}{\left(x-1\right)\left(x+1\right)}=\frac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x}{x+1}\)

Sr còn thiếu

Để \(P\in Z\Leftrightarrow\frac{x}{x+1}=\frac{x+1-1}{x+1}=1-\frac{1}{x+1}\in Z\Rightarrow x+1\inƯ\left(1\right)\)

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

cảm ơn bạn

\(ĐKXĐ:x\ne\pm1\)

a) \(P=\frac{2x+3}{x+1}-\frac{x+2}{x-1}+\frac{3x+5}{x^2-1}\)

\(\Leftrightarrow P=\frac{\left(2x+3\right)\left(x-1\right)-\left(x+2\right)\left(x+1\right)+3x+5}{\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow P=\frac{2x^2+x-3-x^2-3x-2+3x+5}{\left(x-1\right)\left(x+1\right)}\)

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

\(\Leftrightarrow P=\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow P=\frac{x}{x-1}\)

b) Để \(P\inℤ\)

\(\Leftrightarrow x⋮x-1\)

\(\Leftrightarrow x-1+1⋮x-1\)

\(\Leftrightarrow1⋮x-1\)

\(\Leftrightarrow x-1\inƯ\left(1\right)=\left\{\pm1\right\}\)

\(\Leftrightarrow x\in\left\{0;2\right\}\)

Vậy để \(A\inℤ\Leftrightarrow x\in\left\{0;2\right\}\)