Tập hợp các giá trị nguyên để Q nguyên là: {0;1;9;5;6;7}

a) ĐKXĐ: \(\hept{\begin{cases}x+2\ne0\\x^2-4\ne0\\2-x\ne0\end{cases}}\) => \(\hept{\begin{cases}x\ne-2\\x\ne\pm2\\x\ne2\end{cases}}\) => \(x\ne\pm2\)

Ta có:Q = \(\frac{x-1}{x+2}+\frac{4x+4}{x^2-4}+\frac{3}{2-x}\)

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

Q = \(\frac{x^2-2x-x+2+4x+4-3x-6}{\left(x+2\right)\left(x-2\right)}\)

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

b) ĐKXĐ P: x - 3 \(\ne\)0 => x \(\ne\)3

Ta có: P = 3 => \(\frac{x+2}{x-3}=3\)

=> x + 2 = 3(x - 3)

=> x + 2 = 3x - 9

=> x - 3x = -9 - 2

=> -2x = -11

=> x = 11/2 (tm)

Với x = 11/2 thay vào Q => Q = \(\frac{\frac{11}{2}}{\frac{11}{2}+2}=\frac{11}{15}\)

c) Với x \(\ne\)\(\pm\)2; x \(\ne\)3

Ta có: M = PQ = \(\frac{x+2}{x-3}\cdot\frac{x}{x+2}=\frac{x}{x-3}=\frac{x-3+3}{x-3}=1+\frac{3}{x-3}\)

Để M \(\in\)Z <=> 3 \(⋮\)x - 3

=> x - 3 \(\in\)Ư(3) = {1; -1; 3; -3}

Lập bảng:

Vậy ...

a) \(Q=\frac{x+3}{2x+1}-\frac{x-7}{2x+1}\left(ĐK:x\ne-\frac{1}{2}\right)\)

\(=\frac{x+3-x+7}{2x+1}=\frac{10}{2x+1}\)

b) Để Q nguyên \(\Leftrightarrow\frac{10}{2x+1}\in Z\)

=> \(2x+1\inƯ\left(10\right)\)

=> \(2x+1\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\)

Ta có bảng sau:

Vậy \(x\in\left\{0;-1\right\}\)