a) biểu thức nguyên khi 5 chia hết cho 4-x

\(\Rightarrow4-x=Ư\left(5\right)=\left\{-5;-1;1;5\right\}\)

Từ đó suy ra x

b) \(\frac{8x-1}{4x-1}=\frac{2\left(4x-1\right)+1}{4x-1}=\frac{2\left(4x-1\right)}{4x-1}+\frac{1}{4x-1}=2+\frac{1}{4x-1}\) 

Do đó biểu thức nguyên khi 1/(4x-1) nguyên, tương tự câu a

c) Tương tự câu b

a: \(A=\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+x+1}{x+1}\right)\cdot\dfrac{\left(x+1\right)^2}{2x+1}\)

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

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

\(=\dfrac{2x+1}{x-1}\cdot\dfrac{x+1}{2x+1}=\dfrac{x+1}{x-1}\)

b: Thay x=1/2 vào A, ta được:

\(A=\dfrac{\dfrac{1}{2}+1}{\dfrac{1}{2}-1}=\dfrac{3}{2}:\dfrac{-1}{2}=-3\)

c: Để A là số nguyên thì \(x-1+2⋮x-1\)

\(\Leftrightarrow x-1\in\left\{1;-1;2;-2\right\}\)

\(\Leftrightarrow x\in\left\{2;0;3\right\}\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

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 ...

15/10

a/ Để C nguyên thì 3x+1 phải là Ư(5)

\(\Leftrightarrow3x+1\in\left\{1;-1;5;-5\right\}\)

\(\Leftrightarrow x\in\left\{0;-\frac{2}{3};\frac{4}{3};-2\right\}\)

mà x nguyên nên x={0;-2}

b/ \(\frac{x+1}{x-1}=1+\frac{2}{x-1}\)\(\Rightarrow x-1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)

\(\Leftrightarrow x\in\left\{2;0;3;-1\right\}\)

a) Để \(C\inℤ\) 

\(\Rightarrow5⋮3x+1\)

\(\Rightarrow3x+1\inƯ\left(5\right)\)

\(\Rightarrow3x+1\in\left\{1;5;-1;-5\right\}\)

Lập bảng xét các trường hợp ta có : 

Vậy \(x\in\left\{0;2\right\}\)

b) Để \(D\inℤ\)

\(\Rightarrow\left(x+1\right)⋮\left(x-1\right)\)

\(\Rightarrow\left(x-1+2\right)⋮\left(x-1\right)\)

Vì \(\left(x-1\right)⋮\left(x-1\right)\)

\(\Rightarrow2⋮\left(x-1\right)\)

\(\Rightarrow\left(x-1\right)\inƯ\left(2\right)\)

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

Lập bảng xét các trường hợp ta có : 

Vậy \(x\in\left\{2;0;3;-1\right\}\)