Anh thấy ảnh lỗi em ạ

Mik ko thấy ảnh đâu bn ơi

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

A là số nguyên khi: \(\dfrac{3}{x-2}\) nguyên 

3 ⋮ x - 2

\(\Rightarrow x-2\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

\(\Rightarrow x\in\left\{3;1;5;-1\right\}\)

A nguyên

=>10x-15+6 chia hết cho 2x-3

=>\(2x-3\in\left\{1;-1;3;-3\right\}\)

=>\(x\in\left\{2;1;3;0\right\}\)

Để A nguyên thì \(x^2-4x-4⋮x-7\)

\(\Rightarrow x^2+3x-7x-21+17⋮x-7\)

\(\Rightarrow\left(x-7\right)\left(x+3\right)+17⋮x-7\)

Mà \(\left(x-7\right)\left(x+3\right)⋮x-7\)

\(\Rightarrow17⋮x-7\)

\(\Rightarrow x-7\in\left\{1;17;-1;-17\right\}\)

\(\Rightarrow x\in\left\{8;24;6;-10\right\}\)

\(\text{A=}\frac{x^2-4x-4}{x-7}\)

\(=\frac{x^2-4x-21+17}{x-7}\)

\(=\frac{x^2+3x-7x-21}{x-7}+\frac{17}{x-7}\)

\(=\frac{x\left(x+3\right)-7\left(x+3\right)}{x-7}+\frac{17}{x-7}\)

\(=\frac{\left(x-7\right)\left(x+3\right)}{x-7}+\frac{17}{x-7}\)

\(=\left(x+3\right)+\frac{17}{x-7}\)

Vì \(3\in Z\)

\(\Leftrightarrow x+3\in Z\)

\(\Rightarrow\text{A}\in Z\text{ khi }\frac{17}{x-7}\in Z\)

\(\Leftrightarrow\left(x-7\right)\inƯ\left(17\right)=\left\{1;-1;17;-17\right\}\)

\(\Leftrightarrow x=\left\{8;6;24;-10\right\}\)

Vậy với \(x=\left\{-10;6;8;24\right\}\)thì A có giá trị nguyên

Để A là số nguyê thì:

\(x^2-4x-4⋮x-7\)

\(x^2+3x-7x-21+17⋮x-7\)

\(\left(x-7\right)\left(x-3\right)+17⋮x-7\)

Mà \(\left(x-7\right)\left(x-3\right)⋮x-7\)nên

\(17⋮x-7\)

\(x-7\inƯ\left(17\right)=\left\{-1;1;-17;17\right\}\)

x={8;24;6;-10}

\(\frac{x+1}{x-2}\)

Để \(\frac{x+1}{x-2}\inℤ\Rightarrow x+1⋮x-2\Rightarrow\left(x-2\right)+3⋮x-2\Rightarrow3⋮x-2\)

\(\Rightarrow x-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\Rightarrow x\in\left\{3;1;5;-1\right\}\)

\(\frac{12x+1}{30x+2}\)

Gọi \(n=ƯC\left(12x+1;30x+2\right)\)

\(\Rightarrow\hept{\begin{cases}12x+1⋮n\Rightarrow60x+5⋮n\\30x+2⋮n\Rightarrow60x+4⋮n\end{cases}}\)

\(\Rightarrow\left(60x+5\right)-\left(60x+4\right)⋮x\Rightarrow1⋮n\Rightarrow n=1\Rightarrow\frac{12x+1}{30x+2}\)là phân số tối giản

a) Giải:

Để A có giá trị là số nguyên thì \(x+1⋮x-2\)

Ta có:
\(x+1⋮x-2\)

\(\Rightarrow\left(x-2\right)+3⋮x-2\)

\(\Rightarrow3⋮x-2\)

\(\Rightarrow x-2\in\left\{\pm1;\pm3\right\}\)

+) \(x-2=1\Rightarrow x=3\)

+) \(x-2=-1\Rightarrow x=1\)

+) \(x-2=3\Rightarrow x=5\)

+) \(x-2=-3\Rightarrow x=-1\)

Vậy \(x\in\left\{3;1;5;-1\right\}\)

b) Để B có giá trị nguyên thì \(2x-1⋮x+5\)

Ta có:

\(2x-1⋮x+5\)

\(\Rightarrow\left(2x+10\right)-9⋮x+5\)

\(\Rightarrow2.\left(x-5\right)-9⋮x+5\)

\(\Rightarrow-9⋮x+5\)

\(\Rightarrow x+5\in\left\{\pm1;\pm3;\pm9\right\}\)

+) \(x+5=1\Rightarrow x=-4\)

+) \(x+5=-1\Rightarrow x=-6\)

+) \(x+5=3\Rightarrow x=-2\)

+) \(x+5=-3\Rightarrow x=-8\)

+) \(x+5=9\Rightarrow x=4\)

+) \(x+5=-9\Rightarrow x=-14\)

Vậy \(x\in\left\{-4;-;-2;-8;4;-14\right\}\)