\(A=\frac{1-2x}{x+3}=\frac{-2\left(x+3\right)+7}{x+3}=-2+\frac{7}{x+3}\)

Vậy để A nguyên thì: \(x+3\inƯ\left(7\right)\)

Mà Ư(7)={1;-1;7;-7}

=>x+3={1;-1;7;-7}

Ta có bảng sau:

Vậy x={-10;-4;-2;4}

Ta có:

\(A=\frac{1-2x}{x+3}=\frac{7-2x-6}{x+3}=\frac{7-2.\left(x+3\right)}{x+3}=\frac{7}{x+3}-\frac{2.\left(x+3\right)}{x+3}=\frac{7}{x+3}-2\)

Để \(A\in Z\Leftrightarrow\frac{7}{x+3}\in Z\)

\(\Rightarrow x+3\inƯ\left(7\right)\)

\(\Rightarrow x+3\in\left\{1;-1;7;-7\right\}\)

\(\Rightarrow x\in\left\{-2;-4;4;-10\right\}\)

Các giá trị A nguyên tương ứng là: 5; -9; -1; -3

Vậy \(\begin{cases}x=-2\\A=5\end{cases}\); \(\begin{cases}x=-4\\A=-9\end{cases}\); \(\begin{cases}x=4\\A=-1\end{cases}\); \(\begin{cases}x=-10\\A=-3\end{cases}\)

a) Điều kiện : \(x\ne2;x\ne3\)

 \(B=\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+4}{3-x}=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x+3}{x-2}+\frac{2x+4}{x-3}\)

\(=\frac{2x-9-\left(x-3\right)\left(x+3\right)+2\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\frac{2x-9-x^2+9+2x^2-8}{\left(x-2\right)\left(x-3\right)}=\frac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}\)

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

b) Điều kiện \(x\in Z;x\ne2;x\ne3\)

Có \(B=\frac{x+4}{x-3}\in Z\), mà x+4 và x-3 nguyên do x nguyên, nên

\(x+4⋮x-3\Leftrightarrow7⋮x-3\), do đó \(x-3\inƯ\left(7\right)=\left\{1;7;-1;-7\right\}\Rightarrow x\in\left\{4;10;2;-4\right\}\)

mà do x khác 2 (điều kiện) nên ta kết luận \(x\in\left\{4;10;-4\right\}\)

a.ĐKXĐ \(\hept{\begin{cases}x\ne-3\\x\ne2\end{cases}}\)

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

=\(\frac{\left(x+2\right)\left(x-2\right)-5-\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)

=\(\frac{x-4}{x-2}\)

b. Để A >0  thì \(\frac{x-4}{x-2}\) >0 \(\Rightarrow\orbr{\begin{cases}x< 2\\x>4\end{cases}}\)

Kết hợp ĐK thì \(\orbr{\begin{cases}x< 2,x\ne-3\\x>4\end{cases}}\)

c. \(A=\frac{x-4}{x-2}=1+\frac{-2}{x-2}\)

Để A nguyên thì \(x-2\inƯ\left(-2\right)=\left\{-2;-1;1;2\right\}\)

\(\Rightarrow x\in\left\{0,1,3,4\right\}\)

Khi thay vào A, để A dương thì \(x\in\left\{0;1\right\}\)

Vậy để A nguyên dương thì \(x\in\left\{0;1\right\}\)

Câu c, có thể nói kết hợp với điều kiện giải được trong câu b, ta tìm được \(x\in\left\{0;1\right\}\)

a.

\(A=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{9\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\frac{x+2\sqrt{x}+3\sqrt{x}-6-9\sqrt{x}+10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\frac{x-4\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\frac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\frac{\sqrt{x}-2}{\sqrt{x}+2}\)

b. Ta có

\(\sqrt{x}=\sqrt{4-2\sqrt{3}}=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)

Thay vào A ta được

\(A=\frac{\sqrt{x}-2}{\sqrt{x}+2}\\ =\frac{\sqrt{3}-1-2}{\sqrt{3}-1+2}\\ =\frac{\sqrt{3}-3}{\sqrt{3}+1}\\ =\frac{\left(\sqrt{3}-3\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\\ =\frac{6-4\sqrt{3}}{2}=3-2\sqrt{3}\)

c. \(A=\frac{\sqrt{x}-2}{\sqrt{x}+2}=\frac{\sqrt{x}+2-4}{\sqrt{x}+2}=1-\frac{4}{\sqrt{x}+2}\)

Để \(A\in Z\Leftrightarrow4⋮\sqrt{x}+2\Leftrightarrow\sqrt{x}+2\inƯ\left(4\right)\)

Ta thấy \(\sqrt{x}\ge0\forall x\ge0\left(ĐK\right)\Leftrightarrow\sqrt{x}+2\ge2\)

Nên \(\sqrt{x}+2\in\left\{2;4\right\}\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+2=2\\\sqrt{x}+2=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}=2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=4\left(ktm\right)\end{matrix}\right.\)

Vậy x=0 thì A thuộc Z

a) Ta có: A= \(\frac{x}{x-2}+\frac{2-x}{x+2}+\frac{12-10x}{x^2-4}\)

A = \(\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(2-x\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{12-10x}{\left(x-2\right)\left(x+2\right)}\)

A = \(\frac{x^2+2x-x^2+4x-4+12-10x}{\left(x-2\right)\left(x+2\right)}\)

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

A = \(\frac{-4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=-\frac{4}{x+2}\)

b) ĐKXĐ: x \(\ne\) \(\pm\)2

Để A \(\in\)Z <=> \(-\frac{4}{x+2}\in Z\) <=> -4 \(⋮\)x + 2

<=> x + 2 \(\in\)Ư(-4) = {1; -1; 2; -2; 4; -4}

Lập bảng :

a) Rút gọn:

\(A=\frac{x}{x-2}+\frac{2-x}{x+2}+\frac{12-10x}{x^2-4}\)

\(A=\frac{x}{x-2}+\frac{2-x}{x+2}+\frac{12-10x}{\left(x-2\right).\left(x+2\right)}\)

\(A=\frac{x.\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}+\frac{\left(2-x\right).\left(x-2\right)}{\left(x-2\right).\left(x+2\right)}+\frac{12-10x}{\left(x-2\right).\left(x+2\right)}\)

\(A=\frac{x^2+2x}{\left(x-2\right).\left(x+2\right)}+\frac{2x-4-x^2+2x}{\left(x-2\right).\left(x+2\right)}+\frac{12-10x}{\left(x-2\right).\left(x+2\right)}\)

\(A=\frac{x^2+2x}{\left(x-2\right).\left(x+2\right)}+\frac{4x-4-x^2}{\left(x-2\right).\left(x+2\right)}+\frac{12-10x}{\left(x-2\right).\left(x+2\right)}\)

\(A=\frac{x^2+2x+4x-4-x^2+12-10x}{\left(x-2\right).\left(x+2\right)}\)

\(A=\frac{8-4x}{\left(x-2\right).\left(x+2\right)}\)

\(A=\frac{4.\left(2-x\right)}{\left(x-2\right).\left(x+2\right)}\)

\(A=\frac{4}{x+2}.\)

Chúc bạn học tốt!

a) \(\frac{x}{x-2}+\frac{2-x}{x+2}+\frac{12-10x}{x^2-4}\)

=\(\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(2-x\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{12-10x}{\left(x-2\right)\left(x+2\right)}\)

=\(\frac{x\left(x+2\right)+\left(2-x\right)\left(x-2\right)+12-10x}{\left(x-2\right)\left(x+2\right)}\)

=\(\frac{x^2+2x+2x-4-x^2+2x+12-10x}{\left(x-2\right)\left(x+2\right)}\)

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

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

=\(\frac{-4}{x+2}\)

b)(ĐKXĐ của A là x\(\ne\pm2\))

Với x\(\ne\pm2\) ta có:

A\(\in Z\)

\(\Leftrightarrow\frac{-4}{x+2}\in Z\)

\(\Rightarrow x+2\inƯ_{\left(-4\right)}=\left\{\pm1;\pm2;\pm4\right\}\)

Ta có bảng sau :

Vậy để \(A\in Z\) thì x = {-6,-4,-3,-1,0}