ĐK: \(x\ne-3;x\ne2\)
a) \(A=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}\)
\(A=\frac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}=\frac{\left(x+3\right)\left(x-4\right)}{\left(x-2\right)\left(x+3\right)}=\frac{x-4}{x-2}\)
b) \(A=-\frac{3}{4}\Leftrightarrow\frac{x-4}{x-2}=-\frac{3}{4}\Rightarrow4x-16=-3x+6\Leftrightarrow7x=22\Leftrightarrow x=\frac{22}{7}\)
c) \(A=\frac{x-4}{x-2}=\frac{x-2-2}{x-2}=1-\frac{2}{x-2}\Rightarrow A\in Z\Leftrightarrow\frac{2}{x-2}\in Z\left(1\in Z\right)\Leftrightarrow x-2\inƯ\left(2\right)\Leftrightarrow x-2\in\left\{1;-1;2;-2\right\}\)
| x-2 | 1 | -1 | 2 | -2 |
| x | 3(t/m đk) | 1(t/m đk) | 4(t/m đk) | 0(t/m đk) |
=> A nguyên <=>\(x\in\left\{0;1;3;4\right\}\)
d) \(x^2-9=0\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\Rightarrow\)x=3 hoặc x=-3
thay lần lượt x=3, x=-3 vào A =x-4/x-2 là ra thôi.
bài này dài làm mệt quá, bằng mình làm mấy bài khác
