a, \(Đkxđ:x\ne-3;x\ne2\)
b,\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)
\(=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{1}{x-2}\)
\(=\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}\)\(=\dfrac{x-4}{x-2}\)
c,\(A=-\dfrac{3}{4}\) khi \(\dfrac{x-4}{x-2}=-\dfrac{3}{4}\)
\(\Leftrightarrow\left(x-4\right).4=-3\left(x-2\right)\)
\(\Leftrightarrow4x-16=-3x+6\)
\(\Leftrightarrow7x=22\)
\(\Leftrightarrow x=\dfrac{22}{7}\)
Vậy khi \(x=\dfrac{22}{7}\) thì \(A=-\dfrac{3}{4}\)
a) ĐKXĐ : \(\left\{{}\begin{matrix}x+3\ne0\\2-x\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\)
b) \(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}-\dfrac{1}{x-2}\)
\(A=\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)
\(A=\dfrac{-x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\dfrac{x-4}{x-2}\)
c) Để \(A=\dfrac{-3}{4}\) thì :
\(A=\dfrac{x-4}{x-2}=\dfrac{-3}{4}\)
\(\Rightarrow\dfrac{x-4}{x-2}+\dfrac{3}{4}=0\)
\(\Rightarrow\dfrac{4\left(x-4\right)}{4\left(x-2\right)}+\dfrac{3\left(x-2\right)}{4\left(x-2\right)}=0\)
\(\Rightarrow4x-16+3x-6=0\)
\(\Rightarrow7x+22=0\)
\(\Rightarrow x=\dfrac{-22}{7}\)
d) Ta có : \(A=\dfrac{x-4}{x-2}=\dfrac{x-2-2}{x-2}=1-\dfrac{2}{x-2}\)
Vì \(1\in Z\) để \(A\in Z\) thì \(\dfrac{2}{x-2}\in Z\)
\(\Rightarrow x-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Có : \(\left\{{}\begin{matrix}x-2=1=>x=3\\x-2=-1=>x=1\\x-2=2=>x=4\\x-2=-2=>0\end{matrix}\right.\)
Vậy để A nhận gt nguyên thì x \(\in\left\{3;1;4;0\right\}\)
e) \(x^2-9=0\)
\(\Rightarrow\left(x+3\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(loại\right)\\x=3\end{matrix}\right.\)
Thay vào A ta có :
\(A=\dfrac{x-4}{x-2}=\dfrac{3-4}{3-2}=-1\)
d, \(A\) nguyên thì \(\dfrac{x-4}{x-2}\) nguyên.
Ta có: \(\dfrac{x-4}{x-2}=\dfrac{x-2-2}{x-2}=1-\dfrac{2}{x-2}\)
Mà \(1\) nguyên, để \(A\) nguyên thì \(\dfrac{2}{x-2}\in Z\)
\(\Rightarrow x-2\inƯ\left(2\right)=\left\{2;-2;1;-1\right\}\)
Ta có bảng sau:
| \(x-2\) | 2 | -2 | 1 | -1 |
| \(x\) | 4 | 0 | 3 | 1 |
Vậy A nguyên khi \(x\in\left\{4;0;3;1\right\}\)
e, \(x^2-9=0\)
\(\Leftrightarrow x^2-3^2=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-3\left(ktm\right)\end{matrix}\right.\)
Khi \(x=3\) thì \(A=\dfrac{3-4}{3-2}=-1\)
\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)
a)\(ĐK:\left\{{}\begin{matrix}x+3\ne0\\x^2+x-6\ne\\2-x\ne0\end{matrix}\right.0\Leftrightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\)
b)A=\(\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{-\left(x-2\right)}\)
=\(\dfrac{\left(x+2\right)\left(x-2\right)-5-1\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)
=\(\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)
=\(\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)
=\(\dfrac{x-4}{x-2}\)
c)\(A=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{-3}{4}=\dfrac{x-4}{x-2}\)
\(\Leftrightarrow x-4=\left(x-2\right).\dfrac{-3}{4}\Leftrightarrow x-4=\dfrac{-3}{4}x+\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{7}{4}x=\dfrac{11}{2}\)
\(\Leftrightarrow x=\dfrac{22}{7}\)
d)A=\(\dfrac{x-4}{x-2}=1+\dfrac{-2}{x-2}\)
Để \(A\in Z\) thì \(x-2\inƯ\left(-2\right)=\left\{\pm1;\pm2\right\}\)
Xét các trường hợp:
*x-2=1 =>x=3
*x-2=-1 =>x=1
*x-2=2 =>x=4
* x-2=-2 =>x=0
Với x={0;1;3;4} thoả mãn yêu cầu.
e)x2-9=0
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Thay x=3 vào biểu thức A
Ta có:A=\(\dfrac{x-4}{x-2}=\dfrac{3-4}{3-2}=-1\)
