\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)
a) ĐKXĐ:
\(\begin{cases} x+3\ne 0\\ x^2+x-6 \ne 0 \Rightarrow (x+3)(x-2) \ne 0\\ 2-x\ne 0 \end{cases} \\\Leftrightarrow \begin{cases} x\ne -3\\ x\ne 2 \end{cases} \)
b) Với \(x\ne-3;x\ne2\) ta có:
\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)
\(\Leftrightarrow\dfrac{x+2}{x+3}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}+\dfrac{1}{2-x}\)
\(\Leftrightarrow\dfrac{x+2}{x+3}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}-\dfrac{1}{x-2}\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}-\dfrac{x+3}{\left(x-2\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x^2-4-5-\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x-4}{x-2}\)
c) Để \(A=-\dfrac{3}{4}\) thì:
\(\dfrac{x-4}{x-2}=-\dfrac{3}{4}\)
\(\Leftrightarrow\left(x-4\right).4=\left(x-2\right).\left(-3\right)\)
\(\Leftrightarrow4x-16=-3x+6\)
\(\Leftrightarrow4x+3x=6+16\)
\(\Leftrightarrow7x=22\)
\(\Leftrightarrow x=\dfrac{22}{7}\)
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 \(\dfrac{2}{x-2}\in Z\)
\(\Rightarrow x-2\inƯ\left(2\right)=\left\{-1;-2;1;2\right\}\)
Do đó
| \(x-2\) | \(-1\) | \(-2\) | \(1\) | \(2\) |
| \(x\) | 1 | 0 | 3 | 4 |
Vậy để A nhận giá trị nguyên thì \(x\in\left\{0;1;3;4\right\}\)
Làm nốt con e nha :v
x2 - 9 = 0
⇔ ( x + 3)( x - 3) = 0
⇔ x = 3 ( TM) hoặc : x = -3 ( KTM)
Với : x = 3 , ta có :
A = \(\dfrac{3-4}{3-2}=\dfrac{-1}{1}=-1\)
KL.....