b, \(P=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+3x-2x-6}+\dfrac{1}{2-x}\)
\(=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{1}{x-2}\)
\(=\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-4x+3x-12}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{x-4}{x-2}\)
c, Để \(P=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{x-4}{x-2}=-\dfrac{3}{4}\)
\(\Leftrightarrow4\left(x-4\right)=-3\left(x-2\right)\)
\(\Leftrightarrow4x-16+3x-6=0\)
\(\Leftrightarrow7x-22=0\)
\(\Leftrightarrow x=\dfrac{22}{7}\)
d, Để P có giá trị nguyên
\(\Leftrightarrow x-4⋮x-2\)
\(\Leftrightarrow\left(x-2\right)-2⋮x-2\)
\(\Leftrightarrow2⋮x-2\Leftrightarrow x-2\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
| \(x-2\) | 1 | -1 | 2 | -2 |
| x | 3 | 1 | 4 | 0 |
e,
\(x^2-9=0\)
\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Với x=3,có :
\(\dfrac{x-4}{x-2}=\dfrac{3-4}{3-2}=-\dfrac{1}{1}=-1\)
Với x=-3,có :
\(\dfrac{x-4}{x-2}=\dfrac{-3-4}{-3-2}=\dfrac{7}{5}\)