Đáp án:
a)x=$\frac{22}{7}$
b)x∈{0;1;3;4}
c)Khi x=3 thì P=-1
Giải thích các bước giải:
\(
\begin{array}{l}
P = \frac{{x + 2}}{{x + 3}} - \frac{5}{{x^2 + x - 6}} + \frac{1}{{2 - x}} \\
Đk:x \ne - 3;x \ne 2 \\
P = \frac{{x + 2}}{{x + 3}} - \frac{5}{{(x - 2)(x + 3)}} - \frac{1}{{x - 2}} \\
= \frac{{(x + 2)(x - 2)}}{{(x + 3)(x - 2)}} - \frac{5}{{(x - 2)(x + 3)}} - \frac{{x + 3}}{{(x - 2)(x + 3)}} \\
= \frac{{x^2 - 4 - 5 - x - 3}}{{(x - 2)(x + 3)}} \\
= \frac{{x^2 - x - 12}}{{(x - 2)(x + 3)}} \\
= \frac{{(x - 4)(x + 3)}}{{(x - 2)(x + 3)}} \\
= \frac{{x - 4}}{{x - 2}} \\
a)P = \frac{{ - 3}}{4} \Leftrightarrow \frac{{x - 4}}{{x - 2}} = \frac{{ - 3}}{4}(x \ne - 3;x \ne 2) \\
\Leftrightarrow 4x - 16 = - 3x + 6 \\
\Leftrightarrow 7x = 22 \\
\Leftrightarrow x = \frac{{22}}{7} \\
b)P = \frac{{x - 4}}{{x - 2}} = \frac{{x - 2 - 2}}{{x - 2}} = 1 - \frac{2}{{x - 2}}(x \ne - 3;x \ne 2) \\
P \in Z \Leftrightarrow \frac{2}{{x - 2}} \in Z \Leftrightarrow x - 2 \in Ư(2) = {\rm{\{ }} \pm {\rm{1;}} \pm {\rm{2\} }} \\
{\rm{ = > x}} \in {\rm{\{ 0;1;3;4\} (tm)}} \\
{\rm{c)P = }}\frac{{{\rm{x - 4}}}}{{{\rm{x - 2}}}}(x \ne - 3;x \ne 2) \\
{\rm{x}}^{\rm{2}} - 9 = 0 \\
\Leftrightarrow \left[ {\begin{array}{*{20}c}
{x = 3(tm)} \\
{x = - 3(ktm)} \\
\end{array}} \right. \\
+ )x = 3 = > P = \frac{{3 - 4}}{{3 - 2}} = - 1 \\
\end{array}
\)
