Đáp án:
\(\begin{array}{l}
B3:\\
a)\dfrac{{{{\left( {x + 1} \right)}^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
b)dpcm\\
B4:\\
a)\dfrac{{x + 1}}{{x - 3}}\\
b)K = - 1\\
c)x = 7\\
d)x < 3\\
e)\left[ \begin{array}{l}
x = 7\\
x = - 1\\
x = 5\\
x = 1\\
x = 4
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B3:\\
a)A = \dfrac{{x + 2 + x - 2 + {x^2} + 1}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{{x^2} + 2x + 1}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{{{\left( {x + 1} \right)}^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
b)A < 0\\
\to \dfrac{{{{\left( {x + 1} \right)}^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} < 0\\
\to \left( {x - 2} \right)\left( {x + 2} \right) < 0\left( {do:{{\left( {x + 1} \right)}^2} \ge 0\forall x \ne \pm 2} \right)\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 2 > 0\\
x + 2 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 2 < 0\\
x + 2 > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 2\\
x < - 2
\end{array} \right.\left( l \right)\\
- 2 < x < 2
\end{array} \right.\\
\to dpcm\\
B4:\\
a)DK:x \ne \left\{ {2;3} \right\}\\
K = \dfrac{{2x - 9 - \left( {x + 3} \right)\left( {x - 3} \right) + \left( {2x + 1} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x - 3} \right)}}\\
= \dfrac{{2x - 9 - {x^2} + 9 + 2{x^2} - 3x - 2}}{{\left( {x - 2} \right)\left( {x - 3} \right)}}\\
= \dfrac{{{x^2} - x - 2}}{{\left( {x - 2} \right)\left( {x - 3} \right)}}\\
= \dfrac{{\left( {x - 2} \right)\left( {x + 1} \right)}}{{\left( {x - 2} \right)\left( {x - 3} \right)}}\\
= \dfrac{{x + 1}}{{x - 3}}\\
b){x^2} - 3x + 2 = 0\\
\to \left( {x - 2} \right)\left( {x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\left( l \right)\\
x = 1
\end{array} \right.\\
Thay:x = 1\\
\to K = \dfrac{{1 + 1}}{{1 - 3}} = - 1\\
c)K = 2\\
\to \dfrac{{x + 1}}{{x - 3}} = 2\\
\to x + 1 = 2x - 6\\
\to x = 7\\
d)K < 1\\
\to \dfrac{{x + 1}}{{x - 3}} < 1\\
\to \dfrac{{x + 1 - x + 3}}{{x - 3}} < 0\\
\to \dfrac{4}{{x - 3}} < 0\\
\to x < 3\\
e)K = \dfrac{{x + 1}}{{x - 3}} = \dfrac{{x - 3 + 4}}{{x - 3}}\\
= 1 + \dfrac{4}{{x - 3}}\\
K \in Z \to \dfrac{4}{{x - 3}} \in Z\\
\to x - 3 \in U\left( 4 \right)\\
\to \left[ \begin{array}{l}
x - 3 = 4\\
x - 3 = - 4\\
x - 3 = 2\\
x - 3 = - 2\\
x - 3 = 1\\
x - 3 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 7\\
x = - 1\\
x = 5\\
x = 1\\
x = 4\\
x = 2\left( l \right)
\end{array} \right.
\end{array}\)
