Đáp án:
\(\begin{array}{l}
B1:\\
a)x \ne \pm 2\\
b) - \dfrac{2}{{x - 2}}\\
c)A = \dfrac{1}{3}\\
d)\left[ \begin{array}{l}
x = 4\\
x = 0\\
x = 3\\
x = 1
\end{array} \right.\\
e)x < 2\\
B2:\\
a)A = \dfrac{4}{9}\\
b) - \dfrac{2}{{x + 2}}\\
c)x = 3
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)DK:x \ne \pm 2\\
b)A = \dfrac{{3{x^2} - 4 - 2x\left( {x + 2} \right) - x\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{3{x^2} - 4 - 2{x^2} - 4x - {x^2} + 2x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{ - 2x - 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= - \dfrac{2}{{x - 2}}\\
c)\left| {x + 3} \right| = 1\\
\to \left[ \begin{array}{l}
x + 3 = 1\\
x + 3 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 2\left( l \right)\\
x = - 4
\end{array} \right.\\
Thay:x = - 4\\
\to A = - \dfrac{2}{{ - 4 - 2}} = \dfrac{1}{3}\\
d)A \in Z \to - \dfrac{2}{{x - 2}} \in Z\\
\to x - 2 \to U\left( 2 \right)\\
\to \left[ \begin{array}{l}
x - 2 = 2\\
x - 2 = - 2\\
x - 2 = 1\\
x - 2 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 4\\
x = 0\\
x = 3\\
x = 1
\end{array} \right.\\
e)A > 0\\
\to - \dfrac{2}{{x - 2}} > 0\\
\to x - 2 < 0\\
\to x < 2\\
B2:\\
a)Thay:x = 7\\
\to A = \dfrac{{7 - 3}}{{7 + 2}} = \dfrac{4}{9}\\
b)B = \dfrac{{6 - 7x + 3\left( {x - 2} \right) + 2\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{6 - 7x + 3x - 6 + 2x + 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{ - 2x + 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= - \dfrac{2}{{x + 2}}\\
c)M = A:B = \dfrac{{x - 3}}{{x + 2}}: - \dfrac{2}{{x + 2}}\\
= - \dfrac{{x - 3}}{2}\\
\left| M \right| = M \to \left[ \begin{array}{l}
M = M\left( {ld} \right)\\
M = - M
\end{array} \right.\\
\to 2M = 0\\
\to M = 0\\
\to - \dfrac{{x - 3}}{2} = 0\\
\to x - 3 = 0\\
\to x = 3
\end{array}\)
