Đáp án:
a. \(A = \dfrac{{2x}}{{1 - x}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ne \pm 1\\
A = \dfrac{{2x\left( {x + 1} \right)}}{{\left( {1 - x} \right)\left( {1 + x} \right)}} = \dfrac{{2x}}{{1 - x}}\\
b.DK:x \ne \left\{ {1;2} \right\}\\
B = \dfrac{{1 - 2x}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} + \dfrac{{x + 1}}{{x - 2}}\\
= \dfrac{{1 - 2x + {x^2} - 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
= \dfrac{{x\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \dfrac{x}{{x - 1}}\\
P = A.B = \dfrac{{2x}}{{1 - x}}.\dfrac{x}{{x - 1}} = \dfrac{{ - 2{x^2}}}{{{x^2} - 2x + 1}}\\
P = \dfrac{9}{2}\\
\to \dfrac{{ - 2{x^2}}}{{{x^2} - 2x + 1}} = \dfrac{9}{2}\\
\to 9{x^2} - 18x + 9 = - 4{x^2}\\
\to 13{x^2} - 18x + 9 = 0\\
\to {\left( {x\sqrt {13} } \right)^2} - 2x\sqrt {13} .\dfrac{9}{{\sqrt {13} }} + {\left( {\dfrac{9}{{\sqrt {13} }}} \right)^2} + \dfrac{{36}}{{13}} = 0\\
\to {\left( {x\sqrt {13} - \dfrac{9}{{\sqrt {13} }}} \right)^2} + \dfrac{{36}}{{13}} = 0\left( {voly} \right)\\
Do:{\left( {x\sqrt {13} - \dfrac{9}{{\sqrt {13} }}} \right)^2} \ge 0\forall x\\
\to {\left( {x\sqrt {13} - \dfrac{9}{{\sqrt {13} }}} \right)^2} + \dfrac{{36}}{{13}} > 0\\
\to x \in \emptyset \\
c.P < 1\\
\to \dfrac{{ - 2{x^2}}}{{{x^2} - 2x + 1}} < 1\\
\to \dfrac{{ - 2{x^2} - {x^2} + 2x - 1}}{{{x^2} - 2x + 1}} < 0\\
\to \dfrac{{ - 3{x^2} + 2x - 1}}{{{{\left( {x - 1} \right)}^2}}} < 0\\
\to - 3{x^2} + 2x - 1 < 0\left( {do:{{\left( {x - 1} \right)}^2} > 0\forall x \ne 1} \right)\\
Mà: - 3{x^2} + 2x - 1 < 0\left( {ld} \right)\forall x \ne 1\\
KL:P < 1\forall x \ne 1
\end{array}\)
