Đáp án:
\(\begin{array}{l}
B1)\\
a)x < 5\\
b)x \in \left\{ { - 4; - 3; - 2; - 1;0;1;2;3;4} \right\}\\
B2:\\
a)A = - \dfrac{{{x^2}}}{{{{\left( {x - 1} \right)}^2}}}\\
b)A = - \dfrac{9}{{16}}\\
c)x = \dfrac{1}{2}\\
d)x < \dfrac{1}{2};x \ne \left\{ { - 1;0} \right\}\\
e)A < 0
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1)\\
a)\dfrac{{x - 5}}{4} - \dfrac{{2x - 1}}{2} < 3\\
\to \dfrac{{x - 5 - 4x + 2 - 12}}{4} < 0\\
\to - 3x - 15 < 0\\
\to - 5 < x\\
\dfrac{{2x - 1}}{3} < \dfrac{{x + 1}}{2}\\
\to \dfrac{{4x - 2}}{6} < \dfrac{{3x + 3}}{6}\\
\to 4x - 2 < 3x + 3\\
\to x < 5\\
b)\left\{ \begin{array}{l}
x > - 5\\
x < 5
\end{array} \right.\\
Do:x \in Z \to x \in \left\{ { - 4; - 3; - 2; - 1;0;1;2;3;4} \right\}\\
B2:\\
a)DK:x \ne \left\{ { - 1;0;1} \right\}\\
A = \dfrac{1}{{{{\left( {x - 1} \right)}^2}}} - \left[ {\dfrac{{{x^2} - 1}}{{x\left( {{x^2} - 1} \right)}}} \right]:\dfrac{{{{\left( {x - 1} \right)}^2}}}{{x\left( {{x^2} + 1} \right)}}\\
= \dfrac{1}{{{{\left( {x - 1} \right)}^2}}} - \dfrac{1}{x}.\dfrac{{x\left( {{x^2} + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}\\
= \dfrac{1}{{{{\left( {x - 1} \right)}^2}}} - \dfrac{1}{{{{\left( {x - 1} \right)}^2}}} - \dfrac{{{x^2} + 1}}{{{{\left( {x - 1} \right)}^2}}}\\
= - \dfrac{{{x^2}}}{{{{\left( {x - 1} \right)}^2}}}\\
b)\left| {x + 2} \right| = 1\\
\to \left[ \begin{array}{l}
x + 2 = 1\\
x + 2 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\left( l \right)\\
x = - 3
\end{array} \right.\\
Thay:x = - 3\\
\to A = - \dfrac{{{{\left( { - 3} \right)}^2}}}{{{{\left( { - 3 - 1} \right)}^2}}} = - \dfrac{9}{{16}}\\
c)A = - 1\\
\to - \dfrac{{{x^2}}}{{{{\left( {x - 1} \right)}^2}}} = - 1\\
\to \dfrac{{{x^2}}}{{{{\left( {x - 1} \right)}^2}}} = 1\\
\to {x^2} = {x^2} - 2x + 1\\
\to 2x = 1\\
\to x = \dfrac{1}{2}\\
d)A > - 1\\
\to - \dfrac{{{x^2}}}{{{{\left( {x - 1} \right)}^2}}} > - 1\\
\to \dfrac{{{x^2}}}{{{{\left( {x - 1} \right)}^2}}} < 1\\
\to \dfrac{{{x^2} - {x^2} + 2x - 1}}{{{{\left( {x - 1} \right)}^2}}} < 0\\
\to 2x - 1 < 0\left( {do:{{\left( {x - 1} \right)}^2} > 0\forall x \ne 1} \right)\\
\to x < \dfrac{1}{2};x \ne \left\{ { - 1;0} \right\}\\
e)Do:Do:\left\{ \begin{array}{l}
{x^2} > 0\\
{\left( {x - 1} \right)^2} > 0
\end{array} \right.\forall x \ne \left\{ { - 1;0;1} \right\}\\
\to \dfrac{{{x^2}}}{{{{\left( {x - 1} \right)}^2}}} > 0\\
\to - \dfrac{{{x^2}}}{{{{\left( {x - 1} \right)}^2}}} < 0\\
\to A < 0
\end{array}\)
