Đáp án:
B1:
a) \(\dfrac{{x - 1}}{{x + 2}}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)DK:x \ne \left\{ { - 3; - 2;1} \right\}\\
A = \dfrac{{4 - {x^2} - 9 + {x^2} + 2 - x}}{{\left( {x + 3} \right)\left( {x + 2} \right)}}:\dfrac{{x - 1 - x}}{{x - 1}}\\
= \dfrac{{ - 3 - x}}{{\left( {x + 3} \right)\left( {x + 2} \right)}}.\dfrac{{x - 1}}{{ - 1}}\\
= \dfrac{{\left( {x + 3} \right)\left( {x - 1} \right)}}{{\left( {x + 3} \right)\left( {x + 2} \right)}} = \dfrac{{x - 1}}{{x + 2}}\\
b)A = 0\\
\to \dfrac{{x - 1}}{{x + 2}} = 0\\
\to x - 1 = 0\\
\to x = 1\left( l \right)\\
\to x \in \emptyset \\
A > 0\\
\to \dfrac{{x - 1}}{{x + 2}} > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 1 > 0\\
x + 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 1 < 0\\
x + 2 < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 1\\
x > - 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 1\\
x < - 2
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x > 1\\
x < - 2;x \ne - 3
\end{array} \right.\\
B2:\\
a)DK:y \ne \left\{ { - \dfrac{3}{2}; - 1} \right\}\\
B = \dfrac{{3{y^3} - 3{y^2} - 4{y^2} + 4y + y - 1}}{{2{y^3} - 2{y^2} + {y^2} - y - 3y + 3}}\\
= \dfrac{{3{y^2}\left( {y - 1} \right) - 4y\left( {y - 1} \right) + \left( {y + 1} \right)}}{{2{y^2}\left( {y - 1} \right) + y\left( {y - 1} \right) - 3\left( {y - 1} \right)}}\\
= \dfrac{{\left( {y - 1} \right)\left( {3{y^2} - 4y + 1} \right)}}{{\left( {y - 1} \right)\left( {2{y^2} + y - 3} \right)}}\\
= \dfrac{{3{y^2} - 4y + 1}}{{2{y^2} + y - 3}}\\
= \dfrac{{\left( {y - 1} \right)\left( {3y - 1} \right)}}{{\left( {y - 1} \right)\left( {2y + 3} \right)}}\\
= \dfrac{{3y - 1}}{{2y + 3}}\\
c)B \ge 1\\
\to \dfrac{{3y - 1}}{{2y + 3}} \ge 1\\
\to \dfrac{{3y - 1 - 2y - 3}}{{2y + 3}} \ge 0\\
\to \dfrac{{y - 4}}{{2y + 3}} \ge 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
y - 4 \ge 0\\
2y + 3 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
y - 4 \le 0\\
2y + 3 < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
y \ge 4\\
y < - \dfrac{3}{2}
\end{array} \right.
\end{array}\)
