Đáp án:
Giải thích các bước giải:
Câu 1:
\(\begin{array}{l}
a.2x + 10 - {x^2} - 5x = 0\\
\to \left[ \begin{array}{l}
x = 2\\
x = - 5
\end{array} \right.\\
c.{\left( {x - 1} \right)^2} + 4\left( {x + 2} \right) - {x^2} + 3 = 0\\
\to {x^2} - 2x + 1 + 4x + 8 - {x^2} + 3 = 0\\
\to x = - 6\\
C2:\\
a.ĐK:x \ne \pm 3\\
B = \frac{{{x^2} + 6x + 9 - 2{x^2} + 6 + {x^2} - 3x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}:\frac{{6x - 12}}{{2{x^2} - 18}}\\
= \frac{{3x - 15x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\frac{{\left( {x - 3} \right)\left( {x + 3} \right)}}{{6\left( {x - 2} \right)}}\\
= \frac{{3\left( {x - 5} \right)}}{{\left( {3 - x} \right)\left( {3 + x} \right)}}.\frac{{\left( {x - 3} \right)\left( {x + 3} \right)}}{{6\left( {x - 2} \right)}} = \frac{{x - 5}}{{2x - 4}}\\
b.\left[ \begin{array}{l}
x + 1 = 2\\
x + 1 = - 2
\end{array} \right. \to \left[ \begin{array}{l}
x = 1\\
x = - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
B = 2\\
B = \frac{4}{5}
\end{array} \right.
\end{array}\)
