Đáp án:
4) a) \(
{\left[ {\begin{array}{*{20}c}
{x = - 2} \\
{x = 3} \\
\end{array}} \right.}
\)
b)\(
{\left[ {\begin{array}{*{20}c}
{x = 6} \\
{x = - 2} \\
\end{array}} \right.}
\)
6)B=-1
Giải thích các bước giải:
\(
\begin{array}{l}
4) \\
a)x(x + 2) - 3x - 6 = 0 \\
\Leftrightarrow x(x + 2) - 3(x + 2) = 0 \\
\Leftrightarrow (x + 2)(x - 3) = 0 \\
\Leftrightarrow \left[ {\begin{array}{*{20}c}
{x + 2 = 0} \\
{x - 3 = 0} \\
\end{array} \Leftrightarrow \left[ {\begin{array}{*{20}c}
{x = - 2} \\
{x = 3} \\
\end{array}} \right.} \right. \\
b)x^2 - 4x - 12 = 0 \\
\Leftrightarrow x^2 - 6x + 2x - 12 = 0 \\
\Leftrightarrow x(x - 6) + 2(x - 6) = 0 \\
\Leftrightarrow (x - 6)(x + 2) = 0 \\
\Leftrightarrow \left[ {\begin{array}{*{20}c}
{x - 6 = 0} \\
{x + 2 = 0} \\
\end{array} \Leftrightarrow \left[ {\begin{array}{*{20}c}
{x = 6} \\
{x = - 2} \\
\end{array}} \right.} \right. \\
6)B = (\frac{x}{{x^2 - 25}} - \frac{{x - 5}}{{x^2 + 5x}}):\frac{{2x - 5}}{{x^2 + 5x}} + \frac{x}{{5 - x}} \\
(x \ne 0;x \ne \pm 5) \\
= > B = {\rm{[}}\frac{{\rm{x}}}{{{\rm{(x - 5)(x + 5)}}}} - \frac{{x - 5}}{{x(x + 5)}}{\rm{]}}{\rm{.}}\frac{{{\rm{x(x + 5)}}}}{{{\rm{2x - 5}}}} - \frac{x}{{x - 5}} \\
= \frac{{x^2 - (x - 5)^x }}{{x(x - 5)(x + 5)}}.\frac{{x(x + 5)}}{{2x - 5}} - \frac{x}{{x - 5}} \\
= \frac{{x^2 - x^2 + 10x - 25}}{{(x - 5)(2x - 5)}} - \frac{x}{{x - 5}} \\
= \frac{{10x - 25}}{{(x - 5)(2x - 5)}} - \frac{x}{{x - 5}} \\
= \frac{{5(2x - 5)}}{{(x - 5)(2x - 5)}} - \frac{x}{{x - 5}} \\
= \frac{5}{{x - 5}} - \frac{x}{{x - 5}} = \frac{{5 - x}}{{x - 5}} = - 1 \\
\end{array}
\)
