Đáp án:
B2:
\(a = - \dfrac{{479}}{{17}}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)DK:x \ne 5\\
\dfrac{{x\left( {x - 5} \right)}}{{x - 5}} = 5 \to x = 5\left( {KTM} \right)\\
\to x \in \emptyset \\
b)DK:x \ne 1\\
\dfrac{{{x^2} + x + 1 - 3{x^2} - 2x\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = 0\\
\to - 4{x^2} - x + 1 = 0\\
\to - \left( {4{x^2} + 2.2x.\dfrac{1}{4} + \dfrac{1}{{16}} - \dfrac{{17}}{{16}}} \right) = 0\\
\to - {\left( {2x + \dfrac{1}{4}} \right)^2} + \dfrac{{17}}{{16}} = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{ - 1 + \sqrt {17} }}{8}\\
x = \dfrac{{ - 1 - \sqrt {17} }}{8}
\end{array} \right.\\
c)DK:x \ne 0\\
\to \dfrac{{1 + 2x - 3\left( {{x^2} + 1} \right)}}{x} = 0\\
\to \dfrac{{ - 3{x^2} + 2x - 2}}{x} = 0\\
\to - 3{x^2} + 2x - 2 = 0\\
\to - \left( {3{x^2} - 2.x\sqrt 3 .\dfrac{1}{{\sqrt 3 }} + \dfrac{1}{3} + \dfrac{5}{3}} \right) = 0\\
\to - {\left( {x\sqrt 3 - \dfrac{1}{{\sqrt 3 }}} \right)^2} - \dfrac{5}{3} = 0\left( {KTM} \right)\\
\to x \in \emptyset \\
B2:\\
DK:a \ne - 3\\
\dfrac{{10}}{3} - \dfrac{{3a - 1}}{{4a + 12}} - \dfrac{{7a + 2}}{{6a + 18}} = 0\\
\to \dfrac{{10.4\left( {a + 12} \right) - 3\left( {3a - 1} \right) - 2\left( {7a + 2} \right)}}{{12\left( {a + 3} \right)}} = 0\\
\to 40a + 480 - 9a + 3 - 14a - 4 = 0\\
\to 17a + 479 = 0\\
\to a = - \dfrac{{479}}{{17}}
\end{array}\)
