Đáp án:
g) \(\left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ {1;2} \right\}\\
\dfrac{{4\left( {x - 2} \right) - 5\left( {x - 1} \right) + 3\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = 0\\
\to 4x - 8 - 5x + 5 + 3\left( {{x^2} - 3x + 2} \right) = 0\\
\to - x - 3 + 3{x^2} - 9x + 6 = 0\\
\to 3{x^2} - 10x + 3 = 0\\
\to \left( {x - 3} \right)\left( {3x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 3\left( l \right)\\
x = \dfrac{1}{3}
\end{array} \right.\\
c)DK:x \ne \left\{ {1;2;3} \right\}\\
\dfrac{{x + 4}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} + \dfrac{{x + 1 - 2x - 5}}{{\left( {x - 1} \right)\left( {x - 3} \right)}} = 0\\
\to \dfrac{{\left( {x + 4} \right)\left( {x - 2} \right) + \left( { - x - 4} \right)\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}} = 0\\
\to {x^2} + 2x - 8 - {x^2} - 2x + 8 = 0\\
\to - 8 + 8 = 0\left( {ld} \right)
\end{array}\)
⇒ Phương trình có nghiệm với \(x \ne \left\{ {1;2;3} \right\}\)
\(\begin{array}{l}
e)DK:x \ne \left\{ { - 3; - 1} \right\}\\
\dfrac{{8x - 2\left( {{x^2} + 4x + 3} \right)}}{{2\left( {x + 1} \right)\left( {x + 3} \right)}} = \dfrac{{6\left( {2x + 2} \right) - 6\left( {x + 3} \right)}}{{2\left( {x + 1} \right)\left( {x + 3} \right)}}\\
\to - 2{x^2} - 6 = 12x + 12 - 6x - 18\\
\to 2{x^2} + 6x = 0\\
\to 2x\left( {x + 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = - 3\left( l \right)\\
x = 0
\end{array} \right.\\
g)DK:x \ne 1\\
\dfrac{{{x^2} + x + 1 + 2{x^2} - 5 - 4\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = 0\\
\to 3{x^2} - 3x = 0\\
\to x\left( {x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.
\end{array}\)