Đáp án:
6) Phương trình vô nghiệm
Giải thích các bước giải:
\(\begin{array}{l}
5)DK:x \ne \left\{ {2;3;4} \right\}\\
\dfrac{1}{{x - 2}} - \dfrac{2}{{x - 3}} - \dfrac{3}{{x - 4}} = 0\\
\to \dfrac{{\left( {x - 3} \right)\left( {x - 4} \right) - 2\left( {x - 2} \right)\left( {x - 4} \right) - 3\left( {x - 2} \right)\left( {x - 3} \right)}}{{\left( {x - 2} \right)\left( {x - 3} \right)\left( {x - 4} \right)}} = 0\\
\to \dfrac{{{x^2} - 7x + 12 - 2\left( {{x^2} - 6x + 8} \right) - 3\left( {{x^2} - 5x + 6} \right)}}{{\left( {x - 2} \right)\left( {x - 3} \right)\left( {x - 4} \right)}} = 0\\
\to {x^2} - 7x + 12 - 2{x^2} + 12x - 16 - 3{x^2} + 15x - 18 = 0\\
\to - 4{x^2} + 34x - 22 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{17 + \sqrt {201} }}{4}\\
x = \dfrac{{17 - \sqrt {201} }}{4}
\end{array} \right.\\
6)\dfrac{{ - 2}}{{x - 2}} + \dfrac{1}{{x - 3}} - \dfrac{3}{{x - 4}} = 0\\
\to \dfrac{{ - 2\left( {x - 3} \right)\left( {x - 4} \right) + \left( {x - 2} \right)\left( {x - 4} \right) - 3\left( {x - 2} \right)\left( {x - 3} \right)}}{{\left( {x - 2} \right)\left( {x - 3} \right)\left( {x - 4} \right)}} = 0\\
\to - 2\left( {{x^2} - 7x + 12} \right) + {x^2} - 6x + 8 - 3\left( {{x^2} - 5x + 6} \right) = 0\\
\to - 2{x^2} + 14x - 24 + {x^2} - 6x + 8 - 3{x^2} + 15x - 18 = 0\\
\to - 4{x^2} + 23x - 34 = 0
\end{array}\)
⇒ Phương trình vô nghiệm
