Đáp án:
c) Phương trình vô nghiệm
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne - 1\\
\dfrac{{ - 7{x^2} + 4}}{{{x^3} + 1}} = \dfrac{5}{{{x^2} - x + 1}} - \dfrac{1}{{x + 1}}\\
\to \dfrac{{ - 7{x^2} + 4}}{{{x^3} + 1}} = \dfrac{{5\left( {x + 1} \right) - {x^2} + x - 1}}{{{x^3} + 1}}\\
\to - 7{x^2} + 4 = 5x + 5 - {x^2} + x - 1\\
\to 6{x^2} + 6x = 0\\
\to 6x\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\left( {TM} \right)\\
x = - 1\left( l \right)
\end{array} \right.\\
b)DK:x \ne \pm 2\\
\dfrac{{x - 2}}{{x + 2}} - \dfrac{3}{{x - 2}} = \dfrac{{2x - 22}}{{{x^2} - 4}}\\
\to \dfrac{{{{\left( {x - 2} \right)}^2} - 3\left( {x + 2} \right) - 2x + 22}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = 0\\
\to \dfrac{{{x^2} - 4x + 4 - 3x - 6 - 2x + 22}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = 0\\
\to {x^2} - 9x + 20 = 0\\
\to {x^2} - 4x - 5x + 20 = 0\\
\to x\left( {x - 4} \right) - 5\left( {x - 4} \right) = 0\\
\to \left( {x - 4} \right)\left( {x - 5} \right) = 0\\
\to \left[ \begin{array}{l}
x = 4\\
x = 5
\end{array} \right.\\
c)DK:x \ne \left\{ {1;3} \right\}\\
\dfrac{{x + 5}}{{x - 1}} - \dfrac{{x + 1}}{{x - 3}} = \dfrac{{ - 8}}{{{x^2} - 4x + 3}}\\
\to \dfrac{{\left( {x + 5} \right)\left( {x - 3} \right) - \left( {x + 1} \right)\left( {x - 1} \right) + 8}}{{\left( {x - 1} \right)\left( {x - 3} \right)}} = 0\\
\to \dfrac{{{x^2} + 2x - 15 - {x^2} + 1 + 8}}{{\left( {x - 1} \right)\left( {x - 3} \right)}} = 0\\
\to 2x - 6 = 0\\
\to x = 3\left( l \right)
\end{array}\)
⇒ Phương trình vô nghiệm
