Đáp án:
a) Vô nghiệm.
b) Vô nghiệm.
Giải thích các bước giải:
a) \(\dfrac{{2x + 1}}{{{x^2} - 2x + 1}} - \dfrac{{2x + 3}}{{{x^2} + 1}} = 0\)
ĐK: \({x^2} - 2x + 1 \ne 0 \Leftrightarrow {\left( {x - 1} \right)^2} \ne 0 \Leftrightarrow x \ne 1\)
\(\begin{array}{l}\dfrac{{2x + 1}}{{{x^2} - 2x + 1}} - \dfrac{{2x + 3}}{{{x^2} + 1}} = 0\\ \Leftrightarrow \dfrac{{2x + 1}}{{{x^2} - 2x + 1}} = \dfrac{{2x + 3}}{{{x^2} + 1}}\\ \Leftrightarrow \left( {2x + 1} \right)\left( {{x^2} + 1} \right) = \left( {2x + 3} \right)\left( {{x^2} + 2x + 1} \right)\\ \Leftrightarrow 2{x^3} + 2x + {x^2} + 1 = 2{x^3} + 4{x^2} + 2x + 3{x^2} + 6x + 3\\ \Leftrightarrow 2{x^3} + {x^2} + 2x + 1 = 2{x^3} + 7{x^2} + 8x + 3\\ \Leftrightarrow 2{x^3} + {x^2} + 2x + 1 - 2{x^3} - 7{x^2} - 8x - 3 = 0\\ \Leftrightarrow \left( {2{x^3} - 2{x^3}} \right) + \left( {{x^2} - 7{x^2}} \right) + \left( {2x - 8x} \right) + \left( {1 - 3} \right) = 0\\ \Leftrightarrow - 6{x^2} - 6x - 2 = 0\\ \Leftrightarrow 3{x^2} + 3x + 1 = 0\\ \Leftrightarrow 3\left( {{x^2} + x} \right) + 1 = 0\\ \Leftrightarrow 3\left( {{x^2} + 2.x.\dfrac{1}{2} + \dfrac{1}{4}} \right) - \dfrac{3}{4} + 1 = 0\\ \Leftrightarrow 3{\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{1}{4} = 0\end{array}\)
Do \({\left( {x + \dfrac{1}{2}} \right)^2} \ge 0\,\,\forall x \Rightarrow 3{\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{1}{4} > 0\,\,\forall x\).
Vậy phương trình đã cho vô nghiệm.
b) \(\dfrac{{{x^2} + 3x + 2}}{{{x^3} + 2{x^2} - x - 2}} = 0\)
ĐKXĐ:
\(\begin{array}{l}{x^3} + 2{x^2} - x - 2 \ne 0\\ \Leftrightarrow {x^2}\left( {x + 2} \right) - \left( {x + 2} \right) \ne 0\\ \Leftrightarrow \left( {x + 2} \right)\left( {{x^2} - 1} \right) \ne 0\\ \Leftrightarrow \left\{ \begin{array}{l}x \ne - 2\\x \ne \pm 1\end{array} \right.\end{array}\)
\(\begin{array}{l}\dfrac{{{x^2} + 3x + 2}}{{{x^3} + 2{x^2} - x - 2}} = 0\\ \Leftrightarrow {x^2} + 3x + 2 = 0\\ \Leftrightarrow {x^2} + x + 2x + 2 = 0\\ \Leftrightarrow x\left( {x + 1} \right) + 2\left( {x + 1} \right) = 0\\ \Leftrightarrow \left( {x + 1} \right)\left( {x + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x + 1 = 0\\x + 2 = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = - 1\\x = - 2\end{array} \right.\,\,\left( {ktm\,DKXD} \right)\end{array}\)
Vậy phương trình đã cho vô nghiệm.
