Giải các phương trình a,x\2x-6-x\2x-2=2x\(x+1)(x-3) b,5+96\x^2-16=2x-1\x+4+3x-1\x-4 20/11/2021 Bởi Remi Giải các phương trình a,x\2x-6-x\2x-2=2x\(x+1)(x-3) b,5+96\x^2-16=2x-1\x+4+3x-1\x-4
Đáp án: b) x=-40 Giải thích các bước giải: \(\begin{array}{l}a)DK:x \ne \left\{ { – 1;1;3} \right\}\\\dfrac{x}{{2\left( {x – 3} \right)}} – \dfrac{x}{{2\left( {x – 1} \right)}} = \dfrac{{2x}}{{\left( {x + 1} \right)\left( {x – 3} \right)}}\\ \to \dfrac{{x\left( {x – 1} \right)\left( {x + 1} \right) – x\left( {x – 3} \right)\left( {x + 1} \right) – 4x\left( {x – 1} \right)}}{{2\left( {x – 3} \right)\left( {x + 1} \right)\left( {x – 1} \right)}} = 0\\ \to x\left( {{x^2} – 1} \right) – x\left( {{x^2} – 2x – 3} \right) – 4{x^2} + 4x = 0\\ \to {x^3} – x – {x^3} + 2{x^2} + 3x – 4{x^2} + 4x = 0\\ \to – 2{x^2} + 6x = 0\\ \to – 2x\left( {x – 3} \right) = 0\\ \to \left[ \begin{array}{l}x = 0\left( {TM} \right)\\x = 3\left( l \right)\end{array} \right.\\b)DK:x \ne \pm 4\\5 + \dfrac{{96}}{{{x^2} – 16}} = \dfrac{{2x – 1}}{{x + 4}} + \dfrac{{3x – 1}}{{x – 4}}\\ \to \dfrac{{5{x^2} – 80 – \left( {2x – 1} \right)\left( {x – 4} \right) – \left( {3x – 1} \right)\left( {x + 4} \right)}}{{\left( {x – 4} \right)\left( {x + 4} \right)}} = 0\\ \to 5{x^2} – 80 – 2{x^2} + 9x – 4 – 3{x^2} – 11x + 4 = 0\\ \to – 2x = 80\\ \to x = – 40\end{array}\) Bình luận
Đáp án:
b) x=-40
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ { – 1;1;3} \right\}\\
\dfrac{x}{{2\left( {x – 3} \right)}} – \dfrac{x}{{2\left( {x – 1} \right)}} = \dfrac{{2x}}{{\left( {x + 1} \right)\left( {x – 3} \right)}}\\
\to \dfrac{{x\left( {x – 1} \right)\left( {x + 1} \right) – x\left( {x – 3} \right)\left( {x + 1} \right) – 4x\left( {x – 1} \right)}}{{2\left( {x – 3} \right)\left( {x + 1} \right)\left( {x – 1} \right)}} = 0\\
\to x\left( {{x^2} – 1} \right) – x\left( {{x^2} – 2x – 3} \right) – 4{x^2} + 4x = 0\\
\to {x^3} – x – {x^3} + 2{x^2} + 3x – 4{x^2} + 4x = 0\\
\to – 2{x^2} + 6x = 0\\
\to – 2x\left( {x – 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\left( {TM} \right)\\
x = 3\left( l \right)
\end{array} \right.\\
b)DK:x \ne \pm 4\\
5 + \dfrac{{96}}{{{x^2} – 16}} = \dfrac{{2x – 1}}{{x + 4}} + \dfrac{{3x – 1}}{{x – 4}}\\
\to \dfrac{{5{x^2} – 80 – \left( {2x – 1} \right)\left( {x – 4} \right) – \left( {3x – 1} \right)\left( {x + 4} \right)}}{{\left( {x – 4} \right)\left( {x + 4} \right)}} = 0\\
\to 5{x^2} – 80 – 2{x^2} + 9x – 4 – 3{x^2} – 11x + 4 = 0\\
\to – 2x = 80\\
\to x = – 40
\end{array}\)