Giải phương trình a)x+1/x-2-x-1/x+2=2(x^2+2)/x^2-4 b)2x+1/x-1=5(x-1)/x+1 19/11/2021 Bởi Gabriella Giải phương trình a)x+1/x-2-x-1/x+2=2(x^2+2)/x^2-4 b)2x+1/x-1=5(x-1)/x+1
Đáp án: $\begin{array}{l}a)Dkxd:x \ne 2;x \ne – 2\\\frac{{x + 1}}{{x – 2}} – \frac{{x – 1}}{{x + 2}} = \frac{{2\left( {{x^2} + 2} \right)}}{{{x^2} – 4}}\\ \Rightarrow \frac{{x + 1}}{{x – 2}} – \frac{{x – 1}}{{x + 2}} = \frac{{2{x^2} + 4}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\ \Rightarrow \frac{{\left( {x + 1} \right)\left( {x + 2} \right) – \left( {x – 1} \right)\left( {x – 2} \right)}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} = \frac{{2{x^2} + 4}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\ \Rightarrow {x^2} + 3x + 2 – \left( {{x^2} – 3x + 2} \right) = 2{x^2} + 4\\ \Rightarrow 6x = 2{x^2} + 4\\ \Rightarrow {x^2} – 3x + 2 = 0\\ \Rightarrow \left( {x – 2} \right)\left( {x – 1} \right) = 0\\ \Rightarrow \left[ \begin{array}{l}x = 2\left( {ktm} \right)\\x = 1\left( {tm} \right)\end{array} \right.\\Vay\,x = 1\\b)Dkxd:x \ne 1;x \ne – 1\\\frac{{2x + 1}}{{x – 1}} = \frac{{5\left( {x – 1} \right)}}{{x + 1}}\\ \Rightarrow \left( {2x + 1} \right)\left( {x + 1} \right) = \left( {5x – 5} \right)\left( {x – 1} \right)\\ \Rightarrow 2{x^2} + 3x + 1 = 5{x^2} – 5x – 5x + 5\\ \Rightarrow 3{x^2} – 13{x^2} + 4 = 0\\ \Rightarrow \left( {3x – 1} \right)\left( {x – 4} \right) = 0\\ \Rightarrow \left[ \begin{array}{l}x = \frac{1}{3}\left( {tm} \right)\\x = 4\left( {tm} \right)\end{array} \right.\end{array}$ Bình luận
Đáp án:
Giải thích các bước giải:
Đáp án:
$\begin{array}{l}
a)Dkxd:x \ne 2;x \ne – 2\\
\frac{{x + 1}}{{x – 2}} – \frac{{x – 1}}{{x + 2}} = \frac{{2\left( {{x^2} + 2} \right)}}{{{x^2} – 4}}\\
\Rightarrow \frac{{x + 1}}{{x – 2}} – \frac{{x – 1}}{{x + 2}} = \frac{{2{x^2} + 4}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\
\Rightarrow \frac{{\left( {x + 1} \right)\left( {x + 2} \right) – \left( {x – 1} \right)\left( {x – 2} \right)}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} = \frac{{2{x^2} + 4}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\
\Rightarrow {x^2} + 3x + 2 – \left( {{x^2} – 3x + 2} \right) = 2{x^2} + 4\\
\Rightarrow 6x = 2{x^2} + 4\\
\Rightarrow {x^2} – 3x + 2 = 0\\
\Rightarrow \left( {x – 2} \right)\left( {x – 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 2\left( {ktm} \right)\\
x = 1\left( {tm} \right)
\end{array} \right.\\
Vay\,x = 1\\
b)Dkxd:x \ne 1;x \ne – 1\\
\frac{{2x + 1}}{{x – 1}} = \frac{{5\left( {x – 1} \right)}}{{x + 1}}\\
\Rightarrow \left( {2x + 1} \right)\left( {x + 1} \right) = \left( {5x – 5} \right)\left( {x – 1} \right)\\
\Rightarrow 2{x^2} + 3x + 1 = 5{x^2} – 5x – 5x + 5\\
\Rightarrow 3{x^2} – 13{x^2} + 4 = 0\\
\Rightarrow \left( {3x – 1} \right)\left( {x – 4} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{1}{3}\left( {tm} \right)\\
x = 4\left( {tm} \right)
\end{array} \right.
\end{array}$