Rút gọn $\frac{2}{3x}$ – $\frac{2}{x+1}$ . ($\frac{x+1}{3x}$ – x-1): $\frac{x-1}{x}$ 30/07/2021 Bởi Iris Rút gọn $\frac{2}{3x}$ – $\frac{2}{x+1}$ . ($\frac{x+1}{3x}$ – x-1): $\frac{x-1}{x}$
Đáp án: $\begin{array}{l}\frac{2}{{3x}} – \frac{2}{{x + 1}}.\left( {\frac{{x + 1}}{{3x}} – x – 1} \right):\frac{{x – 1}}{x}\\ = \frac{2}{{3x}} – \frac{2}{{x + 1}}.\left( {x + 1} \right).\left( {\frac{1}{{3x}} – 1} \right).\frac{x}{{x – 1}}\\ = \frac{2}{{3x}} – 2.\frac{{1 – 3x}}{{3x}}.\frac{x}{{x – 1}}\\ = \frac{2}{{3x}} – \frac{2}{{3x}}.\frac{{x\left( {1 – 3x} \right)}}{{x – 1}}\\ = \frac{2}{{3x}}.\left( {1 – \frac{{x\left( {1 – 3x} \right)}}{{x – 1}}} \right)\\ = \frac{2}{{3x}}.\frac{{x – 1 – x + 3{x^2}}}{{x – 1}}\\ = \frac{2}{{3x}}.\frac{{3{x^2} – 1}}{{x – 1}}\\ = \frac{{6{x^2} – 2}}{{3{x^2} – 3x}}\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
\frac{2}{{3x}} – \frac{2}{{x + 1}}.\left( {\frac{{x + 1}}{{3x}} – x – 1} \right):\frac{{x – 1}}{x}\\
= \frac{2}{{3x}} – \frac{2}{{x + 1}}.\left( {x + 1} \right).\left( {\frac{1}{{3x}} – 1} \right).\frac{x}{{x – 1}}\\
= \frac{2}{{3x}} – 2.\frac{{1 – 3x}}{{3x}}.\frac{x}{{x – 1}}\\
= \frac{2}{{3x}} – \frac{2}{{3x}}.\frac{{x\left( {1 – 3x} \right)}}{{x – 1}}\\
= \frac{2}{{3x}}.\left( {1 – \frac{{x\left( {1 – 3x} \right)}}{{x – 1}}} \right)\\
= \frac{2}{{3x}}.\frac{{x – 1 – x + 3{x^2}}}{{x – 1}}\\
= \frac{2}{{3x}}.\frac{{3{x^2} – 1}}{{x – 1}}\\
= \frac{{6{x^2} – 2}}{{3{x^2} – 3x}}
\end{array}$