Bài 1 Cho A=(1/3+(-3)/x^2-3x):(x^2/27-3x^2+1/x+3) ?)Rút gọn A 26/07/2021 Bởi Brielle Bài 1 Cho A=(1/3+(-3)/x^2-3x):(x^2/27-3x^2+1/x+3) ?)Rút gọn A
Đáp án: $\begin{array}{l}\left( {dk:x \ne 0;x \ne 3;x \ne – 3} \right)\\A = \left( {\frac{1}{3} + \frac{{ – 3}}{{{x^2} – 3x}}} \right):\left( {\frac{{{x^2}}}{{27 – 3{x^2}}} + \frac{1}{{x + 3}}} \right)\\ = \frac{{{x^2} – 3x – 9}}{{3\left( {{x^2} – 3x} \right)}}:\left( {\frac{{{x^2}}}{{ – 3\left( {{x^2} – 9} \right)}} + \frac{1}{{x + 3}}} \right)\\ = \frac{{{x^2} – 3x – 9}}{{3x\left( {x – 3} \right)}}:\frac{{{x^2} – 3\left( {x – 3} \right)}}{{ – 3\left( {x – 3} \right)\left( {x + 3} \right)}}\\ = \frac{{{x^2} – 3x – 9}}{{3x\left( {x – 3} \right)}}.\frac{{ – 3\left( {x – 3} \right)\left( {x + 3} \right)}}{{{x^2} – 3x + 9}}\\ = – \frac{{{x^2} – 3x – 9}}{{{x^2} – 3x + 9}}\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
\left( {dk:x \ne 0;x \ne 3;x \ne – 3} \right)\\
A = \left( {\frac{1}{3} + \frac{{ – 3}}{{{x^2} – 3x}}} \right):\left( {\frac{{{x^2}}}{{27 – 3{x^2}}} + \frac{1}{{x + 3}}} \right)\\
= \frac{{{x^2} – 3x – 9}}{{3\left( {{x^2} – 3x} \right)}}:\left( {\frac{{{x^2}}}{{ – 3\left( {{x^2} – 9} \right)}} + \frac{1}{{x + 3}}} \right)\\
= \frac{{{x^2} – 3x – 9}}{{3x\left( {x – 3} \right)}}:\frac{{{x^2} – 3\left( {x – 3} \right)}}{{ – 3\left( {x – 3} \right)\left( {x + 3} \right)}}\\
= \frac{{{x^2} – 3x – 9}}{{3x\left( {x – 3} \right)}}.\frac{{ – 3\left( {x – 3} \right)\left( {x + 3} \right)}}{{{x^2} – 3x + 9}}\\
= – \frac{{{x^2} – 3x – 9}}{{{x^2} – 3x + 9}}
\end{array}$