Toán Lớp 9: 2x/x+3-x+1/3-x-3-11x/x^2-9 a) Rút gọn, giúp em giải bài này ạ, em cảm ơn thầy cô và các bạn nhiều.
\[\begin{array}{l}\frac{{2x}}{{x + 3}} – \frac{{x + 1}}{{3 – x}} – \frac{{3 – 11x}}{{{x^2} – 9}}\\dk:x \ne \pm 3\\= \frac{{2x}}{{x + 3}} + \frac{{x + 1}}{{x – 3}} – \frac{{3 – 11x}}{{(x – 3)(x + 3)}}\\= \frac{{2x(x – 3) + (x + 1)(x + 3) – 3 + 11x}}{{(x – 3)(x + 3)}}\\= \frac{{2{x^2} – 6x + {x^2} + 3x + x + 3 – 3 + 11x}}{{(x – 3)(x + 3)}}\\= \frac{{3{x^2} + 9x}}{{(x – 3)(x + 3)}}\\= \frac{{3x(x + 3)}}{{(x – 3)(x + 3)}}\\= \frac{{3x}}{{x – 3}}\end{array}\] Trả lời
\frac{{2x}}{{x + 3}} – \frac{{x + 1}}{{3 – x}} – \frac{{3 – 11x}}{{{x^2} – 9}}\\
dk:x \ne \pm 3\\
= \frac{{2x}}{{x + 3}} + \frac{{x + 1}}{{x – 3}} – \frac{{3 – 11x}}{{(x – 3)(x + 3)}}\\
= \frac{{2x(x – 3) + (x + 1)(x + 3) – 3 + 11x}}{{(x – 3)(x + 3)}}\\
= \frac{{2{x^2} – 6x + {x^2} + 3x + x + 3 – 3 + 11x}}{{(x – 3)(x + 3)}}\\
= \frac{{3{x^2} + 9x}}{{(x – 3)(x + 3)}}\\
= \frac{{3x(x + 3)}}{{(x – 3)(x + 3)}}\\
= \frac{{3x}}{{x – 3}}
\end{array}\]