Rút gọn: \(\left(\frac{3a+1}{a^2-3a}+\frac{3a-1}{a^2+3a}\right):\frac{a^2+1}{a^2-9}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
1 tháng 2 2022
\(=\left(\dfrac{2a+1}{2\left(a+2\right)}-\dfrac{a}{3\left(a-2\right)}-\dfrac{2a^2}{3\left(a-2\right)\left(a+2\right)}\right):\dfrac{13a+6}{24-12a}\)
\(=\dfrac{3\left(2a+1\right)\left(a-2\right)-2a\left(a+2\right)-4a^2}{6\left(a-2\right)\left(a+2\right)}:\dfrac{13a+6}{-12\left(a-2\right)}\)
\(=\dfrac{3\left(2a^2-3a-2\right)-2a\left(a+2\right)-4a^2}{6\left(a-2\right)\left(a+2\right)}\cdot\dfrac{-12\left(a-2\right)}{13a+6}\)
\(=\dfrac{6a^2-9a-6-2a^2-4a-4a^2}{a+2}\cdot\dfrac{-2}{13a+6}\)
\(=\dfrac{-\left(13a+6\right)}{a+2}\cdot\dfrac{-2}{13a+6}=\dfrac{2}{a+2}\)
30 tháng 7 2016
\(\left(\frac{3a}{a^2-4}+\frac{1}{2-a}-\frac{2}{a+2}\right):\left(1-\frac{a^2+4}{a^2-4}\right)\)điều kiện : a khác {-2,2}
=\(\left(\frac{3a}{a^2-4}-\frac{a+2}{a^2-4}-\frac{2a-4}{a^2-4}\right):\left(-\frac{8}{a^2-4}\right)\)
=\(\left(\frac{3a-a-2-2a+4}{a^2-4}\right).\left(\frac{a^2-4}{-8}\right)\)
=\(-\frac{1}{4}\)
KL
30 tháng 7 2016
\(=\left[\frac{3a}{\left(a-2\right)\left(a+2\right)}-\frac{1}{\left(a-2\right)}-\frac{2}{\left(a+2\right)}\right]:\left(\frac{a^2-4-a^2-4}{a^2-4}\right)=\left(\frac{3a-a-2-2a+4}{\left(a-2\right)\left(a+2\right)}\right).\frac{\left(a-2\right)\left(a+2\right)}{-8}=\frac{2}{\left(a-2\right)\left(a+2\right)}.\frac{\left(a-2\right)\left(a+2\right)}{-8}\)
\(=\frac{-1}{4}\)
4 tháng 1 2017
a/ đk: a\(\ne b\), b\(\ne0,a\ne-b\)
= \(\frac{a\left(a-b\right)-a^2-b^2}{a-b}.\frac{a+b+2b}{b\left(a+b\right)}\)
= \(\frac{a^2-ab-a^2-b^2}{a-b}.\frac{a+3b}{b\left(a+b\right)}\)
= \(\frac{-ab-b^2}{a-b}.\frac{a+3b}{b\left(a+b\right)}\)
= \(\frac{-b\left(a+b\right)\left(a+3b\right)}{b\left(a+b\right)\left(a-b\right)}\)
= \(\frac{-a-3b}{a-b}\)
b/ đk: a\(\ne0,a\ne\pm3\)
= \(\left[\frac{3a+1}{a\left(a-3\right)}+\frac{3a-1}{a\left(a+3\right)}\right].\frac{\left(a-3\right)\left(a+3\right)}{a^2+1}\)
= \(\frac{\left(3a+1\right)\left(a+3\right)+\left(3a-1\right)\left(a-3\right)}{a\left(a-3\right)\left(a+3\right)}.\frac{\left(a-3\right)\left(a+3\right)}{a^2+1}\)
= \(\frac{6a^2+6}{a\left(a-3\right)\left(a+3\right)}.\frac{\left(a-3\right)\left(a+3\right)}{a^2+1}\)
= \(\frac{6\left(a^2+1\right)\left(a-3\right)\left(a+3\right)}{a\left(a^2+1\right)\left(a-3\right)\left(a+3\right)}\)
= \(\frac{6}{a}\)
\(\left(\frac{3a+1}{a^2-3a}+\frac{3a-1}{a^2+3a}\right)\):\(\frac{a^2+1}{a^2-9}\)
=\(\left[\frac{3a+1}{a\left(a-3\right)}+\frac{3a-1}{a\left(a+3\right)}\right]\): \(\frac{a^2+1}{\left(a-3\right)\left(a+3\right)}\)
=\(\left[\frac{\left(3a+1\right)\left(a+3\right)}{a\left(a-3\right)\left(a+3\right)}+\frac{\left(3a-1\right)\left(a-3\right)}{a\left(a+3\right)\left(a-3\right)}\right]\): \(\frac{a^2+1}{\left(a-3\right)\left(a+3\right)}\)
=\(\frac{3a^2+9a+a+3+3a^2-9a-a+3}{a\left(a-3\right)\left(a+3\right)}\): \(\frac{a^2+1}{\left(a-3\right)\left(a+3\right)}\)
=\(\frac{6a^2+6}{a\left(a-3\right)\left(a+3\right)}\): \(\frac{a^2+1}{\left(a-3\right)\left(a+3\right)}\)
=\(\frac{6\left(a^2+1\right)}{a\left(a-3\right)\left(a+3\right)}\).\(\frac{\left(a-3\right)\left(a+3\right)}{a^2+1}\)
=\(\frac{6}{a}\)