Những câu hỏi liên quan
DN
Xem chi tiết
NT
1 tháng 2 2022 lúc 19:57

\(=\left(\dfrac{2a+1}{2\left(a+2\right)}-\dfrac{a}{3\left(a-2\right)}-\dfrac{2a^2}{3\left(a-2\right)\left(a+2\right)}\right):\dfrac{13a+6}{24-12a}\)

\(=\dfrac{3\left(2a+1\right)\left(a-2\right)-2a\left(a+2\right)-4a^2}{6\left(a-2\right)\left(a+2\right)}:\dfrac{13a+6}{-12\left(a-2\right)}\)

\(=\dfrac{3\left(2a^2-3a-2\right)-2a\left(a+2\right)-4a^2}{6\left(a-2\right)\left(a+2\right)}\cdot\dfrac{-12\left(a-2\right)}{13a+6}\)

\(=\dfrac{6a^2-9a-6-2a^2-4a-4a^2}{a+2}\cdot\dfrac{-2}{13a+6}\)

\(=\dfrac{-\left(13a+6\right)}{a+2}\cdot\dfrac{-2}{13a+6}=\dfrac{2}{a+2}\)

Bình luận (0)
DN
Xem chi tiết
H24
30 tháng 7 2016 lúc 8:33

\(\left(\frac{3a}{a^2-4}+\frac{1}{2-a}-\frac{2}{a+2}\right):\left(1-\frac{a^2+4}{a^2-4}\right)\)điều kiện : a khác {-2,2}

=\(\left(\frac{3a}{a^2-4}-\frac{a+2}{a^2-4}-\frac{2a-4}{a^2-4}\right):\left(-\frac{8}{a^2-4}\right)\)

=\(\left(\frac{3a-a-2-2a+4}{a^2-4}\right).\left(\frac{a^2-4}{-8}\right)\)

=\(-\frac{1}{4}\)

Bình luận (0)
KL
30 tháng 7 2016 lúc 8:36

\(=\left[\frac{3a}{\left(a-2\right)\left(a+2\right)}-\frac{1}{\left(a-2\right)}-\frac{2}{\left(a+2\right)}\right]:\left(\frac{a^2-4-a^2-4}{a^2-4}\right)=\left(\frac{3a-a-2-2a+4}{\left(a-2\right)\left(a+2\right)}\right).\frac{\left(a-2\right)\left(a+2\right)}{-8}=\frac{2}{\left(a-2\right)\left(a+2\right)}.\frac{\left(a-2\right)\left(a+2\right)}{-8}\)

\(=\frac{-1}{4}\)

Bình luận (0)
NT
30 tháng 7 2016 lúc 8:40

\(\left(\frac{3a}{a^2-4}+\frac{1}{2-a}-\frac{2}{a+2}\right):\left(1-\frac{a^2+4}{a^2-4}\right)\)

\(=\left(\frac{3a}{\left(a-2\right)\left(a+2\right)}-\frac{1}{a-2}-\frac{2}{a+2}\right):\left(\frac{a^2-4}{a^2-4}-\frac{a^2+4}{a^2-4}\right)\) 

\(=\frac{3a-a-2-2a+4}{\left(a-2\right)\left(a+2\right)}:\frac{\left(-8\right)}{a^2-4}\)

\(=\frac{2}{\left(a-2\right)\left(a+2\right)}.\frac{\left(a-2\right)\left(a+2\right)}{\left(-8\right)}\)

\(=-\frac{1}{4}\)

 

Bình luận (0)
DN
Xem chi tiết
NT
4 tháng 1 2017 lúc 15:31

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\(\ne0,a\ne\pm3\)

= \(\left[\frac{3a+1}{a\left(a-3\right)}+\frac{3a-1}{a\left(a+3\right)}\right].\frac{\left(a-3\right)\left(a+3\right)}{a^2+1}\)

= \(\frac{\left(3a+1\right)\left(a+3\right)+\left(3a-1\right)\left(a-3\right)}{a\left(a-3\right)\left(a+3\right)}.\frac{\left(a-3\right)\left(a+3\right)}{a^2+1}\)

= \(\frac{6a^2+6}{a\left(a-3\right)\left(a+3\right)}.\frac{\left(a-3\right)\left(a+3\right)}{a^2+1}\)

= \(\frac{6\left(a^2+1\right)\left(a-3\right)\left(a+3\right)}{a\left(a^2+1\right)\left(a-3\right)\left(a+3\right)}\)

= \(\frac{6}{a}\)

Bình luận (0)
NT
Xem chi tiết
DH
Xem chi tiết
LD
Xem chi tiết
LD
Xem chi tiết
H24
Xem chi tiết
MT
Xem chi tiết