Rút gọn : \(\left(\frac{1+2a}{4+2a}-\frac{a}{3a-6}+\frac{2a^2}{12-3a^2}\right)\div\frac{13a+6}{24-12a}\)
K
Khách
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DT
1
DT
26 tháng 6 2016
\(\left(\frac{3a+1}{a^2-3a}+\frac{3a-1}{a^2+3a}\right)\):\(\frac{a^2+1}{a^2-9}\)
=\(\left[\frac{3a+1}{a\left(a-3\right)}+\frac{3a-1}{a\left(a+3\right)}\right]\): \(\frac{a^2+1}{\left(a-3\right)\left(a+3\right)}\)
=\(\left[\frac{\left(3a+1\right)\left(a+3\right)}{a\left(a-3\right)\left(a+3\right)}+\frac{\left(3a-1\right)\left(a-3\right)}{a\left(a+3\right)\left(a-3\right)}\right]\): \(\frac{a^2+1}{\left(a-3\right)\left(a+3\right)}\)
=\(\frac{3a^2+9a+a+3+3a^2-9a-a+3}{a\left(a-3\right)\left(a+3\right)}\): \(\frac{a^2+1}{\left(a-3\right)\left(a+3\right)}\)
=\(\frac{6a^2+6}{a\left(a-3\right)\left(a+3\right)}\): \(\frac{a^2+1}{\left(a-3\right)\left(a+3\right)}\)
=\(\frac{6\left(a^2+1\right)}{a\left(a-3\right)\left(a+3\right)}\).\(\frac{\left(a-3\right)\left(a+3\right)}{a^2+1}\)
=\(\frac{6}{a}\)
3 tháng 2 2022
\(=\left[\dfrac{\left(a-1\right)^2}{a^2+a+1}+\dfrac{2a^2-4a-1}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{1}{a-1}\right]:\dfrac{2a}{3}\)
\(=\dfrac{a^3-3a^2+3a-1+2a^2-4a-1+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}\)
\(=\dfrac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}=\dfrac{3}{2a}\)
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
\(=\left[\frac{3a}{\left(a-2\right)\left(a+2\right)}-\frac{1}{\left(a-2\right)}-\frac{2}{\left(a+2\right)}\right]:\left(\frac{a^2-4-a^2-4}{a^2-4}\right)=\left(\frac{3a-a-2-2a+4}{\left(a-2\right)\left(a+2\right)}\right).\frac{\left(a-2\right)\left(a+2\right)}{-8}=\frac{2}{\left(a-2\right)\left(a+2\right)}.\frac{\left(a-2\right)\left(a+2\right)}{-8}\)
\(=\frac{-1}{4}\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 11 2019
\(\frac{cosa}{1+sina}+\frac{sina}{cosa}=\frac{cos^2a+sina\left(1+sina\right)}{cosa\left(1+sina\right)}=\frac{1+sina}{cosa\left(1+sina\right)}=\frac{1}{cosa}\)
\(\frac{sin^2a+cos^2a+2sina.cosa}{\left(sina-cosa\right)\left(sina+cosa\right)}=\frac{\left(sina+cosa\right)^2}{\left(sina-cosa\right)\left(sina+cosa\right)}=\frac{sina+cosa}{sina-cosa}=\frac{\frac{sina}{cosa}+1}{\frac{sina}{cosa}-1}=\frac{tana+1}{tana-1}\)
\(\left(sin^2a\right)^3+\left(cos^2a\right)^3=\left(sin^2a+cos^2a\right)^3-3sin^2a.cos^2a\left(sin^2a+cos^2a\right)\)
\(=1-3sin^2a.cos^2a\)
\(sin^2a-tan^2a=tan^4a\left(\frac{sin^2a}{tan^4a}-\frac{1}{tan^2a}\right)=tan^4a\left(sin^2a.\frac{cos^2a}{sin^2a}-\frac{1}{tan^2a}\right)\)
\(=tan^4a\left(cos^2a-cot^2a\right)\) bạn ghi sai đề câu này
\(\frac{tan^3a}{sin^2a}-\frac{1}{sina.cosa}+\frac{cot^3a}{cos^2a}=tan^3a\left(1+cot^2a\right)-\frac{1}{sina.cosa}+cot^3a\left(1+tan^2a\right)\)
\(=tan^3a+tana-\frac{1}{sina.cosa}+cot^3a+cota\)
\(=tan^3a+cot^3a+\frac{sina}{cosa}+\frac{cosa}{sina}-\frac{1}{sina.cosa}\)
\(=tan^3a+cot^3a+\frac{sin^2a+cos^2a-1}{sina.cosa}=tan^3a+cot^3a\)
\(=\left(\dfrac{2a+1}{2\left(a+2\right)}-\dfrac{a}{3\left(a-2\right)}-\dfrac{2a^2}{3\left(a-2\right)\left(a+2\right)}\right):\dfrac{13a+6}{24-12a}\)
\(=\dfrac{3\left(2a+1\right)\left(a-2\right)-2a\left(a+2\right)-4a^2}{6\left(a-2\right)\left(a+2\right)}:\dfrac{13a+6}{-12\left(a-2\right)}\)
\(=\dfrac{3\left(2a^2-3a-2\right)-2a\left(a+2\right)-4a^2}{6\left(a-2\right)\left(a+2\right)}\cdot\dfrac{-12\left(a-2\right)}{13a+6}\)
\(=\dfrac{6a^2-9a-6-2a^2-4a-4a^2}{a+2}\cdot\dfrac{-2}{13a+6}\)
\(=\dfrac{-\left(13a+6\right)}{a+2}\cdot\dfrac{-2}{13a+6}=\dfrac{2}{a+2}\)