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\(\dfrac{sin^3\alpha+3cos^3\alpha}{27sin^3\alpha-25cos^3\alpha}\)
\(=\dfrac{\dfrac{sin^3\alpha}{c\text{os}^3\alpha}+\dfrac{3cos^3\alpha}{c\text{os}^3\alpha}}{\dfrac{27sin^3\alpha}{c\text{os}^3\alpha}-\dfrac{25cos^3\alpha}{c\text{os}^3\alpha}}\)
\(=\dfrac{tan\alpha+3}{27tan\alpha-25}\)
\(=\dfrac{\dfrac{2}{3}+3}{27.\dfrac{2}{3}-25}\)
\(=-\dfrac{11}{21}\)
\(M=\frac{\frac{sina}{cosa}+\frac{cosa}{cosa}}{\frac{sina}{cosa}-\frac{cosa}{cosa}}=\frac{tana+1}{tana-1}=\frac{\frac{3}{5}+1}{\frac{3}{5}-1}=...\)
\(N=\frac{\frac{sina.cosa}{cos^2a}}{\frac{sin^2a}{cos^2a}-\frac{cos^2a}{cos^2a}}=\frac{tana}{tan^2a-1}=...\) (thay số bấm máy)
\(P=\frac{\frac{sin^3a}{cos^3a}+\frac{cos^3a}{cos^3a}}{\frac{2sina.cos^2a}{cos^3a}+\frac{cosa.sin^2a}{cos^3a}}=\frac{tan^3a+1}{2tana+tan^2a}=...\)
sin a=3/5
=>cos a=4/5
tan a=3/5:4/5=3/4; cot a=1:3/4=4/3
M=(4/3+3/4):(4/3-3/4)=25/7
Lời giải:
$\tan a=3\neq 0$ nên $\sin a, \cos a\neq 0$
$\frac{1}{M}=\frac{\sin ^2a-\cos ^2a}{\sin a\cos a}=\frac{\sin a}{\cos a}-\frac{\cos a}{\sin a}$
$=\tan a-\frac{1}{\tan a}=3-\frac{1}{3}=\frac{8}{3}$
Lời giải:
\(M=\frac{\frac{\sin a}{\cos a}+1}{\frac{\sin a}{\cos a}-1}=\frac{\tan a+1}{\tan a-1}=\frac{\frac{3}{5}+1}{\frac{3}{5}-1}=-4\)
\(N = \frac{\frac{\sin a\cos a}{\cos ^2a}}{\frac{\sin ^2a-\cos ^2a}{\cos ^2a}}=\frac{\frac{\sin a}{\cos a}}{(\frac{\sin a}{\cos a})^2-1}=\frac{\tan a}{\tan ^2a-1}=\frac{\frac{3}{5}}{\frac{3^2}{5^2}-1}=\frac{-15}{16}\)
\(M=\frac{\sin^3a+3\cos^3a}{27\sin^3a-25\cos^3a}\)
\(M=\frac{\frac{\sin^3a+3\cos^3a}{\cos^3a}}{\frac{27\sin^3a-25\cos^3a}{\cos^3a}}\)
\(M=\frac{\tan^3a+3}{27\tan^3a-25}\)
\(M=\frac{\frac{8}{27}+3}{27.\frac{8}{27}-25}\)
\(M=\frac{\frac{89}{27}}{-17}\)
\(M=-\frac{89}{459}\)
P/s haphuong