\(A=\dfrac{\dfrac{3sina}{sina}-\dfrac{cosa}{sina}}{\dfrac{2sina}{sina}+\dfrac{cosa}{sina}}=\dfrac{3-cota}{2+cota}=\dfrac{3-3}{2+3}=0\)

\(B=\dfrac{\dfrac{sin^2a}{sin^2a}-\dfrac{3sina.cosa}{sin^2a}+\dfrac{2}{sin^2a}}{\dfrac{2sin^2a}{sin^2a}+\dfrac{sina.cosa}{sin^2a}+\dfrac{cos^2a}{sin^2a}}=\dfrac{1-3cota+2\left(1+cot^2a\right)}{2+cota+cot^2a}=\dfrac{1-3.3+2\left(1+3^2\right)}{2+3+3^2}=...\)

a. \(A=\dfrac{3sin\alpha-cos\alpha}{2sin\alpha+cos\alpha}=\dfrac{3\dfrac{sin\alpha}{cos\alpha}-1}{2\dfrac{sin\alpha}{cos\alpha}+1}=\dfrac{3.\dfrac{1}{3}-1}{2.\dfrac{1}{3}+1}=0\)

b.\(B=\dfrac{sin^2\alpha-3sin\alpha.cos\alpha+2}{2sin^2\alpha+sin\alpha.cos\alpha+cos^2\alpha}\)\(=\dfrac{1-\dfrac{3cos\alpha}{sin\alpha}+\dfrac{2}{sin^2\alpha}}{2+\dfrac{cos\alpha}{sin\alpha}+\dfrac{cos^2\alpha}{sin^2\alpha}}=\dfrac{1-3.3+\dfrac{2}{sin^2\alpha}}{2+3+3^2}\)

Mà \(\dfrac{cos\alpha}{sin\alpha}=3,cos^2\alpha+sin^2\alpha=1\Rightarrow sin^2\alpha=\dfrac{1}{10}\)

\(B=\dfrac{1-3.3+\dfrac{2}{\dfrac{1}{10}}}{2+3+3^2}=\dfrac{6}{7}\)

Dạ em cảm ơn thầy và mọi người ạ! 

\(\dfrac{3sin\alpha-4cos\alpha}{2sin\alpha+3cos\alpha}=\dfrac{\dfrac{3sin\alpha}{cos\alpha}-\dfrac{4cos\alpha}{cos\alpha}}{\dfrac{2sin\alpha}{cos\alpha}+\dfrac{3cos\alpha}{cos\alpha}}=\dfrac{3tan\alpha-4}{2tan\alpha+3}\)

Biết tanα=\(-\dfrac{1}{4}\) nên ta có:

\(\dfrac{3\cdot\dfrac{-1}{4}-4}{2\cdot\dfrac{-1}{4}+3}=\dfrac{-\dfrac{3}{4}-4}{-\dfrac{1}{2}+3}=\dfrac{-19}{10}\)

Những biểu thức này đều không tính toán ra được giá trị cụ thể nên không phù hợp với yêu cầu "tính". Mình nghĩ bạn nên xem xét lại yêu cầu đề.

Lời giải:

Biểu thức $A$ dạng như vậy là gọn rồi bạn ạ. Biến đổi thêm cũng không có ý nghĩa.

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\(B=\sin ^2a+\sin 2a-3\cos ^3a\)

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\(C=\frac{\sin ^2a-\sin a\cos a-\cos ^2a}{2\sin a\cos a}=\frac{\sin a}{2\cos a}-\frac{1}{2}-\frac{\cos a}{2\sin a}\)

\(=\frac{\tan a-1-\cot a}{2}\)

hỏi tí chớ \(TanB=2\) hay \(Tan\alpha=2\) vậy .

Lời giải:
\(\frac{1+\cos a}{1-\cos a}-\frac{1-\cos a}{1+\cos a}=\frac{(1+\cos a)^2-(1-\cos a)^2}{(1-\cos a)(1+\cos a)}=\frac{1+2\cos a+\cos ^2a-(1-2\cos a+\cos ^2a)}{1-\cos ^2a}\)

\(=\frac{4\cos a}{\sin ^2a}=\frac{\frac{4\cos a}{\sin a}}{\sin a}=\frac{4\cot a}{\sin a}\) (đpcm)

\(\frac{2sina+3cosa}{4sina-5cosa}=\frac{\frac{2sina}{cosa}+\frac{3cosa}{cosa}}{\frac{4sina}{cosa}-\frac{5cosa}{cosa}}=\frac{2tana+3}{4tana-5}=\frac{6+3}{12-5}=\frac{9}{7}\)

\(\frac{3sina-2cosa}{5sina+4cos^3a}=\frac{\frac{3sina}{cosa}-\frac{2cosa}{cosa}}{\frac{5sina}{cosa}+\frac{4cos^3a}{cosa}}=\frac{3tana-2}{5tana+4cos^2a}=\frac{3tana-2}{5tana+\frac{4}{1+tan^2a}}=\frac{9-2}{15+\frac{4}{10}}=\frac{5}{11}\)

\(\dfrac{sin2\alpha+sin5\alpha-sin3\alpha}{1+cos\alpha-2sin^22\alpha}\)

\(=\dfrac{2cos\dfrac{5\alpha}{2}.sin\left(-\dfrac{\alpha}{2}\right)+2sin\dfrac{5\alpha}{2}.cos\dfrac{5\alpha}{2}}{cos4\alpha+cos\alpha}\)

\(=\dfrac{2cos\dfrac{5\alpha}{2}.\left(sin\dfrac{5\alpha}{2}-sin\dfrac{\alpha}{2}\right)}{2cos\dfrac{5\alpha}{2}.cos\dfrac{3\alpha}{2}}\)

\(=\dfrac{4cos\dfrac{5\alpha}{2}.cos\dfrac{3\alpha}{2}.sin\alpha}{2cos\dfrac{5\alpha}{2}.cos\dfrac{3\alpha}{2}}\)

\(=2sin\alpha\)

a: \(VT=\dfrac{\left(sina+cosa\right)^3-3\cdot sina\cdot cosa\left(sina+cosa\right)}{sina+cosa}\)

=(sina+cosa)^2-3*sina*cosa

=sin^2a+cos^2a-sina*cosa

=1-sina*cosa=VP

c: VT=(sin^2a+cos^2a)^2-2*sin^2a*cos^2a-(sin^2a+cos^2a)^3+3*sin^2a*cos^2a*(sin^2a+cos^2a)

=1-2sin^2a*cos^2a-1+3*sin^2a*cos^2a

=sin^2a*cos^2a=VP