\(A=\dfrac{\dfrac{4sin\alpha}{sin\alpha}+\dfrac{5cos\alpha}{sin\alpha}}{\dfrac{2sin\alpha}{sin\alpha}-\dfrac{3cos\alpha}{sin\alpha}}\)

\(A=\dfrac{4+5cot\alpha}{2-3cot\alpha}\)

Biết cotα=\(\dfrac{1}{2}\) nên ta có:

\(A=\dfrac{4+5\cdot\dfrac{1}{2}}{2-3\cdot\dfrac{1}{2}}\)

\(A=\dfrac{4+\dfrac{5}{2}}{2-\dfrac{3}{2}}\)

A= 13

\(\dfrac{4sin\alpha+5cos\alpha}{2sin\alpha-3cos\alpha}=\dfrac{\dfrac{4sin\alpha}{cos\alpha}+\dfrac{5cos\alpha}{cos\alpha}}{\dfrac{2sin\alpha}{cos\alpha}-\dfrac{3cos\alpha}{cos\alpha}}=\dfrac{4tan\alpha+5}{2tan\alpha-3}\)

Biết \(tan\)=\(\dfrac{1}{3}\) nên ta có:

\(\dfrac{4\times\dfrac{1}{2}+5}{2\times\dfrac{1}{2}-3}=\dfrac{2+5}{2-3}=\dfrac{7}{-2}=\dfrac{-7}{2}\)

Ta  có :\(sin^2a+cos^2a=1\)

Thay số: \(\left(\frac{2}{3}\right)^2\)\(+cos^2a=1\)\(\Rightarrow cos^2a=\frac{5}{9}\)

A=\(2sin^2a+5cos^2a\)\(\Rightarrow2.\frac{4}{9}+5.\frac{5}{9}\)\(\Rightarrow A=\frac{11}{3}\)

\(A=1-2sin^2\alpha-5cos^2\alpha=1-2\left(sin^2\alpha+cos^2\alpha\right)-3cos^2\alpha\) 

\(=1-2-3.\left(\dfrac{2}{3}\right)^2=-1-3.\dfrac{4}{9}=-1-\dfrac{4}{3}=-\dfrac{7}{3}\)

Đề là \(A=1-2sin^2a+5cos^2a\) hay \(A=1-2sin^2a-5cos^2a\) vậy nhỉ?

\(\hept{\begin{cases}sin^2a+c\text{os}^2a=1\\sina=2cosa\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}sina=\frac{2}{\sqrt{5}}\\c\text{os}a=\frac{1}{\sqrt{5}}\end{cases}}\)hoặc \(\orbr{\begin{cases}sina=-\frac{2}{\sqrt{5}}\\c\text{os}a=-\frac{1}{\sqrt{5}}\end{cases}}\)

Thế vô đi

Ta có: \(tan\alpha=2\Leftrightarrow\dfrac{sin\alpha}{cos\alpha}=2\Leftrightarrow sin\alpha=2cos\alpha\)

A = \(\dfrac{16cos^2\alpha+6cos^2\alpha}{20cos^2\alpha-2cos^2\alpha}=\dfrac{22cos^2\alpha}{18cos^2\alpha}=\dfrac{11}{9}\)

Bạn chia cả tủ và mẫu cho cos² x bạn nhé

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

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

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