Đáp án:
\(\begin{array}{l}
1,\\
A = \dfrac{{11}}{9}\\
2,\\
A = 13\\
3,\\
C = \dfrac{6}{7}\\
4,\\
B = \dfrac{4}{5}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
\tan x = 6 \Leftrightarrow \dfrac{{\sin x}}{{\cos x}} = 6 \Leftrightarrow \sin x = 6\cos x\\
A = \dfrac{{2\sin x - \cos x}}{{\sin x + 3\cos x}} = \dfrac{{2.6\cos x - \cos x}}{{6\cos x + 3\cos x}}\\
= \dfrac{{12\cos x - \cos x}}{{9\cos x}} = \dfrac{{11\cos x}}{{9\cos x}} = \dfrac{{11}}{9}\\
2,\\
\cot x = \dfrac{1}{3} \Leftrightarrow \dfrac{{\cos x}}{{\sin x}} = \dfrac{1}{3} \Leftrightarrow \sin x = 3\cos x\\
A = \dfrac{{3\sin x + 4\cos x}}{{2\sin x - 5\cos x}} = \dfrac{{3.3\cos x + 4\cos x}}{{2.3\cos x - 5\cos x}}\\
= \dfrac{{9\cos x + 4\cos x}}{{6\cos x - 5\cos x}} = \dfrac{{13\cos x}}{{\cos x}} = 13\\
3,\\
\cot x = 4 \Leftrightarrow \dfrac{{\cos x}}{{\sin x}} = 4 \Leftrightarrow \cos x = 4\sin x\\
C = \dfrac{{\cos x + 2\sin x}}{{2\cos x - \sin x}} = \dfrac{{4\sin x + 2\sin x}}{{2.4\sin x - \sin x}}\\
= \dfrac{{6\sin x}}{{8\sin x - \sin x}} = \dfrac{{6\sin x}}{{7\sin x}} = \dfrac{6}{7}\\
4,\\
\tan x = 2 \Leftrightarrow \dfrac{{\sin x}}{{\cos x}} = 2 \Leftrightarrow \sin x = 2\cos x\\
B = \dfrac{{2\cos x + \sin x}}{{3\sin x - \cos x}} = \dfrac{{2\cos x + 2\cos x}}{{3.2\cos x - \cos x}}\\
= \dfrac{{4\cos x}}{{6\cos x - \cos x}} = \dfrac{{4\cos x}}{{5\cos x}} = \dfrac{4}{5}
\end{array}\)
