1. Cho cos x= -4/5. Tính A= tan x + cot x/ tan x - cot x 2. Cho cot x = -2. Tính B= 2/ $sin^{2}$ x -3sinxcosx + 2 $cos^{2}$ x Mn giúp mình nha, thank you ^^

Giải thích các bước giải:

$\begin{array}{l}
1)A = \dfrac{{\tan x + \cot x}}{{\tan x - \cot x}}\\
 = \dfrac{{{{\tan }^2}x + 1}}{{{{\tan }^2}x - 1}}\\
 = \dfrac{{\dfrac{1}{{{{\cos }^2}x}}}}{{\dfrac{1}{{{{\cos }^2}x}} - 2}}\\
 = \dfrac{1}{{1 - 2{{\cos }^2}x}}\\
 = \dfrac{1}{{1 - 2.{{\left( {\dfrac{{ - 4}}{5}} \right)}^2}}}\\
 = \dfrac{{ - 25}}{7}\\
2)B = \dfrac{2}{{{{\sin }^2}x - 3\sin x\cos x + 2{{\cos }^2}x}}\\
 = \dfrac{{2\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}}{{{{\sin }^2}x - 3\sin x\cos x + 2{{\cos }^2}x}}\\
 = \dfrac{{2\left( {1 + {{\left( {\dfrac{{\cos x}}{{\sin x}}} \right)}^2}} \right)}}{{1 - 3.\dfrac{{\cos x}}{{\sin x}} + 2{{\left( {\dfrac{{\cos x}}{{\sin x}}} \right)}^2}}}\\
 = \dfrac{{2\left( {1 + {{\cot }^2}x} \right)}}{{1 - 3\cot x + 2{{\cot }^2}x}}\\
 = \dfrac{{2\left( {1 + {{\left( { - 2} \right)}^2}} \right)}}{{1 - 3.\left( { - 2} \right) + {{\left( { - 2} \right)}^2}}}\\
 = \dfrac{{10}}{{11}}
\end{array}$