Giải thích các bước giải:

\(\begin{array}{l}
a,\\
\sin x + \cos x = m\\
 \Leftrightarrow {\left( {\sin x + \cos x} \right)^2} = {m^2}\\
 \Leftrightarrow {\sin ^2}x + 2.\sin x.\cos x + {\cos ^2}x = {m^2}\\
 \Leftrightarrow \left( {{{\sin }^2}x + {{\cos }^2}x} \right) + 2\sin x.\cos x = {m^2}\\
 \Leftrightarrow 1 + 2\sin x.\cos x = {m^2}\\
 \Leftrightarrow \sin x.\cos x = \dfrac{{{m^2} - 1}}{2}\\
b,\\
{\left( {\sin x - \cos x} \right)^2} = {\sin ^2}x - 2.\sin x.\cos x + {\cos ^2}x\\
 = \left( {{{\sin }^2}x + {{\cos }^2}x} \right) - 2.\sin x.\cos x\\
 = 1 - 2.\dfrac{{{m^2} - 1}}{2} = 1 - \left( {{m^2} - 1} \right) = 2 - {m^2}\\
90^\circ  < x < 180^\circ  \Rightarrow \left\{ \begin{array}{l}
\sin x > 0\\
\cos x < 0
\end{array} \right. \Rightarrow \sin x - \cos x > 0\\
 \Rightarrow \sin x - \cos x = \sqrt {2 - {m^2}} \\
c,\\
{\sin ^3}x + {\cos ^3}x\\
 = \left( {\sin x + \cos x} \right).\left( {{{\sin }^2}x - \sin x.\cos x + {{\cos }^2}x} \right)\\
 = \left( {\sin x + \cos x} \right).\left[ {\left( {{{\sin }^2}x + {{\cos }^2}x} \right) - \sin x.\cos x} \right]\\
 = m.\left[ {1 - \dfrac{{{m^2} - 1}}{2}} \right]\\
 = m.\dfrac{{2 - \left( {{m^2} - 1} \right)}}{2}\\
 = \dfrac{{m.\left( {3 - {m^2}} \right)}}{2}\\
d,\\
{\sin ^4}x + {\cos ^4}x\\
 = \left( {{{\sin }^4}x + 2{{\sin }^2}x.{{\cos }^2}x + {{\cos }^4}x} \right) - 2{\sin ^2}x.{\cos ^2}x\\
 = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2.{\left( {\sin x.\cos x} \right)^2}\\
 = 1 - 2.{\left( {\dfrac{{{m^2} - 1}}{2}} \right)^2}\\
 = 1 - 2.\dfrac{{{m^4} - 2{m^2} + 1}}{4}\\
 = 1 - \dfrac{{{m^4} - 2{m^2} + 1}}{2}\\
 = \dfrac{{2 - \left( {{m^4} - 2{m^2} + 1} \right)}}{2}\\
 = \dfrac{{ - {m^4} + 2{m^2} + 1}}{2}\\
e,\\
{\sin ^6}x + {\cos ^6}x\\
 = \left( {{{\sin }^2}x + {{\cos }^2}x} \right).\left( {{{\sin }^4}x - {{\sin }^2}x.{{\cos }^2}x + {{\cos }^4}x} \right)\\
 = 1.\left[ {\left( {{{\sin }^4}x + {{\cos }^4}x} \right) - {{\left( {\sin x.\cos x} \right)}^2}} \right]\\
 = \dfrac{{ - {m^4} + 2{m^2} + 1}}{2} - {\left( {\dfrac{{{m^2} - 1}}{2}} \right)^2}\\
 = \dfrac{{ - {m^4} + 2{m^2} + 1}}{2} - \dfrac{{{m^4} - 2{m^2} + 1}}{4}\\
 = \dfrac{{2.\left( { - {m^4} + 2{m^2} + 1} \right) - \left( {{m^4} - 2{m^2} + 1} \right)}}{4}\\
 = \dfrac{{ - 3{m^4} + 6{m^2} + 1}}{4}\\
f,\\
{\tan ^2}x + {\cot ^2}x\\
 = \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + \dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}\\
 = \dfrac{{{{\sin }^4}x + {{\cos }^4}x}}{{{{\cos }^2}x.{{\sin }^2}x}}\\
 = \dfrac{{ - {m^4} + 2{m^2} + 1}}{2}:{\left( {\dfrac{{{m^2} - 1}}{2}} \right)^2}\\
 = \dfrac{{ - {m^4} + 2{m^2} + 1}}{2}:\dfrac{{{m^4} - 2{m^2} + 1}}{4}\\
 = \dfrac{{ - {m^4} + 2{m^2} + 1}}{2}.\dfrac{4}{{{m^4} - 2{m^2} + 1}}\\
 = \dfrac{{2.\left( { - {m^4} + 2{m^2} + 1} \right)}}{{{m^4} - 2{m^2} + 1}}
\end{array}\)