Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\dfrac{{{{\sin }^4}x - {{\cos }^4}x}}{{\sin x + \cos x}}\\
= \dfrac{{{{\left( {{{\sin }^2}x} \right)}^2} - {{\left( {{{\cos }^2}x} \right)}^2}}}{{\sin x + \cos x}}\\
= \dfrac{{\left( {{{\sin }^2}x - {{\cos }^2}x} \right).\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}}{{\sin x + \cos x}}\\
= \dfrac{{\left( {{{\sin }^2}x - {{\cos }^2}x} \right).1}}{{\sin x + \cos x}}\\
= \dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{\sin x + \cos x}}\\
= \dfrac{{\left( {\sin x - \cos x} \right)\left( {\sin x + \cos x} \right)}}{{\sin x + \cos x}}\\
= \sin x - \cos x\\
b,\\
{\sin ^6}x + {\cos ^6}x + 3{\sin ^2}x.\cos {x^2}\\
= {\sin ^6}x + {\cos ^6}x + 3{\sin ^2}x.\cos {x^2}.1\\
= {\sin ^6}x + {\cos ^6}x + 3{\sin ^2}x.\cos {x^2}.\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\\
= {\sin ^6}x + {\cos ^6}x + 3{\sin ^4}x.\cos {x^2} + 3{\sin ^2}x.{\cos ^4}x\\
= {\sin ^6}x + 3{\sin ^4}x.{\cos ^2}x + 3{\sin ^2}x.{\cos ^4}x + {\cos ^6}x\\
= {\left( {{{\sin }^2}x} \right)^3} + 3.{\left( {{{\sin }^2}x} \right)^2}.{\cos ^2}x + 3.{\sin ^2}x.{\left( {{{\cos }^2}x} \right)^2} + {\left( {{{\cos }^2}x} \right)^3}\\
= {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^3}\\
= {1^3}\\
= 1
\end{array}\)
