Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
c,\\
\cos 2x = 2{\cos ^2}x - 1 \Rightarrow {\cos ^2}x = \dfrac{{\cos 2x + 1}}{2}\\
\tan \dfrac{x}{2} = \dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}} = \dfrac{{\sin \dfrac{x}{2}.\cos \dfrac{x}{2}}}{{{{\cos }^2}\dfrac{x}{2}}} = \dfrac{{\dfrac{1}{2}\sin x}}{{\dfrac{{\cos x + 1}}{2}}} = \dfrac{{\sin x}}{{\cos x + 1}}\\
\dfrac{{\sqrt 2 - \sin x - \cos x}}{{\sin x - \cos x}} - \tan \left( {\dfrac{x}{2} - \dfrac{\pi }{8}} \right)\\
= \dfrac{{\sqrt 2 - \sin x - \cos x}}{{\sin x - \cos x}} - \dfrac{{\sin \left( {x - \dfrac{\pi }{4}} \right)}}{{\cos \left( {x - \dfrac{\pi }{4}} \right) + 1}}\\
= \dfrac{{\sqrt 2 - \sin x - \cos x}}{{\sin x - \cos x}} - \dfrac{{\sin x.\cos \dfrac{\pi }{4} - \cos x.\sin \dfrac{\pi }{4}}}{{\left( {\cos x.\cos \dfrac{\pi }{4} + \sin x.\sin \dfrac{\pi }{4}} \right) + 1}}\\
= \dfrac{{\sqrt 2 - \sin x - \cos x}}{{\sin x - \cos x}} - \dfrac{{\sin x.\dfrac{{\sqrt 2 }}{2} - \cos x.\dfrac{{\sqrt 2 }}{2}}}{{\cos x.\dfrac{{\sqrt 2 }}{2} + \sin x.\dfrac{{\sqrt 2 }}{2} + 1}}\\
= \dfrac{{\sqrt 2 - \sin x - \cos x}}{{\sin x - \cos x}} - \dfrac{{\sin x - \cos x}}{{\cos x + \sin x + \sqrt 2 }}\\
= \dfrac{{\left( {\sqrt 2 - \sin x - \cos x} \right)\left( {\sqrt 2 + \sin x + \cos x} \right) - {{\left( {\sin x - \cos x} \right)}^2}}}{{\left( {\sin x - \cos x} \right)\left( {\cos x + \sin x + \sqrt 2 } \right)}}\\
= \dfrac{{{{\sqrt 2 }^2} - {{\left( {\sin x + \cos x} \right)}^2} - {{\left( {\sin x - \cos x} \right)}^2}}}{{\left( {\sin x - \cos x} \right)\left( {\cos x + \sin x + \sqrt 2 } \right)}}\\
= \dfrac{{2 - \left( {{{\sin }^2}x + 2\sin x.\cos x + {{\cos }^2}x} \right) - \left( {{{\sin }^2}x - 2\sin x.\cos x + {{\cos }^2}x} \right)}}{{\left( {\sin x - \cos x} \right)\left( {\cos x + \sin x + \sqrt 2 } \right)}}\\
= \dfrac{{2 - 2\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}}{{\left( {\sin x - \cos x} \right)\left( {\cos x + \sin x + \sqrt 2 } \right)}}\\
= \dfrac{{2 - 2.1}}{{\left( {\sin x - \cos x} \right)\left( {\cos x + \sin x + \sqrt 2 } \right)}}\\
= 0
\end{array}\)