$\begin{array}{l} {\cos ^2}x - \sin \left( {\dfrac{\pi }{6} + x} \right)\sin \left( {\dfrac{\pi }{6} - x} \right)\\ = {\cos ^2}x - \dfrac{1}{2}\left[ {\cos \left( {\dfrac{\pi }{6} + x - \dfrac{\pi }{6} + x} \right) - \cos \left( {\dfrac{\pi }{6} - x + \dfrac{\pi }{6} + x} \right)} \right]\\ = {\cos ^2}x - \dfrac{1}{2}\cos 2x + \dfrac{1}{2}.\cos \dfrac{\pi }{3}\\ = {\cos ^2}x - \dfrac{1}{2}\left( {2{{\cos }^2}x - 1} \right) + \dfrac{1}{4}\\ = \dfrac{1}{2} + \dfrac{1}{4} = \dfrac{3}{4} \end{array}$
b)
$\begin{array}{l} \dfrac{{2{{\sin }^2}\dfrac{x}{2} + \sin 2x - 1}}{{2\sin x - 1}} = \dfrac{{1 - \cos x + \sin 2x - 1}}{{2\sin x - 1}} + \sin x\\ = \dfrac{{\cos x - \sin 2x}}{{2\sin x - 1}} + \sin x = \dfrac{{2\sin x.\cos x - \cos x}}{{2\sin x - 1}} + \sin x\\ = \dfrac{{\cos x\left( {2\sin x - 1} \right)}}{{2\sin x - 1}} + \sin x = \cos x + \sin x\\ = \sqrt 2 \left( {\dfrac{{\sqrt 2 }}{2}\cos x + \dfrac{{\sqrt 2 }}{2}.\sin x} \right)\\ = \sqrt 2 \left( {\sin \dfrac{\pi }{4}.\cos x + \cos \dfrac{\pi }{4}.\sin x} \right)\\ = \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) \end{array}$
c) Ta đi chứng minh ${\cos ^2}A + {\cos ^2}B + {\cos ^2}C = 1 - 2\cos A\cos B\cos C$
$\begin{array}{l} VT = \dfrac{{1 + \cos 2A}}{2} + \dfrac{{1 + \cos 2B}}{2} + {\cos ^2}C\\ VT = 1 + \dfrac{1}{2}\left( {\cos 2A + \cos 2B} \right) + {\cos ^2}C\\ VT = 1 + \dfrac{1}{2}.2\cos \left( {A + B} \right)\cos \left( {A - B} \right) + {\cos ^2}C\\ VT = 1 + \cos \left( {A + B} \right)\cos \left( {A - B} \right) + {\cos ^2}C\\ VT = 1 - \cos C.\cos \left( {A - B} \right) + \cos C.\cos C\\ VT = 1 - \cos C.\cos \left( {A - B} \right) - \cos \left( {A - B} \right).\cos C\\ VT = 1 - \cos C\left[ {\cos \left( {A - B} \right) + \cos \left( {A + B} \right)} \right]\\ VT = 1 - 2\cos A.\cos B.\cos C \end{array}$
Theo đề ta có $\cos^2A+\cos^2B+\cos^2C=1$ nên $2\cos A\cos B\cos C = 0$
$\left[ \begin{array}{l} \cos A = 0\\ \cos B = 0\\ \cos C = 0 \end{array} \right. \Rightarrow \left[ \begin{array}{l} A = \dfrac{\pi }{2}\\ B = \dfrac{\pi }{2}\\ C = \dfrac{\pi }{2} \end{array} \right.$
