Giải thích các bước giải:
Ta có:
$\dfrac{2\sin^4x+\cos^2x-\cos^4x}{2(1-\cos x)}$
$=\dfrac{2\sin^4x+\cos^2x(1-\cos^2x)}{2(1-\cos x)}$
$=\dfrac{2\sin^4x+\cos^2x\sin^2x}{2(1-\cos x)}$
$=\dfrac{\sin^2x(2\sin^2x+\cos^2x)}{2(1-\cos x)}$
$=\dfrac{(1-\cos^2x)(2(\sin^2x+\cos^2x)-\cos^2x)}{2(1-\cos x)}$
$=\dfrac{(1-\cos^2x)(2-\cos^2x)}{2(1-\cos x)}$
$=\dfrac{(1-\cos x)(1+\cos x)(2-\cos^2x)}{2(1-\cos x)}$
$=\dfrac12(1+\cos x)(2-\cos^2x)$
$=2-\cos ^2\left(x\right)+2\cos \left(x\right)-\cos ^3\left(x\right)$
