Đáp án: $C$
Giải thích các bước giải:
* Trắc nghiệm:
$x=0^o$
$\to \sin0^o=0; \cos0^o=1$
$\to C=2(0+1+1.0)^2-(0+1)=1$
* Tự luận:
$C=2(\sin^4x+\cos^4x+\sin^2x\cos^2x)^2-(\sin^8x+\cos^8x)$
$=2(1-\sin^2x\cos^2x)^2-[(\sin^4x+\cos^4x)^2-2\sin^4x\cos^4x]$
$=2(1-2\sin^2x\cos^2x+\sin^4x\cos^4x)-(\sin^4x+\cos^4x)^2+2\sin^4x\cos^4x$
$=2-(1-2\sin^2x\cos^2x)^2-4\sin^2x\cos^2x+4\sin^4x\cos^4x$
$=2-1+4\sin^2x\cos^2x-4\sin^4x\cos^4x-4\sin^2x\cos^4x+4\sin^4x\cos^4x$
$=2-1=1$
