Câu hỏi của Trùm Trường

Question

1) Rút gọn biểu thức :

$M=2\left(sin^4x+cos^4x+cos^2.sin^2x\right)^2-\left(sin^8x+cos^8x\right)$

Nguyễn Việt Lâm · Accepted Answer

$\left(sin^4x+cos^4x+cos^2x.sin^2x\right)^2-sin^8x$

$=\left(sin^4x+cos^2x\left(cos^2x+sin^2x\right)\right)^2-sin^8x$

$=\left(sin^4x+cos^2x\right)^2-sin^8x=\left(sin^4x+cos^2x-sin^4x\right)\left(sin^4x+cos^2x+sin^4x\right)$

$=cos^2x\left(2sin^4x+cos^2x\right)=2sin^4x.cos^2x+cos^4x$

Tương tự: $\left(sin^4x+cos^4x+sin^2xcos^2x\right)^2-cos^8x$

$=\left(cos^4x+sin^2x\left(sin^2x+cos^2x\right)\right)^2-cos^8x$

$=\left(cos^4x+sin^2x\right)^2-cos^8x$

$=\left(cos^4x+sin^2x-cos^4x\right)\left(cos^4x+sin^2x+cos^4x\right)$

$=sin^2x\left(2cos^4x+sin^2x\right)=2sin^2x.cos^4x+sin^4x$

$\Rightarrow M=2sin^2x.cos^4x+2sin^2x.cos^2x+sin^2x+cos^4x$

$M=2sin^2x.cos^2x\left(cos^2x+sin^2x\right)+sin^4x+cos^4x$

$M=2sin^2x.cos^2x+sin^4x+cos^4x$

$M=\left(sin^2x+cos^2x\right)^2=1$Câu trả lời: $\left(sin^4x+cos^4x+cos^2x.sin^2x\right)^2-sin^8x$