Câu hỏi của Cẩm Hân Dương Hồng

Question

Biểu thức C = 2 ( sin4x + cos4x + sin2x cos2x )2 - ( sin8x + cos8x ) có giá trị không đổi bằng bao nhiêu?

Hung nguyen · Accepted Answer

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

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

$=2\left(1-sin^2xcos^2x\right)^2-\left(\left(\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x\right)^2-2sin^4xcos^4x\right)$

$=2\left(1-2sin^2xcos^2x+sin^4xcos^4x\right)-\left(1-4sin^2xcos^2x+4sin^4xcos^4x-2sin^4xcos^4x\right)$

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

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

$=2\left(1-sin^2xcos^2x\right)^2-\left(\left(\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x\right)^2-2sin^4xcos^4x\right)$

$=2\left(1-2sin^2xcos^2x+sin^4xcos^4x\right)-\left(1-4sin^2xcos^2x+4sin^4xcos^4x-2sin^4xcos^4x\right)$

$=1$

Cẩm Hân Dương Hồng · Answer

co cach giai trac nghiem cau nay nhanh k akCâu trả lời: co cach giai trac nghiem cau nay nhanh k ak