Câu hỏi của nhannhan

a) \(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x=1-2sin^2xcos^2x\left(đpcm\right)\)b) \(sin^6x+cos^6x=\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2xcos^2x+sin^4x\right)=1.\left(sin^4x+cos^4x-sin^2xcos^2x\right)\)\(=1-2sin^2xcos^2x-sin^2xcos^2x=1-3sin^2xcos^2x\left(do.cmt\right)\)\(\Rightarrowđpcm\)Câu trả lời: a) \(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x=1-2sin^2xcos^2x\left(đpcm\right)\)b) \(sin^6x+cos^6x=\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2xcos^2x+sin^4x\right)=1.\left(sin^4x+cos^4x-sin^2xcos^2x\right)\)\(=1-2sin^2xcos^2x-sin^2xcos^2x=1-3sin^2xcos^2x\left(do.cmt\right)\)\(\Rightarrowđpcm\)