Đáp án+Giải thích các bước giải:

 Bài `27:`

`a)` Với góc `\alpha` là góc nhọn nên:

`A=(sin\alpha+cos\alpha)^{2}+(sin^{2}\alpha-cos^{2}\alpha)^{2}`

`=(sin\alpha+cos\alpha)^{2}+[(sin\alpha-cos\alpha).(sin\alpha+cos\alpha)]^{2}`

`=(sin\alpha+cos\alpha)^{2}.[1+(sin\alpha-cos\alpha)^{2}]`

`=(sin^{2}\alpha+2sin\alphacos\alpha+cos^{2}\alpha).(1+sin^{2}\alpha-2sin\alphacos\alpha+cos^{2}\alpha)`

`=[(sin^{2}\alpha+cos^{2}\alpha)+2sin\alphacos\alpha].[1+(sin^{2}\alpha+cos^{2}\alpha)-2sin\alphacos\alpha)`

`=(1+sin2\alpha).(2+sin2\alpha)`

`=2+sin2\alpha+2sin2\alpha+(sin2\alpha)^{2}`

`=2+3sin2\alpha+(sin2\alpha)^{2}`

`=2+3sin2\alpha+4sin^{2}\alphacos^{2}\alpha`

Vậy `A=2+3sin2\alpha+4sin^{2}\alphacos^{2}\alpha`

`b)B=sin^{6}\alpha+cos^{6}\alpha+3sin^{2}\alpha.cos^{2}\alpha`

`(đk:` Góc `\alpha` là góc nhọn)

`=[(sin^{2}\alpha)^{3}+(cos^{2}\alpha)^{3}]+3sin^{2}\alphacos^{2}\alpha`

`=(sin^{2}\alpha+cos^{2}\alpha).[(sin^{2}\alpha)^{2}-sin^{2}\alpha.cos^{2}\alpha+(cos^{2}\alpha)^{2}]+3sin^{2}\alphacos^{2}\alpha`

`=1.[(sin^{2}\alpha)^{2}-sin^{2}\alpha.cos^{2}\alpha+(cos^{2}\alpha)^{2}]+3sin^{2}\alpha.cos^{2}\alpha`

`=(sin^{2}\alpha)^{2}+2.sin^{2}\alpha. cos^{2}\alpha+(cos^{2}\alpha)^{2}`

`=(sin^{2}\alpha+cos^{2}\alpha)^{2}`

`=1^{2}`

`=1`

Vậy `B=1`