b: \(B=sin^2x\left(sin^2x+cos^2x\right)+cos^2x\)

\(=sin^2x+cos^2x=1\)

c: \(=cos^2x\left(cos^2x+sin^2x\right)+cos^2x\)

=cos^2x+cos^2x

=2*cos^2x có phụ thuộc vào x nha bạn

Đề bài ko chính xác, biểu thức này không rút gọn được (có thể coi việc biến đổi khả dĩ duy nhất \(1+2sina.cosa=\left(sina+cosa\right)^2\) không phải là hành động rút gọn)

chỉnh lại đề 1 chút: \(A=\dfrac{1+2sin\alpha.cos\alpha}{cos^2\alpha-sin^2\alpha}=\dfrac{cos^2\alpha+sin^2\alpha+2sin\alpha.cos\alpha}{\left(cos\alpha-sin\alpha\right)\left(cos\alpha+sin\alpha\right)}\)

\(=\dfrac{\left(cos\alpha+sin\alpha\right)^2}{\left(cos\alpha-sin\alpha\right)\left(cos\alpha+sin\alpha\right)}=\dfrac{cos\alpha+sin\alpha}{cos\alpha-sin\alpha}\)

`A=sin^4x+cos^4x+2sin^2x+cos^2x`

`=(sin^2x+cos^2x)^2-2sin^2xcos^2x+sin^2x+(sin^2x+cos^2x)`

`=1-1/2 sin^2 2x + sin^2 x+1`

`=2-1/2 sin^2 2x + sin^2x`

ta có : \(sin^6x+cos^6x-2sin^4x-cos^4x+sin^2x\)

\(=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)-2sin^4x-cos^4x+sin^2x\)

\(=1-3sin^2x.cos^2x-2sin^4x-cos^4x+sin^2x\)

\(=1-2sin^2x.cos^2x-2sin^4x-sin^2x.cos^2x+sin^2x-cos^4 x\)

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

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

\(=1-2sin^2x+sin^2x-cos^2x=1-sin^2x-cos^2x\)

\(=1-1=0\) (không phụ thuộc vào biến \(x\)) (đpcm)

\(=3\left(sin^4x+cos^4x\right)\left(sin^2x-cos^2x\right)+4cos^6x-8sin^6x+6sin^4x\)

\(=3\left(sin^4x+cos^4x\right)\left(sin^2x-cos^2x\right)+4cos^6x-2sin^6x+6sin^4x\left(1-sin^2x\right)\)

\(=sin^6x+3sin^4x.cos^2x+3cos^2x.sin^4x+cos^6x\)

\(=\left(sin^2x+cos^2x\right)^3=1\)

a) P = cos(\(\frac{\Pi}{2}\) + x) + cos(2π - x) + cos(3π + x)   = -sinx + cosx - cosx = -sinx

\(A=\frac{2sin^2x-5sinx.cosx+cos^2x}{2sin^2x+sinx.cosx+cos^2x}=\frac{\frac{2sin^2x}{cos^2x}-\frac{5sinx.cosx}{cos^2x}+\frac{cos^2x}{cos^2x}}{\frac{2sin^2x}{cos^2x}+\frac{sinx.cosx}{cos^2x}+\frac{cos^2x}{cos^2x}}\)

\(=\frac{2tan^2x-5tanx+1}{2tan^2x+tanx+1}=\frac{2.3^2-5.3+1}{2.3^2+3+1}=...\)

1,\(A=3\left(sin^4x+cos^4x\right)-2\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)\)

\(=3\left(sin^4x+cos^4x\right)-2\left(sin^4x-sin^2x.cos^4x+cos^4x\right)\)

\(=sin^4x+2sin^2x.cos^2x+cos^4x=\left(sin^2x+cos^2x\right)^2=1\)

Vậy...

2,\(B=cos^6x+2sin^4x\left(1-sin^2x\right)+3\left(1-cos^2x\right)cos^4x+sin^4x\)

\(=-2cos^6x+3sin^4x-2sin^6x+3cos^4x\)

\(=-2\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3\left(cos^4x+sin^4x\right)\)

\(=-2\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3\left(cos^4x+sin^4x\right)\)\(=cos^4x+sin^4x+2sin^2x.cos^2x=1\)

Vậy...

3,\(C=\dfrac{1}{2}\left[cos\left(-\dfrac{7\pi}{12}\right)+cos\left(2x-\dfrac{\pi}{12}\right)\right]+\dfrac{1}{2}\left[cos\left(-\dfrac{7\pi}{12}\right)+cos\left(2x+\dfrac{11\pi}{12}\right)\right]\)

\(=cos\left(-\dfrac{7\pi}{12}\right)+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)+cos\left(2x+\dfrac{11\pi}{12}\right)\right]\)\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)+cos\left(2x-\dfrac{\pi}{12}+\pi\right)\right]\)

\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)-cos\left(2x-\dfrac{\pi}{12}\right)\right]\)\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}\)

Vậy...

4, \(D=cos^2x+\left(-\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx\right)^2+\left(-\dfrac{1}{2}.cosx+\dfrac{\sqrt{3}}{2}.sinx\right)^2\)

\(=cos^2x+\dfrac{1}{4}cos^2x+\dfrac{\sqrt{3}}{4}cosx.sinx+\dfrac{3}{4}sin^2x+\dfrac{1}{4}cos^2x-\dfrac{\sqrt{3}}{4}cosx.sinx+\dfrac{3}{4}sin^2x\)

\(=\dfrac{3}{2}\left(cos^2x+sin^2x\right)=\dfrac{3}{2}\)

Vậy...

5, Xem lại đề

6,\(F=-cosx+cosx-tan\left(\dfrac{\pi}{2}+x\right).cot\left(\pi+\dfrac{\pi}{2}-x\right)\)

\(=tan\left(\pi-\dfrac{\pi}{2}-x\right).cot\left(\dfrac{\pi}{2}-x\right)\)\(=tan\left(\dfrac{\pi}{2}-x\right).cot\left(\dfrac{\pi}{2}-x\right)\)\(=cotx.tanx=1\)

Vậy...