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

\(+\left(sin^2x+cos^2x\right)^2-2sin^2x\cdot cos^2x+5\cdot sin^2x\cdot cos^2x\)

\(=sin^4x+cos^4x-sin^2x\cdot cos^2x+1-2\cdot sin^2x\cdot cos^2x+5\cdot sin^2x\cdot cos^2x\)

\(=1-2\cdot sin^2x\cdot cos^2x-sin^2x\cdot cos^2x+1-2\cdot sin^2x\cdot cos^2x+5\cdot sin^2x\cdot cos^2x\)

\(=2\)

Cảm ơn bạn !!!

Bạn kiểm tra lại đề bài câu 1, câu này chỉ có thể rút gọn đến \(2cot^2x+2cotx+1\) nên biểu thức ko hợp lý

Đồng thời kiểm tra luôn đề câu 2, trong cả 2 căn thức đều xuất hiện \(6sin^2x\) rất không hợp lý, chắc chắn phải có 1 cái là \(6cos^2x\)

Mình sửa lại đề rồi á

\(\sqrt{sin^4x+cos^2x}+\sqrt{sin^2x+cos^4x}\)

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

\(=\sqrt{1-cos^2x+cos^4x}+\sqrt{sin^2x+cos^4x}\)

\(=\sqrt{sin^2x+cos^4x}+\sqrt{sin^2x+cos^4x}\)

\(=2\sqrt{sin^2x+cos^4x}\)

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...

Giả sử các biểu thức đều có nghĩa

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

\(A=2\left(sin^2x+cos^2x\right)\left(\left(sin^2x+cos^2x\right)^2-3sin^2xcos^2x\right)-3\left(\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x\right)\)

\(A=2\left(1-3sin^2xcos^2x\right)-3\left(1-2sin^2xcos^2x\right)\)

\(A=2-6sin^2xcos^2x-3+6sin^2xcos^2x=-1\)

b/ \(B=\dfrac{1+cotx}{1-cotx}-\dfrac{2}{tanx-1}=\dfrac{1+cotx}{1-cotx}-\dfrac{2}{\dfrac{1}{cotx}-1}\)

\(B=\dfrac{1+cotx}{1-cotx}-\dfrac{2cotx}{1-cotx}=\dfrac{1+cotx-2cotx}{1-cotx}=\dfrac{1-cotx}{1-cotx}=1\)

c/ \(C=cos^4x-sin^4x+cos^4x+sin^2xcos^2x+3sin^2x\)

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

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

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

a: \(A=2\left(\sin^6x+cos^6x\right)-3\cdot\left(\sin^4x+cos^4x\right)\)

\(=2\cdot\left\lbrack\left(\sin^2x+cos^2x\right)^3-3\cdot\sin^2x\cdot cos^2x\cdot\left(\sin^2x+cos^2x\right)\right\rbrack-3\cdot\left\lbrack\left(sin^2x+cos^2x\right)^2-2\cdot\sin^2x\cdot cos^2x\right\rbrack\)

\(=2\left\lbrack1-3\cdot sin^2x\cdot cos^2x\right\rbrack-3\cdot\left\lbrack1-2\cdot\sin^2x\cdot cos^2x\right\rbrack\)

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

=2-3

=-1

c: \(C=\frac{\sin^2x}{1+\cot x}+\frac{cos^2x}{1+\tan x}+\sin x\cdot cosx\)

\(=\frac{\sin^2x}{1+\frac{cosx}{\sin x}}+\frac{cos^2x}{1+\frac{\sin x}{cosx}}+\sin x\cdot cosx=\sin^2x:\frac{\sin x+cosx}{\sin x}+cos^2x:\frac{\sin x+cosx}{cosx}+\sin x\cdot cosx\)

\(=\frac{\sin^3x+cos^3x}{\sin x+cosx}+\sin x\cdot cosx\)

\(=\frac{\left(\sin x+cosx\right)\left(\sin^2x-\sin x\cdot cosx+cos^2x\right)}{\sin x+cosx}+\sin x\cdot cosx\)

\(=\sin^2x-\sin x\cdot cosx+cos^2x+\sin x\cdot cosx\)

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

d: \(D=\frac{\cot^2x-cos^2x}{cot^2x}+\frac{\sin x\cdot cosx}{\cot x}\)

\(=\left(\frac{cos^2x}{\sin^2x}-cos^2x\right):\frac{cos^2x}{sin^2x}+\frac{\sin x\cdot cosx}{\frac{cosx}{\sin x}}\)

\(=cos^2x\left(\frac{1}{\sin^2x}-1\right)\cdot\frac{\sin^2x}{cos^2x}+\frac{\sin x\cdot cosx\cdot\sin x}{cosx}\)

\(=\frac{1-\sin^2x}{\sin^2x}\cdot\sin^2x+\sin^2x=1-\sin^2x+\sin^2x=1\)

Chọn C.

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