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

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

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

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

là giá trị không phụ thuộc vào biến (đpcm)

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

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

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

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

\(=-\sin ^2x+\sin ^2x=0\)

là giá trị không phụ thuộc vào biến (đpcm)

\(C=(\sin ^4x+\cos ^4x-1)(\tan ^2x+\cot ^2x+2)=(\sin ^4x+\cos ^4x-1)(\frac{\sin ^2x}{\cos ^2x}+\frac{\cos ^2x}{\sin ^2x}+2)\)

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

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

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

là giá trị không phụ thuộc vào biến $x$

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\(D=\frac{1}{\cos ^6x}-\tan ^6x-\frac{\tan ^2x}{\cos ^2x}=\frac{1}{\cos ^6x}-\frac{\sin ^6x}{\cos ^6x}-\frac{\sin ^2x}{\cos ^4x}\)

\(=\frac{1-\sin ^6x-\sin ^2x\cos ^2x}{\cos ^6x}=\frac{(\sin ^2x+\cos ^2x)^3-\sin ^6x-\sin ^2x\cos ^2x}{\cos ^6x}\)

\(=\frac{\sin ^6x+\cos ^6x+3\sin ^2x\cos ^2x(\sin ^2x+\cos ^2x)-\sin ^6x-\sin ^2x\cos ^2x}{\cos ^6x}\)

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

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

Giá trị biểu thức này vẫn phụ thuộc vào $x$. Bạn xem lại đề.

Ta có: \(A=\sin^6x+3\cdot\sin^4x\cdot\cos^2x+3\cdot\sin^2x\cdot\cos^4x+\cos^6x\)

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

=1

a: Sửa đề: 6x^3

\(\dfrac{6x^3-4x^2+3x-2}{3x-2}=\dfrac{2x^2\left(3x-2\right)+3x-2}{3x-2}=2x^2+1\)

b: \(\dfrac{6x^3-3x^2+4x+2}{3x^2+2}\)

\(=\dfrac{6x^3+4x-3x^2-2+4}{3x^2+2}\)

\(=2x-1+\dfrac{4}{3x^2+2}\)

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...

\(M=\sin^6x+\cos^6x\)(1)

Có công thức: \(y=\sin^{2n}x+\cos^{2n}x\)

\(\Rightarrow\max\limits_y=1;\min\limits_y=\dfrac{1}{2^{n-1}}\)

\(\Rightarrow\left(1\right)\) có \(\max\limits_M=1\)và \(\min\limits_M=\dfrac{1}{2^2}=\dfrac{1}{4}\)

e) \(=x^2\left(x+1\right)-2x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x^2-2x+3\right)\)

g) \(=x^2\left(3x-1\right)-x\left(3x-1\right)+4\left(3x-1\right)=\left(3x-1\right)\left(x^2-x+4\right)\)

h) \(=3x^2\left(2x+1\right)-x\left(2x+1\right)+\left(2x+1\right)=\left(2x+1\right)\left(3x^2-x+1\right)\)

i) \(=2x^2\left(2x+1\right)+2x\left(2x+1\right)+\left(2x+1\right)=\left(2x+1\right)\left(2x^2+2x+1\right)\)