b, ta có

8\((x)^{9}\)-\(9(x)^{8} +1 \)= (8x^9 -8x^8)-(x^8-1)

=8x^8(x-1)-(x-1)(x^7+x^6+x^5+...+x+1)

=(x-1)(8x^8-x^7-x^6-......-x-1)

=(x-1)[(x^8-x^7)+(x^8-x^6)+.....+(x^8-1)]

=(x-1)[x^7(x-1)+ x^6(x^2-1)+.......+(x-1).(x^7+x^6+.....+x+1)]

=(x-1)^2.[x^7+x^6(x+1)+x^5(x^2+x+1)+.....+(x^7+x^6+...+x+1)]

\(\Rightarrow\) C chia hết cho D(dpcm)

1,\(VT=\dfrac{sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}{cos\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}+\dfrac{cos\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}{sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}\)\(=\dfrac{sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)^2+cos^2\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}{cos\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}\)

\(=\dfrac{1}{\dfrac{1}{2}.sin\left(\dfrac{\pi}{2}+x\right)}=\dfrac{2}{cosx}=VP\)

2,\(VT=\left(sin^4x-cos^4x\right)\left(sin^4x+cos^4x\right)=\left(sin^2x+cos^2x\right)\left(sin^2x-cos^2x\right)\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)

\(=\left(sin^2-cos^2x\right)\left(1-2sin^2x.cos^2x\right)\)\(=-cos2x\left(1-\dfrac{1}{2}sin^22x\right)\)\(=-\dfrac{cos2x\left(2-sin^22x\right)}{2}=-\dfrac{cos2x\left(1+cos^22x\right)}{2}\)

\(VP=-\left(\dfrac{7}{8}cos2x+\dfrac{1}{8}cos6x\right)=-\dfrac{7}{8}cos2x-\dfrac{1}{8}\left[4cos^32x-3cos2x\right]=-\dfrac{7}{8}.cos2x-\dfrac{1}{2}cos^32x+\dfrac{3}{8}cos2x\)

\(=-\dfrac{1}{2}cos2x-\dfrac{1}{2}cos^32x=\dfrac{-cos2x\left(1+cos^22x\right)}{2}\)

\(\Rightarrow VT=VP\)(đpcm)

3, \(VT=3-4\left(1-2sin^2x\right)+1-2sin^22x=8sin^2x-2sin^22x=8sin^2x-8.sin^2x.cos^2x=8sin^2x\left(1-cos^2x\right)=8sin^4x=VP\)

4,\(VP=\dfrac{1}{2}\left[sin\left(x+\dfrac{\pi}{2}\right)+sin\left(3x+\dfrac{\pi}{6}\right)\right]-\dfrac{1}{2}\left[cos\left(3x-\dfrac{\pi}{3}\right)+cos\left(x+\pi\right)\right]\)

\(=\dfrac{1}{2}\left(cosx+sin3x.\dfrac{\sqrt{3}}{2}+\dfrac{cos3x}{2}\right)-\dfrac{1}{2}\left(\dfrac{cos3x}{2}+sin3x.\dfrac{\sqrt{3}}{2}-cosx\right)\)

\(=\dfrac{1}{2}.2cosx=cosx=VP\)

5, \(VP=4cos\left(2x-\dfrac{\pi}{6}\right).\left(sinx.\dfrac{\sqrt{3}}{2}+\dfrac{cosx}{2}\right)^2\)\(=cos\left(2x-\dfrac{\pi}{6}\right).\left(sinx.\sqrt{3}+cosx\right)^2\)

\(VT=2.cos\left(2x-\dfrac{\pi}{6}\right)+2.sin\left(2x-\dfrac{\pi}{6}\right).cos\left(2x-\dfrac{\pi}{6}\right)=2cos\left(2x-\dfrac{\pi}{6}\right)\left[1+sin\left(2x-\dfrac{\pi}{6}\right)\right]\)

\(=2cos\left(2x-\dfrac{\pi}{6}\right)\left(1+\dfrac{sin2x.\sqrt{3}}{2}-\dfrac{cos2x}{2}\right)\)\(=2cos\left(2x-\dfrac{\pi}{6}\right)\left(sin^2x+cos^2x+sinx.cosx.\sqrt{3}-\dfrac{cos^2x-sin^2x}{2}\right)\)

\(=2cos\left(2x-\dfrac{\pi}{6}\right)\left(sin^2x.\dfrac{3}{2}+sinx.cosx.\sqrt{3}+\dfrac{cos^2x}{2}\right)\)\(=cos\left(2x-\dfrac{\pi}{6}\right)\left(sin^2x.3+2sinx.cosx.\sqrt{3}+cos^2x\right)\)

\(=cos\left(2x-\dfrac{\pi}{6}\right)\left(sinx.\sqrt{3}+cosx\right)^2\)

\(\Rightarrow VT=VP\) (dpcm)

a) Điều kiện xác định \(16x+8\ge0\Leftrightarrow x\ge-\frac{1}{2}.\)

Theo bất đẳng thức Cô-Si cho 4 số ta được 

\(4\sqrt[4]{16x+8}=4\sqrt[4]{2\cdot2\cdot2\cdot\left(2x+1\right)}\le2+2+2+2x+1=2x+7\)

Do vậy mà \(4x^3+4x^2-5x+9\le2x+7\Leftrightarrow\left(2x-1\right)^2\left(x+2\right)\le0\).

Vì \(x\ge-\frac{1}{2}\to x+2>0\to\left(2x-1\right)^2\le0\to x=\frac{1}{2}.\) 

b. Ta viết phương trình dưới dạng sau đây  \(9x^4-21x^3+27x^2+16x+16=0\Leftrightarrow3x^2\left(3x^2-7x+7\right)+4\left(x+2\right)^2=0\)

Vì \(3x^2-7x+7=\frac{36x^2-2\cdot6x\cdot7+49+35}{12}=\frac{\left(6x-7\right)^2+35}{12}>0\) nên vế trái dương, suy ra phương trinh vô nghiệm.