a.

\(y'=-\dfrac{3}{2}x^3+\dfrac{6}{5}x^2-x+5\)

b.

\(y'=\dfrac{\left(x^2+4x+5\right)'}{2\sqrt{x^2+4x+5}}=\dfrac{2x+4}{2\sqrt{x^2+4x+5}}=\dfrac{x+2}{\sqrt{x^2+4x+5}}\)

c.

\(y=\left(3x-2\right)^{\dfrac{1}{3}}\Rightarrow y'=\dfrac{1}{3}\left(3x-2\right)^{-\dfrac{2}{3}}=\dfrac{1}{3\sqrt[3]{\left(3x-2\right)^2}}\)

d.

\(y'=2\sqrt{x+2}+\dfrac{2x-1}{2\sqrt{x+2}}=\dfrac{6x+7}{2\sqrt{x+2}}\)

e.

\(y'=3sin^2\left(\dfrac{\pi}{3}-5x\right).\left[sin\left(\dfrac{\pi}{3}-5x\right)\right]'=-15sin^2\left(\dfrac{\pi}{3}-5x\right).cos\left(\dfrac{\pi}{3}-5x\right)\)

g.

\(y'=4cot^3\left(\dfrac{\pi}{6}-3x\right)\left[cot\left(\dfrac{\pi}{3}-3x\right)\right]'=12cot^3\left(\dfrac{\pi}{6}-3x\right).\dfrac{1}{sin^2\left(\dfrac{\pi}{3}-3x\right)}\)

a: cos3x=8

mà -1<=cos3x<=1

nên \(x\in\varnothing\)

b; \(-2\cdot cosx+\sqrt{3}=0\)

=>\(-2\cdot cosx=-\sqrt{3}\)

=>\(cosx=\dfrac{\sqrt{3}}{2}\)

=>x=pi/6+k2pi hoặc x=-pi/6+k2pi

c: cos(3x-pi/6)=0

=>3x-pi/6=pi/2+k2pi

=>3x=2/3pi+k2pi

=>x=2/9pi+k2pi/3

d: cos(x+2/3pi)=cos(pi/5)

=>x+2/3pi=pi/5+k2pi hoặc x+2/3pi=-pi/5+k2pi

=>x=-7/15pi+k2pi hoặc x=-13/15pi+k2pi

e: cos^2(3x)=4

=>cos3x=2(loại) hoặc cos3x=-2(loại)

Để giải phương trình này, chúng ta sẽ sử dụng các công thức chuyển đổi của hàm lượng giác để làm cho phương trình có dạng đơn giản hơn.Trước tiên, chúng ta sẽ sử dụng công thức chuyển đổi:sin(π/3 - 3x) = sin(π/3)cos(3x) - cos(π/3)sin(3x)= (√3/2)cos(3x) - (1/2)sin(3x)Sau đó, phương trình trở thành:cos(3x + π/6) - (√3/2)cos(3x) + (1/2)sin(3x) = √3Tiếp theo, chúng ta sẽ sử dụng công thức cộng hai cosin và sin:cos(a + b) = cos(a)cos(b) - sin(a)sin(b)sin(a + b) = sin(a)cos(b) + cos(a)sin(b)Áp dụng công thức này, phương trình trở thành:cos(3x)cos(π/6) - sin(3x)sin(π/6

\(cos\left(3x+\dfrac{pi}{6}\right)-sin\left(\dfrac{pi}{3}-3x\right)=\sqrt{3}\)

=>\(cos\left(3x+\dfrac{pi}{6}\right)-cos\left(\dfrac{pi}{2}-\dfrac{pi}{3}+3x\right)=\sqrt{3}\)

=>\(cos\left(3x+\dfrac{pi}{6}\right)-cos\left(3x+\dfrac{pi}{6}\right)=\sqrt{3}\)

=>0x=căn 3(vô lý)

a) Pt \(\Leftrightarrow3.cos4x-\left(cos6x+1\right)=1\)

\(\Leftrightarrow3cos4x-cos6x-2=0\)

Đặt \(t=2x\)

Pttt:\(3cos2t-cos3t-2=0\)

\(\Leftrightarrow3\left(2cos^2t-1\right)-\left(4cos^3t-3cost\right)-2=0\)

\(\Leftrightarrow-4cos^3t+6cos^2t+3cost-5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cost=1\\cost=\dfrac{1+\sqrt{21}}{4}\left(vn\right)\\cost=\dfrac{1-\sqrt{21}}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=k2\pi\\t=\pm arc.cos\left(\dfrac{1-\sqrt{21}}{4}\right)+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\dfrac{1}{2}.arccos\left(\dfrac{1-\sqrt{21}}{4}\right)+k\pi\end{matrix}\right.\) (\(k\in Z\))

Vậy...

a2) \(2cos2x-8cosx+7=\dfrac{1}{cosx}\) (ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\))

\(\Leftrightarrow2.\left(2cos^2x-1\right)-8cosx+7=\dfrac{1}{cosx}\)

\(\Leftrightarrow2.\left(2cos^2x-1\right)cosx-8cos^2x+7cosx=1\)

\(\Leftrightarrow4cos^3x-8cos^2x+5cosx-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\) (tm) (\(k\in Z\))

Vậy...

a3) Đk: \(x\ne-\dfrac{\pi}{4}+k\pi;x\ne\dfrac{\pi}{2}+k\pi\)

Pt \(\Leftrightarrow\dfrac{\left(1+sinx+1-2sin^2x\right).\dfrac{1}{\sqrt{2}}\left(sinx+cosx\right)}{1+\dfrac{sinx}{cosx}}=\dfrac{1}{\sqrt{2}}cosx\)

\(\Leftrightarrow\dfrac{\left(-2sin^2x+sinx+2\right).\left(sinx+cosx\right)cosx}{cosx+sinx}=cosx\)

\(\Leftrightarrow\left(2+sinx-2sin^2x\right).cosx=cosx\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\2+sinx-2sin^2x=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\) (\(k\in Z\))

Vậy...

a4) Pt \(\Leftrightarrow9sinx+6cosx-6sinx.cosx+1-2sin^2x=8\)

\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sin^2x-9sinx+7\right)=0\)

\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sinx-7\right)\left(sinx-1\right)=0\)

\(\Leftrightarrow\left(1-sinx\right)\left(6cosx+2sinx+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\6cosx+2sinx=7\left(vn\right)\end{matrix}\right.\) (\(6cosx+2sinx=7\) vô nghiệm do \(6^2+2^2< 7^2\))

\(\Rightarrow sinx=1\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi;k\in Z\)

Vậy...

6:ĐKXĐ: x>=0; x<>1/25

BPT=>\(\dfrac{3\sqrt{x}}{5\sqrt{x}-1}+3< =0\)

=>\(\dfrac{3\sqrt{x}+15\sqrt{x}-5}{5\sqrt{x}-1}< =0\)

=>\(\dfrac{18\sqrt{x}-5}{5\sqrt{x}-1}< =0\)

=>\(\dfrac{1}{5}< \sqrt{x}< =\dfrac{5}{18}\)

=>\(\dfrac{1}{25}< x< =\dfrac{25}{324}\)

7:

ĐKXĐ: x>=0

BPT \(\Leftrightarrow\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}>\dfrac{8}{3}:\dfrac{8}{3}=1\)

=>\(\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}-1>=0\)

=>\(\dfrac{\sqrt{x}+1-2\sqrt{x}-3}{2\sqrt{x}+3}>=0\)

=>\(-\sqrt{x}-2>=0\)(vô lý)

8:

ĐKXĐ: x>=0; x<>9/4

BPT \(\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0\)

=>\(\dfrac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0\)

=>\(\dfrac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0\)

TH1: 9căn x-14>0 và 2căn x-3<0

=>căn x>14/9 và căn x<3/2

=>14/9<căn x<3/2

=>196/81<x<9/4

TH2: 9căn x-14<0 và 2căn x-3>0

=>căn x>3/2 hoặc căn x<14/9

mà 3/2<14/9

nên trường hợp này Loại

9: 

ĐKXĐ: x>=0

\(BPT\Leftrightarrow\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}< =-\dfrac{1}{3}\)

=>\(\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}+\dfrac{1}{3}< =0\)

=>\(\dfrac{6\sqrt{x}+9+5\sqrt{x}+7}{3\left(5\sqrt{x}+7\right)}< =0\)

=>\(\dfrac{11\sqrt{x}+16}{3\left(5\sqrt{x}+7\right)}< =0\)(vô lý)

10: 

ĐKXĐ: x>=0; x<>1/49

\(BPT\Leftrightarrow\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}+6>0\)

=>\(\dfrac{6\sqrt{x}-2+42\sqrt{x}-6}{7\sqrt{x}-1}>0\)

=>\(\dfrac{48\sqrt{x}-8}{7\sqrt{x}-1}>0\)

=>\(\dfrac{6\sqrt{x}-1}{7\sqrt{x}-1}>0\)

TH1: 6căn x-1>0 và 7căn x-1>0

=>căn x>1/6 và căn x>1/7

=>căn x>1/6

=>x>1/36

TH2: 6căn x-1<0 và 7căn x-1<0

=>căn x<1/6 và căn x<1/7

=>căn x<1/7

=>0<=x<1/49

a, Ta có : \(\sin\left(3x+60\right)=\dfrac{1}{2}\)

\(\Rightarrow3x+60=30+2k180\)

\(\Rightarrow3x=2k180-30\)

\(\Leftrightarrow x=120k-10\)

Vậy ...

b, Ta có : \(\cos\left(2x-\dfrac{\pi}{3}\right)=-\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow2x-\dfrac{\pi}{3}=\dfrac{3}{4}\pi+k2\pi\)

\(\Leftrightarrow x=\dfrac{13}{24}\pi+k\pi\)

Vậy ...

c, Ta có : \(tan\left(x+\dfrac{\pi}{6}\right)=\sqrt{3}\)

\(\Rightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{3}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\)

Vậy ...

d, Ta có : \(\cot\left(2x+\pi\right)=-1\)

\(\Rightarrow2x+\pi=\dfrac{3}{4}\pi+k\pi\)

\(\Leftrightarrow x=-\dfrac{1}{8}\pi+\dfrac{k}{2}\pi\)

Vậy ...

a) \(sin\left(3x+60^0\right)=\dfrac{1}{2}\)

\(\Leftrightarrow sin\left(3x+\dfrac{\pi}{3}\right)=sin\dfrac{\pi}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\3x+\dfrac{\pi}{3}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)(\(k\in Z\))\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{18}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\end{matrix}\right.\)(\(k\in Z\))

Vậy...

b) Pt\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=cos\dfrac{3\pi}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{3\pi}{4}+k2\pi\\2x-\dfrac{\pi}{3}=-\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)(\(k\in Z\))\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13\pi}{24}+k\pi\\x=-\dfrac{5\pi}{24}+k\pi\end{matrix}\right.\)(\(k\in Z\))

Vậy...

c) Pt \(\Leftrightarrow tan\left(x+\dfrac{\pi}{6}\right)=tan\dfrac{\pi}{3}\)

\(\Leftrightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{3}+k\pi,k\in Z\)\(\Leftrightarrow x=\dfrac{\pi}{6}+k\pi,k\in Z\)

Vậy...

d) Pt \(\Leftrightarrow tan\left(2x+\pi\right)=-1\)

\(\Leftrightarrow2x+\pi=-\dfrac{\pi}{4}+k\pi,k\in Z\)

\(\Leftrightarrow x=-\dfrac{5\pi}{8}+\dfrac{k\pi}{2},k\in Z\)

Vậy...

a, ĐK: \(x\ne\dfrac{5\pi}{6}+k2\pi;x\ne\dfrac{\pi}{6}+k2\pi\)

\(\dfrac{2sin^2\left(\dfrac{3x}{2}-\dfrac{\pi}{4}\right)+\sqrt{3}cos^3x\left(1-3tan^2x\right)}{2sinx-1}=-1\)

\(\Leftrightarrow2sin^2\left(\dfrac{3x}{2}-\dfrac{\pi}{4}\right)+\sqrt{3}cos^3x\left(1-3tan^2x\right)=1-2sinx\)

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

\(\Leftrightarrow-sin3x+\sqrt{3}cosx.\left(cos^2x-3sin^2x\right)=-2sinx\)

\(\Leftrightarrow-sin3x+\sqrt{3}cosx.\left(4cos^2x-3\right)=-2sinx\)

\(\Leftrightarrow-sin3x+\sqrt{3}cos3x=-2sinx\)

\(\Leftrightarrow\dfrac{1}{2}sin3x-\dfrac{\sqrt{3}}{2}cos3x-sinx=0\)

\(\Leftrightarrow sin\left(3x-\dfrac{\pi}{3}\right)-sinx=0\)

\(\Leftrightarrow2cos\left(2x-\dfrac{\pi}{6}\right)sin\left(x-\dfrac{\pi}{6}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(2x-\dfrac{\pi}{6}\right)=0\\sin\left(x-\dfrac{\pi}{6}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+k\pi\\x-\dfrac{\pi}{6}=k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)

Đối chiếu điều kiện ta được:

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k\pi\\x=\dfrac{7\pi}{6}+k2\pi\\x=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)

a.

\(\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=3sinx+cosx+2\)

\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)

\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0\)

\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)

\(\Leftrightarrow\left(2cosx-3\right)\left(sinx+cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{3}{2}\left(vn\right)\\sinx+cosx+1=0\end{matrix}\right.\)

\(\Rightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-1\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(cosx\ne\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{3}+k2\pi\\x\ne-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\dfrac{\left(2-\sqrt{3}\right)cosx-2sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2cosx-1}=1\)

\(\Rightarrow\left(2-\sqrt{3}\right)cosx+cos\left(x-\dfrac{\pi}{2}\right)=2cosx\)

\(\Leftrightarrow-\sqrt{3}cosx+sinx=0\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=0\)

\(\Rightarrow x-\dfrac{\pi}{3}=k\pi\)

\(\Rightarrow x=\dfrac{\pi}{3}+k\pi\)

Kết hợp ĐKXĐ \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)

c.

\(2\sqrt{2}cos\left(\dfrac{5\pi}{12}-x\right)sinx=1\)

\(\Leftrightarrow\sqrt{2}\left(sin\left(\dfrac{5\pi}{12}\right)+sin\left(2x-\dfrac{5\pi}{12}\right)\right)=1\)

\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=\dfrac{-\sqrt{6}+\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=sin\left(-\dfrac{\pi}{12}\right)\)

\(\Leftrightarrow...\)