C1:D                  C2:C                C3:C                   C4:A                     C5:A

C6:A                  C7:A                 C8:D                  C9:B                     C10:C

C11:A                C12:B               C13:C                C14:C                   C15:B

C16:D               C17:D              C18:D                 C19:B                   C20:B

C21:A               C22:C              C23:D                 C24:A                    C25D

\(k\in Z\)

a.

\(cos\left(x-2\right)=\dfrac{2}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=arccos\left(\dfrac{2}{5}\right)+k2\pi\\x-2=-arccos\left(\dfrac{2}{5}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2+arccos\left(\dfrac{2}{5}\right)+k2\pi\\x=2-arcos\left(\dfrac{2}{5}\right)+k2\pi\end{matrix}\right.\)

d.

\(\Leftrightarrow\left[{}\begin{matrix}cosx=-\dfrac{1}{2}\\cosx=3>1\left(vô-nghiệm\right)\end{matrix}\right.\)

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

f.

\(\Leftrightarrow cosx=-\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\pi}{3}+k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)

h.

\(cos\left(3x+10^0\right)=-1\)

\(\Leftrightarrow3x+10^0=180^0+k360^0\)

\(\Leftrightarrow3x=170^0+k360^0\)

\(\Leftrightarrow x=\dfrac{1}{3}.170^0+k120^0\)

j.

\(cos\left[cos\left(x+2\right)\right]=1\)

\(\Leftrightarrow cos\left(x+2\right)=k2\pi\)

Do \(-1\le cos\left(x+2\right)\le1\Rightarrow-1\le k2\pi\le1\)

\(\Rightarrow k=0\)

\(\Rightarrow cos\left(x+2\right)=0\)

\(\Rightarrow x+2=\dfrac{\pi}{2}+n\pi\)

\(\Rightarrow x=-2+\dfrac{\pi}{2}+n\pi\)

\(cos^2x=\dfrac{1}{2}\Leftrightarrow2cos^2x-1=0\)

\(\Leftrightarrow cos2x=0\)

\(\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

TH1: `m=0 `

`2x>0 <=> x>0`

`=>` Không thỏa mãn.

TH2: `m>0`

Bất PT có tập nghiệm là `RR <=> \Delta'<0`

`<=> (m-1)^2-m.4m<0`

`<=> m<-1 ; 1/3 <m`

Vậy `m in (0;+∞)` thỏa mãn.

`m \in (1/3 ; +∞)`.

ĐK: ` x \ne 0; x \ne1`

`(x-1)/x>=(3x-1)/(x-1)`

`<=>((x-1)^2-x(3x-1))/(x(x-1))>=0`

`<=> -((2x-1)(x+1))/(x(x-1)) >= 0`

`<=> ((2x-1)(x+1))/(x(x-1)) <= 0`

Bảng xét dấu bạn tự kẻ nkaaaaa.

Vậy `S=[-1;0) \cup [1/2 ;1)`.

Đoán đề: \(\dfrac{x^2-1}{\left(x+1\right)\left(x^2-x-6\right)}\ge0\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x-3\right)\left(x+2\right)}\ge0\)

Xét x-1=0 <=> x=1

x+1=0 <=> x=-1

x-3=0 <=> x=3

x+2=0 <=>x=-2

Bảng xét dấu:

Để VT \(\ge0\) <=> x\(\in\left(-2;-1\right)\cup\left(3;+\infty\right)\cup\left\{1\right\}\) 

\(tan\left(\dfrac{3\pi}{2}-\alpha\right)+cot\left(3\pi-\alpha\right)-cos\left(\dfrac{\pi}{2}-\alpha\right)+2.sin\left(\pi+\alpha\right)\)

\(=tan\left(\pi+\dfrac{\pi}{2}-\alpha\right)+cot\left(-\alpha\right)-sin\alpha+2\left(sin\pi.cos\alpha+cos\pi.sin\alpha\right)\)

\(=tan\left(\dfrac{\pi}{2}-\alpha\right)-cot\alpha-sin\alpha+2.-sin\alpha\)

\(=cot\alpha-cot\alpha-3sin\alpha\)

\(=-3sin\alpha\)

`A=sin(π-α)+cos(π+α)+cos(-α)`

`= sinα-cosα+cosα=sinα=3/5`