1.

Đường thẳng d' song song với d cần tìm có dạng: \(x-2y+m=0\left(m\in R\right)\)

Mà \(d'\) đi qua \(A=\left(1;0\right)\Rightarrow1+m=0\Leftrightarrow m=-1\)

\(\Rightarrow d:x-2y-1=0\)

2.

Đường thẳng d' vuông góc với d có dạng: \(2x+y+m=0\left(m\in R\right)\)

Mà d' đi qua \(B=\left(-1;2\right)\Rightarrow-2+2+m=0\Leftrightarrow m=0\)

\(\Rightarrow d':2x+y=0\)

\(k\in Z\)

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.

TH1 là m=0 thì TH2 là \(m\ne0\)

Bpt có tập nghiệm là R <=> \(\left\{{}\begin{matrix}a>0\\\Delta'< 0\end{matrix}\right.\)

Đáp án: m\(\in\left(\dfrac{1}{3};+\infty\right)\)

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:

x -2 -1 1 3 -vc +vc x-1 x+2 x-3 x+1 VT 0 0 0 0 0 + + + + - - - - + + + + + - - - - - - + + + - + - - +

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`