a: Để A giao B bằng rỗng thì \(\left[{}\begin{matrix}m>+\infty\left(lọa\right)\\2< n\end{matrix}\right.\Leftrightarrow n>2\)

\(\left|\overrightarrow{AB}+\overrightarrow{AD}\right|=AC\)

\(\left|\overrightarrow{OA}-\overrightarrow{BO}\right|=\left|\overrightarrow{OA}-\overrightarrow{OD}\right|=DA=a\)

2:

a: pi/2<a<pi

=>cosa<0

sin^2a+cos^2a=1

=>cos^2a=1-4/9=5/9

=>cosa=-căn 5/3

cos2a=2*cos^2a-1=2*5/9-1=10/9-1=1/9

sin(2a-pi/6)

=sin2a*cospi/6-cos2a*sinpi/6

=2*sina*cosa*(căn 3/2)-1/9*1/2

\(=2\cdot\dfrac{2}{3}\cdot\dfrac{-\sqrt{5}}{3}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{1}{18}=\dfrac{-4\sqrt{15}-1}{18}\)

b; tan a=2

=>sin a=2*cosa

\(A=\dfrac{3\cdot\left(2\cdot cosa\right)^2-cos^2a+2}{5\cdot\left(2\cdot cosa\right)^2+3cosa\cdot2cosa}\)

\(=\dfrac{12\cdot cos^2a-cos^2a+2}{20cos^2a+6cos^2a}\)

\(=\dfrac{11cos^2a+2\left(4cos^2a+cos^2a\right)}{26cos^2a}=\dfrac{21}{26}\)

4:

a: (C): x^2+y^2-4x+2y-4=0

=>x^2-4x+4+y^2+2y+1=9

=>(x-2)^2+(y+1)^2=9

=>I(2;-1); R=3

b: Gọi (d) là phương trình cần tìm

(d)//4x+3y-1=0

=>(d): 4x+3y+c=0

I(2;-1);R=3

Theo đề, ta có: d(I;(d))=R=3

=>\(\dfrac{\left|4\cdot2+3\cdot\left(-1\right)+c\right|}{\sqrt{4^2+3^2}}=3\)

=>|c+5|=15

=>c=10 hoặc c=-20

Câu 1:

\(\left(4x+3\right)\left(3x^2+x-2\right)\left(2x^2-3x-5\right)=0\\ \Leftrightarrow\left(4x+3\right)\left(3x-2\right)\left(x+1\right)\left(2x-5\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=-1\\x=\dfrac{2}{3}\\x=\dfrac{5}{2}\end{matrix}\right.\\ \Leftrightarrow A=\left\{-1;-\dfrac{3}{4};\dfrac{2}{3};\dfrac{5}{2}\right\}\)

Câu 2:

\(\left(x^2-4\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=3\end{matrix}\right.\Leftrightarrow A=\left\{-2;2;3\right\}\\ \left|5x\right|-11\le0\Leftrightarrow\left|5x\right|\le11\Leftrightarrow-11\le5x\le11\\ \Leftrightarrow-\dfrac{11}{5}\le x\le\dfrac{11}{5}\\ \Leftrightarrow B=\left[-\dfrac{11}{5};\dfrac{11}{5}\right]\)

\(\Leftrightarrow A\cap B=\left\{-2;2\right\}\\ A\cup B=\left[-\dfrac{11}{5};3\right]\\ A\B=\left\{3\right\}\)

5:

a: sin x=2*cosx

\(A=\dfrac{6cosx+2cosx-4\cdot8\cdot cos^3x}{cos^3x-2cosx}\)

\(=\dfrac{8-32cos^2x}{cos^2x-2}\)

b: VT=sin^4(pi/2-x)+cos^4(x+pi/2)+6*1/2*sin^22x+1/2*cos4x

=cos^4x+sin^4x+3*sin^2(2x)+1/2*(1-2*sin^2(2x))

=1-2*sin^2x*cos^2x+3*sin^2(2x)+1/2-sin^2(2x)

==3/2=VP