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5:
(d) vuông góc 2x-y-2018=0
=>(d): x+2y+c=0
(C): x^2+4x+4+y^2-6y+9-25=0
=>(x+2)^2+(y-3)^2=25
=>R=5; I(-2;3)
Theo đề, ta có: d(I;(d))=5
=>\(\dfrac{\left|1\cdot\left(-2\right)+2\cdot3+c\right|}{\sqrt{5}}=5\)
=>|c+4|=5căn 5
=>c=5căn5-4 hoặc c=-5căn 5-4
4:
a: Δ=(2m+4)^2-4*2*(3m+2)
=4m^2+16m+16-24m-16
=4m^2-8m
Để f(x)>0 với mọi x thì 4m^2-8m<0
=>0<m<2
b: Δ=m^2-4*(-3)*m=m^2+12m
f(x)<0 với mọi x thì m^2+12m<0
=>-12<m<0
13:
\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}sin\left(\dfrac{pi}{33}\right)\cdot cos\left(\dfrac{pi}{33}\right)\cdot cos\left(\dfrac{2pi}{33}\right)\cdot cos\left(\dfrac{4pi}{33}\right)\cdot cos\left(\dfrac{8pi}{33}\right)\cdot cos\left(\dfrac{16pi}{33}\right)\)
\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{2}\cdot sin\dfrac{2}{33}pi\cdot cos\left(\dfrac{2}{33}pi\right)cos\left(\dfrac{4}{33}pi\right)\cdot cos\left(\dfrac{8}{33}pi\right)\cdot cos\left(\dfrac{16}{33}pi\right)\)
\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{2}\cdot sin\dfrac{2}{33}pi\cdot cos\left(\dfrac{2}{33}pi\right)cos\left(\dfrac{4}{33}pi\right)\cdot cos\left(\dfrac{8}{33}pi\right)\cdot cos\left(\dfrac{16}{33}pi\right)\)\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{4}\cdot sin\dfrac{4}{33}pi\cdot cos\left(\dfrac{4}{33}pi\right)\cdot cos\left(\dfrac{8}{33}pi\right)\cdot cos\left(\dfrac{16}{33}pi\right)\)
\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{8}\cdot sin\dfrac{8}{33}pi\cdot cos\left(\dfrac{8}{33}pi\right)\cdot cos\left(\dfrac{16}{33}pi\right)\)
\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{16}\cdot sin\dfrac{16}{33}pi\cdot cos\left(\dfrac{16}{33}pi\right)\)
\(=\dfrac{1}{sin\left(\dfrac{pi}{3}\right)}\cdot\dfrac{1}{32}\cdot sin\dfrac{32}{33}pi\)
=1/32
10:
\(=\dfrac{1}{2}\left[cos100+cos60\right]+\dfrac{1}{2}\cdot\left[cos100+cos20\right]\)
=cos100+1/2*cos20+1/4
6:
sin6*cos12*cos24*cos48
=1/cos6*cos6*sin6*cos12*cos24*cos48
=1/cos6*1/2*sin12*cos12*cos24*cos48
=1/cos6*1/4*sin24*cos24*cos48
=1/cos6*1/8*sin48*cos48
=1/cos6*1/16*sin96
=1/16
Bài 10:
a: \(\overrightarrow{AB}+\overrightarrow{BO}+\overrightarrow{OA}\)
\(=\overrightarrow{AO}+\overrightarrow{OA}=\overrightarrow{0}\)
b: \(\overrightarrow{OA}+\overrightarrow{BC}+\overrightarrow{DO}+\overrightarrow{CD}\)
\(=\overrightarrow{OA}+\overrightarrow{DO}+\overrightarrow{BD}\)
\(=\overrightarrow{OA}+\overrightarrow{BO}=\overrightarrow{BA}\)
4:
a: -90<a<0
=>cos a>0
cos^2a=1-(-4/5)^2=9/25
=>cosa=3/5
\(sin\left(45-a\right)=sin45\cdot cosa-cos45\cdot sina=\dfrac{\sqrt{2}}{2}\left(cosa-sina\right)\)
\(=\dfrac{\sqrt{2}}{2}\left(\dfrac{3}{5}-\dfrac{4}{5}\right)=\dfrac{-\sqrt{2}}{10}\)
b: pi/2<a<pi
=>cosa<0
cos^2a+sin^2a=0
=>cos^2a=16/25
=>cosa=-4/5
tan a=3/5:(-4/5)=-3/4
\(tan\left(a+\dfrac{pi}{3}\right)=\dfrac{tana+\dfrac{tanpi}{3}}{1-tana\cdot tan\left(\dfrac{pi}{3}\right)}\)
\(=\dfrac{-\dfrac{3}{4}+\sqrt{3}}{1-\dfrac{-3}{4}\cdot\sqrt{3}}=\dfrac{48-25\sqrt{3}}{11}\)
c: 3/2pi<a<pi
=>cosa>0
cos^2a+sin^2a=1
=>cos^2a=25/169
=>cosa=5/13
cos(pi/3-a)
\(=cos\left(\dfrac{pi}{3}\right)\cdot cosa+sin\left(\dfrac{pi}{3}\right)\cdot sina\)
\(=\dfrac{5}{13}\cdot\dfrac{1}{2}+\dfrac{-12}{13}\cdot\dfrac{\sqrt{3}}{2}=\dfrac{5-12\sqrt{3}}{26}\)
4b.
\(\dfrac{\pi}{2}< a< \pi\Rightarrow cosa< 0\Rightarrow cosa=-\sqrt{1-sin^2a}=-\dfrac{4}{5}\)
\(\Rightarrow tana=\dfrac{sina}{cosa}=-\dfrac{3}{4}\)
\(tan\left(a+\dfrac{\pi}{3}\right)=\dfrac{tana+tan\left(\dfrac{\pi}{3}\right)}{1-tana.tan\left(\dfrac{\pi}{3}\right)}=\dfrac{-\dfrac{3}{4}+\sqrt{3}}{1-\left(-\dfrac{3}{4}\right).\sqrt{3}}=...\)
c.
\(\dfrac{3\pi}{2}< a< 2\pi\Rightarrow cosa>0\Rightarrow cosa=\sqrt{1-sin^2a}=\dfrac{5}{13}\)
\(cos\left(\dfrac{\pi}{3}-a\right)=cos\left(\dfrac{\pi}{3}\right).cosa+sin\left(\dfrac{\pi}{3}\right).sina=\dfrac{1}{2}.\dfrac{5}{13}+\left(-\dfrac{12}{13}\right).\dfrac{\sqrt{3}}{2}=...\)
1: vecto AC=(-1;-7)
=>VTPT là (-7;1)
PTTS là:
x=3-t và y=6-7t
Phương trình AC là:
-7(x-3)+1(y-6)=0
=>-7x+21+y-6=0
=>-7x+y+15=0
2: Tọa độ M là:
x=(3+2)/2=2,5 và y=(6-1)/2=2,5
PTTQ đường trung trực của AC là:
-7(x-2,5)+1(y-2,5)=0
=>-7x+17,5+y-2,5=0
=>-7x+y+15=0
3: \(AB=\sqrt{\left(-1-3\right)^2+\left(3-6\right)^2}=5\)
Phương trình (A) là:
(x-3)^2+(y-6)^2=AB^2=25