\(cos\dfrac{\pi}{7}-cos\dfrac{2\pi}{7}+cos\dfrac{3\pi}{7}\)

\(=-2.sin\dfrac{3\pi}{14}.sin\dfrac{-\pi}{14}+cos\left(\dfrac{\pi}{2}-\dfrac{\pi}{14}\right)\)

\(=2.sin\dfrac{3\pi}{14}.sin\dfrac{\pi}{14}+sin\dfrac{\pi}{14}\)

\(=sin\dfrac{\pi}{14}\left(1+2sin\dfrac{3\pi}{14}\right)\)\(=sin\dfrac{\pi}{14}\left(1+2.sin\left(\dfrac{\pi}{2}-\dfrac{2\pi}{7}\right)\right)\)\(=sin\dfrac{\pi}{14}\left(1+2.cos\dfrac{2\pi}{7}\right)\)

\(=sin\dfrac{\pi}{14}\left(2cos^2\dfrac{\pi}{7}+cos\dfrac{2\pi}{7}\right)\)\(=\dfrac{sin\dfrac{\pi}{7}}{2.cos\dfrac{\pi}{14}}\left(2.cos^2\dfrac{\pi}{7}+cos\dfrac{2\pi}{7}\right)\)

\(=\dfrac{1}{2}.\dfrac{1}{cos\dfrac{\pi}{14}}\left(2.sin\dfrac{\pi}{7}.cos^2\dfrac{\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{2\pi}{7}\right)\)

\(=\dfrac{1}{2}.\dfrac{1}{cos\dfrac{\pi}{14}}\left(sin\dfrac{2\pi}{7}.cos\dfrac{\pi}{7}+cos\dfrac{2\pi}{7}.sin\dfrac{\pi}{2}\right)\)

\(=\dfrac{1}{2}.\dfrac{1}{cos\dfrac{\pi}{14}}.sin\dfrac{3\pi}{7}\)\(=\dfrac{1}{2}.\dfrac{1}{cos\dfrac{\pi}{14}}.cos\dfrac{\pi}{14}\)

\(=\dfrac{1}{2}\)(đpcm)

C2:B

C3:D

C4:B

C5:B

C6:D

C7:A

C8:B

\(P=-3x^2-4x\sqrt{y}+16x--y+12\sqrt{y}+1999\)

\(=-2\left(x^2+2x\sqrt{y}+y\right)+12\left(x+\sqrt{y}\right)-18-x^2+4x-4+2021\)

\(=-2\left(x+\sqrt{y}\right)^2+12\left(x+\sqrt{y}\right)-18-\left(x-2\right)^2+2021\)

\(=-2\left(x+\sqrt{y}-3\right)^2-\left(x-2\right)^2+2021\)\(\le2021\) với mọi x và y không âm

Dấu = xảy ra <=> x=2 và y=1

Vậy maxP=2021 

\(2sin\left(x+y\right)=sinx+siny\)

\(\Leftrightarrow2.2.sin\dfrac{x+y}{2}.cos\dfrac{x+y}{2}=2.sin\dfrac{x+y}{2}.cos\dfrac{x-y}{2}\)

\(\Leftrightarrow2cos\dfrac{x+y}{2}=cos\dfrac{x-y}{2}\)

\(\Leftrightarrow2\left(cos\dfrac{x}{2}.cos\dfrac{y}{2}-sin\dfrac{x}{2}.sin\dfrac{y}{2}\right)=cos\dfrac{x}{2}.cos\dfrac{y}{2}+sin\dfrac{x}{2}.sin\dfrac{y}{2}\)

\(\Leftrightarrow cos\dfrac{x}{2}.cos\dfrac{y}{2}=3.sin\dfrac{x}{2}.sin\dfrac{y}{2}\)

\(\Leftrightarrow\left(sin\dfrac{x}{2}:cos\dfrac{x}{2}\right).\left(sin\dfrac{y}{2}:cos\dfrac{y}{2}\right)=\dfrac{1}{3}\)

\(\Leftrightarrow tan\dfrac{x}{2}.tan\dfrac{y}{2}=\dfrac{1}{3}\)

Đường tròn (C): \(\left\{{}\begin{matrix}I\left(-1;1\right)\\R=5\end{matrix}\right.\)

Gọi \(\Delta:ax+by+c=0\) (\(a;b\ne0\))

\(M\in\Delta\Rightarrow9a-4b+c=0\)

\(\Leftrightarrow c=4b-9a\)

Do \(\Delta\) là tiếp tuyến của (C) \(\Rightarrow d_{\left(I,\Delta\right)}=R\)

\(\Leftrightarrow\dfrac{\left|-a+b+c\right|}{\sqrt{a^2+b^2}}=5\) \(\Leftrightarrow\dfrac{\left|-a+b+4b-9a\right|}{\sqrt{a^2+b^2}}=5\)

\(\Leftrightarrow\left(5b-10a\right)^2=25\left(a^2+b^2\right)\)

\(\Leftrightarrow-3a^2+4ab=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3a=4b\\a=0\left(L\right)\end{matrix}\right.\)

3a=4b

Chọn a=4;b=3 =>c=-24

=>\(\Delta:4x+3y-24=0\)

\(d_{\left(N;\Delta\right)}=\dfrac{\left|4.2020+3.2021-24\right|}{\sqrt{4^2+3^2}}=2823,8\)

\(2sinx+cosx=0\)

\(\Leftrightarrow\dfrac{2}{\sqrt{5}}.sinx+\dfrac{1}{\sqrt{5}}cosx=0\)

Đặt \(cos\alpha=\dfrac{2}{\sqrt{5}}\) và \(sin\alpha=\dfrac{1}{\sqrt{5}}\) (vì \(\left(\dfrac{2}{\sqrt{5}}\right)^2+\left(\dfrac{1}{\sqrt{5}}\right)^2=1\))

pttt: \(sinx.cos\alpha+cosx.sin\alpha=0\)

\(\Leftrightarrow sin\left(x+\alpha\right)=0\)

\(\Rightarrow x=-arc.sin\alpha+k\pi\left(k\in Z\right)\)
(phải không?)

Gọi mẫu số của phân số đó là \(a\) (a>21)

=> Tử số của phân số đó là: \(a-21\)

Vì 3 lần mẫu số bằng 4 lần tử số:

\(\Rightarrow3a=4\left(a-21\right)\) \(\Leftrightarrow a=84\left(tm\right)\)

=> Tử số là: 84-21=63

Vậy phân số đó là \(\dfrac{63}{84}\)

\(\left(P\right):y=-x^2\) 

Gọi pt đường thẳng (d): y=x.m+b

Có \(M\in\left(d\right)\Rightarrow-2=-m+b\)

\(\Leftrightarrow b=-2+m\)

Pt (d) trở thành: \(y=x.m+m-2\)

Xét pt hoành độ gđ của (P) và (d) có:

\(-x^2=x.m+m-2\)  (*)

\(\Leftrightarrow-x^2-m.x-m+2=0\)

\(\Delta=m^2-4\left(-1\right)\left(-m+2\right)\)

\(=m^2-4m+8=\left(m-2\right)^2+4>0\forall m\)

=> Pt (*) luôn có hai nghiệm pb hay (d) luôn cắt (P) tại hai điểm pb với mọi m

Đk:\(a>0\)

\(\left|\dfrac{1}{2\sqrt{a}}\right|\le\dfrac{1}{4}\)\(\Leftrightarrow\dfrac{1}{2\sqrt{a}}-\dfrac{1}{4}\le0\) \(\Leftrightarrow\dfrac{2-\sqrt{a}}{4\sqrt{a}}\le0\)\(\Leftrightarrow2-\sqrt{a}\le0\)(vì \(4\sqrt{a}>0\forall a>0\))

\(\Leftrightarrow2\le\sqrt{a}\) \(\Leftrightarrow4\le a\)

Vậy a\(\ge4\)

\(\dfrac{4x}{\sqrt{x}-3}=-1\)( Đk:\(x\ge0;x\ne9\))
\(\Rightarrow4x=-\sqrt{x}+3\)

\(\Leftrightarrow4x-3\sqrt{x}+4\sqrt{x}-3=0\)

\(\Leftrightarrow\left(4\sqrt{x}-3\right)\left(\sqrt{x}+1\right)=0\)

\(\Rightarrow4\sqrt{x}-3=0\) (vì \(\sqrt{x}+1\)>0 mọi x)

\(\Leftrightarrow x=\dfrac{9}{16}\) (thỏa mãn)