bài 1:

a)\(\sin x=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\pi-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.k\in Z}\)

b)\(\cos x=\dfrac{\sqrt{2}}{2}\Leftrightarrow x=\pm\dfrac{\pi}{4}+k2\pi\left(k\in Z\right)\)

Bài 2:

a)\(\sqrt{3}\cos3x-\sin3x=-1\)

\(\Leftrightarrow\dfrac{1}{2}\left(\sqrt{3}\cos3x-\sin3x\right)=-\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}\cos3x-\dfrac{1}{2}\sin3x=\dfrac{-1}{2}\)

\(\Leftrightarrow\cos\dfrac{\pi}{6}\cos3x-\sin\dfrac{\pi}{6}\sin3x=\dfrac{-1}{2}\)

\(\Leftrightarrow\cos\left(\dfrac{\pi}{6}+3x\right)=\dfrac{-1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{6}+x=\dfrac{2\pi}{3}+k2\pi\\\dfrac{\pi}{6}+x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=-\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.k\in Z}\)

b)\(2\sin x+\cos x=1\)

\(\Leftrightarrow\dfrac{2}{\sqrt{2^2+1^2}}\cos x+\dfrac{1}{\sqrt{2^2+1^2}}\sin x=\dfrac{1}{\sqrt{2^2+1^2}}\)

\(\Leftrightarrow\dfrac{2}{\sqrt{5}}\cos x+\dfrac{1}{\sqrt{5}}\sin x=\dfrac{1}{\sqrt{5}}\)(*)

Đặt \(\dfrac{2}{\sqrt{5}}=\cos\alpha,\dfrac{1}{\sqrt{5}}=\sin\alpha\)

Từ phương trình (*) ta có:

\(\cos\alpha\cos x+\sin\alpha\sin x=\sin\alpha\)

\(\Leftrightarrow\cos\left(\alpha-x\right)=\sin\alpha\)

\(\Leftrightarrow\cos\left(\alpha-x\right)=\cos\left(\dfrac{\pi}{2}-\alpha\right)\)

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

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

Bài 3:

(C) \(\left(x+2\right)^2+\left(y-3\right)^2=16\)

\(\Leftrightarrow\left\{{}\begin{matrix}R=4\\I\left(-2,3\right)\end{matrix}\right.\)

\(V_{\left(A,\dfrac{3}{2}\right)}C\rightarrow C'\Rightarrow\left\{{}\begin{matrix}R'=\dfrac{3}{2}R=6\\V_{\left(A,\dfrac{3}{2}\right)}I\rightarrow I'\circledast\end{matrix}\right.\)

Từ \(\circledast\Rightarrow I'\left(-\dfrac{7}{2},\dfrac{7}{2}\right)\)là tâm của đường tròn C'

Vậy phương trình đường tròn (C') cần tìm là:

\(\left(x+\dfrac{7}{2}\right)^2+\left(y-\dfrac{7}{2}\right)=36\)

`(C)` có `I(1;-2)` và `R=\sqrt{5}`

Gọi `(C')` có tâm `I'` và bk `R'` là ảnh của `(C)` qua `V_{(O,-2)}`

`V_{(O,-2)} (I)=I'<=>\vec{OI'}=-2\vec{OI}=-2(1;-2)=(-2;4)`

           `=>I'(-2;4)`

`R'=|-2|.R=2sqrt{5}`

   `=>` Ptr đường tròn `(C')` là: `(x+2)^2 +(x-4)^2=20`

Cách sắp xếp học sinh nam có: C\(^2_6\) cách

Cách sắp xếp học sinh nữ có: C\(^3_6\) cách

Theo quy tắc nhân ta có: C\(^2_6\) . C\(^3_6\) = 300 cách sắp xếp

tích đúng 5 sao cho mình nhé. cảm ơn bạn

\(2\left(1-sin^23x\right)+cos3x+1=0\)

\(\Leftrightarrow2cos^23x+cos3x+1=0\)

\(\Leftrightarrow2\left(cos3x+\dfrac{1}{4}\right)^2+\dfrac{7}{8}=0\)

Pt đã cho vô nghiệm

Ta có: \(u_n>0\)

Mặt khác: 

\(u_n=\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{n+n}< \dfrac{1}{n}+\dfrac{1}{n}+...+\dfrac{1}{n}\\ \Leftrightarrow u_n< n.\dfrac{1}{n}=1\)

\(0< u_n< 1\) nên \(u_n\) bị chặn

Xét tính đơn điệu dãy \(u_n\)

\(u_{n+1}=\dfrac{1}{\left(n+1\right)+1}+\dfrac{1}{\left(n+1\right)+2}+...+\dfrac{1}{\left(n+1\right)+\left(n+1-1\right)}+\dfrac{1}{\left(n+1\right)+\left(n+1\right)}\\ =\dfrac{1}{n+2}+\dfrac{1}{n+3}+...+\dfrac{1}{2n+1}+\dfrac{1}{2n+2}\)

\(u_{n+1}-u_n=\dfrac{1}{2n+1}+\dfrac{1}{2n+2}-\dfrac{1}{n+1}\\ =\dfrac{1}{2n+1}+\dfrac{1}{2\left(n+1\right)}-\dfrac{1}{n+1}\\ =\dfrac{1}{2n+1}-\dfrac{1}{2\left(n+1\right)}\\ =\dfrac{2\left(n+1\right)-\left(2n+1\right)}{2\left(2n+1\right)\left(n+1\right)}=\dfrac{1}{2\left(2n+1\right)\left(n+1\right)}>0\)

\(\Rightarrow u_{n+1}>u_n\)

Dãy \(u_n\) là dãy tăng.

Vậy \(u_n\) bị chặn và tăng nghiêm ngặt nên \(u_n\) hội tụ. đpcm