12.

\(y=\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)\le\sqrt[]{2}\)

\(\Rightarrow M=\sqrt{2}\)

13.

Pt có nghiệm khi:

\(5^2+m^2\ge\left(m+1\right)^2\)

\(\Leftrightarrow2m\le24\)

\(\Rightarrow m\le12\)

14.

\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=-\dfrac{5}{3}\left(loại\right)\end{matrix}\right.\)

\(\Leftrightarrow x=k2\pi\)

15.

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=arctan\left(3\right)+k\pi\end{matrix}\right.\)

Đáp án A

16.

\(\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx=\dfrac{1}{2}\)

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

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

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

\(\left[{}\begin{matrix}2\pi\le\dfrac{\pi}{3}+k2\pi\le2018\pi\\2\pi\le\pi+k2\pi\le2018\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}1\le k\le1008\\1\le k\le1008\end{matrix}\right.\)

Có \(1008+1008=2016\) nghiệm

2.

\(\Leftrightarrow cos2x-cos8x-sin3x+cos5x-2sin5x.cos5x=0\)

\(\Leftrightarrow2sin5x.sin3x-sin3x+cos5x-2sin5x.cos5x=0\)

\(\Leftrightarrow sin3x\left(2sin5x-1\right)-cos5x\left(2sin5x-1\right)=0\)

\(\Leftrightarrow\left(sin3x-cos5x\right)\left(2sin5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos5x=sin3x=cos\left(\dfrac{\pi}{2}-3x\right)\\sin5x=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{\pi}{2}-3x+k2\pi\\5x=3x-\dfrac{\pi}{2}+k2\pi\\5x=\dfrac{\pi}{6}+k2\pi\\5x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+\dfrac{k\pi}{4}\\x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{30}+\dfrac{k2\pi}{5}\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{5}\end{matrix}\right.\)

3.

\(\Leftrightarrow1+sinx=cosx-cos3x+2sinx.cosx+1-2sin^2x\)

\(\Leftrightarrow sinx=2sin2x.sinx+2sinx.cosx-2sin^2x\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\1=2sin2x+2cosx-2sinx\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow4sinx.cosx+2cosx-2sinx-1=0\)

\(\Leftrightarrow2cosx\left(2sinx+1\right)-\left(2sinx+1\right)=0\)

\(\Leftrightarrow\left(2cosx+1\right)\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\cosx=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)

17.

Gọi số vi khuẩn ban đầu là x

Sau 5 phút số vi khuẩn là: \(x.2^5=64000\Rightarrow x=2000\)

Sau k phút:

\(2000.2^k=2048000\Rightarrow2^k=1024=2^{10}\)

\(\Rightarrow k=10\)

18.

\(S_{2019}=\left(\dfrac{1}{2}\right)^1+1+\left(\dfrac{1}{2}\right)^2+1+...+\left(\dfrac{1}{2}\right)^{2019}+1\)

\(=\left(\dfrac{1}{2}\right)^1+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{2019}+2019\)

Xét \(S=\left(\dfrac{1}{2}\right)^1+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{2019}\) là tổng cấp số nhân với \(\left\{{}\begin{matrix}u_1=\dfrac{1}{2}\\q=\dfrac{1}{2}\\n=2019\end{matrix}\right.\)

\(\Rightarrow S=\dfrac{1}{2}.\dfrac{\left(\dfrac{1}{2}\right)^{2019}-1}{\dfrac{1}{2}-1}=1-\dfrac{1}{2^{2019}}\)

\(\Rightarrow S_{2020}=2019+S=2020-\dfrac{1}{2^{2019}}\)

19. C là khẳng định sai, ví dụ: \(u_n=2\) ; \(v_n=-\dfrac{1}{n}\)

Câu nào bạn, nếu mà cả thì đăng tách ra đi :)

Ok bạn =))

1.

\(sin^2x-4sinx.cosx+3cos^2x=0\)

\(\Rightarrow\dfrac{sin^2x}{cos^2x}-\dfrac{4sinx}{cosx}+\dfrac{3cos^2x}{cos^2x}=0\)

\(\Rightarrow tan^2x-4tanx+3=0\)

2.

\(\Leftrightarrow\dfrac{1}{2}cos2x+\dfrac{\sqrt{3}}{2}sin2x=\dfrac{1}{2}\)

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

3.

\(\Leftrightarrow2^2+m^2\ge1\)

\(\Leftrightarrow m^2\ge-3\) (luôn đúng)

Pt có nghiệm với mọi m (đề bài sai)

4.

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=1\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=1\)

\(\Leftrightarrow x-\dfrac{\pi}{3}=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\dfrac{5\pi}{6}+k2\pi\)

6.

ĐKXĐ: \(cosx\ne0\)

Nhân 2 vế với \(cos^2x\)

\(sin^2x-4cosx+5cos^2x=0\)

\(\Leftrightarrow1-cos^2x-4cosx+5cos^2x=0\)

\(\Leftrightarrow\left(2cosx-1\right)^2=0\)

\(\Leftrightarrow cosx=\dfrac{1}{2}\Rightarrow x=\pm\dfrac{\pi}{3}+k2\pi\)

6.

\(cos^2x+\sqrt{3}sinx.cosx-1=0\)

\(\Leftrightarrow-sin^2x+\sqrt{3}sinx.cosx=0\)

\(\Leftrightarrow sinx\left(sinx-\sqrt{3}cosx\right)=0\)

\(\Leftrightarrow sinx\left(\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx\right)=0\)

\(\Leftrightarrow sinx.sin\left(x-\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sin\left(x-\dfrac{\pi}{3}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)

7.

\(\sqrt{3}sinx-cosx=2\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx=1\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=1\)

\(\Leftrightarrow x-\dfrac{\pi}{3}=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\dfrac{5\pi}{6}+k2\pi\)

Đặt \(cosx-sinx=t\Rightarrow-\sqrt{2}\le t\le\sqrt{2}\)

\(t^2=1-2sinx.cosx\Rightarrow sinx.cosx=\dfrac{1-t^2}{2}\)

Pt trở thành:

\(t\left(1+\dfrac{1-t^2}{2}\right)+1=0\)

\(\Leftrightarrow t^3-3t-2=0\)

\(\Leftrightarrow\left(t-2\right)\left(t+1\right)^2=0\Rightarrow\left[{}\begin{matrix}t=2\left(loại\right)\\t=-1\end{matrix}\right.\)

\(\Rightarrow cosx-sinx=-1\)

\(\Leftrightarrow\sqrt[]{2}cos\left(x+\dfrac{\pi}{4}\right)=-1\)

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{4}\right)=cos\left(\dfrac{3\pi}{4}\right)\)

\(\Leftrightarrow...\)

Dạ em cảm ơn ạ!! ^^

b.

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cos2x-\dfrac{1}{2}sin2x=-cosx\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{6}+k2\pi\\x=-\dfrac{7\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

c.

\(\Leftrightarrow2cos4x.sin3x=2sin4x.cos4x\)

\(\Leftrightarrow cos4x\left(sin4x-sin3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\sin4x=sin3x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{\pi}{2}+k\pi\\4x=3x+k2\pi\\4x=\pi-3x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\\x=k2\pi\\x=\dfrac{\pi}{7}+\dfrac{k2\pi}{7}\end{matrix}\right.\)

2.

\(f\left(x\right)=\dfrac{1}{2}-\dfrac{1}{2}cos2x-\dfrac{\sqrt{3}}{2}sin2x-5\)

\(=-\dfrac{9}{2}-\left(\dfrac{1}{2}cos2x+\dfrac{\sqrt{3}}{2}sin2x\right)\)

\(=-\dfrac{9}{2}-cos\left(2x-\dfrac{\pi}{3}\right)\)

Do \(-1\le-cos\left(2x-\dfrac{\pi}{3}\right)\le1\Rightarrow-\dfrac{11}{2}\le y\le-\dfrac{7}{2}\)

\(y_{min}=-\dfrac{11}{2}\) khi \(cos\left(2x-\dfrac{\pi}{3}\right)=1\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\)

\(y_{max}=-\dfrac{7}{2}\) khi \(cos\left(2x-\dfrac{\pi}{3}\right)=-1\Rightarrow x=\dfrac{2\pi}{3}+k\pi\)