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

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.

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

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

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

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

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

34 và 35 cách làm giống nhau, mình làm câu 34, câu 35 bạn tự làm

Gọi G là giao điểm BM và CN \(\Rightarrow\) G là trọng tâm tam giác

Tọa độ G thỏa mãn: \(\left\{{}\begin{matrix}2x-y+1=0\\x+y-4=0\end{matrix}\right.\) \(\Rightarrow G\left(1;3\right)\) \(\Rightarrow\overrightarrow{AG}=\left(3;0\right)\)

Gọi P là trung điểm BC \(\Rightarrow\overrightarrow{AG}=\dfrac{2}{3}\overrightarrow{AP}\) \(\Rightarrow P\left(\dfrac{5}{2};3\right)\)

B thuộc BM nên B có dạng: \(B\left(b;2b+1\right)\Rightarrow C\left(5-b;5-2b\right)\)

Do C thuộc CN \(\Rightarrow5-b+5-2b-4=0\Rightarrow b=2\) \(\Rightarrow B\left(2;5\right)\)

\(\Rightarrow\overrightarrow{PB}=\left(-\dfrac{1}{2};2\right)=\dfrac{1}{2}\left(-1;4\right)\)

Đường thẳng BC nhận \(\left(4;1\right)\) là 1 vtpt

Phương trình:

\(4\left(x-2\right)+1\left(y-5\right)=0\Leftrightarrow4x+y-13=0\)

1.

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

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

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

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

\(\Leftrightarrow\left(\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx\right).\left(cos^2x-sin^2x\right)=0\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right).cos2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{3}\right)=0\\cos2x=0\end{matrix}\right.\)

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

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

2.

\(sin3x-\sqrt{3}cos3x+2=4cos^2x\)

\(\Leftrightarrow4cos^2x-2+\sqrt{3}cos3x-sin3x=0\)

\(\Leftrightarrow2cos^2x-1+\dfrac{\sqrt{3}}{2}cos3x-\dfrac{1}{2}sin3x=0\)

\(\Leftrightarrow cos2x+cos\left(3x+\dfrac{\pi}{6}\right)=0\)

\(\Leftrightarrow2cos\left(\dfrac{5x}{2}+\dfrac{\pi}{12}\right).cos\left(\dfrac{x}{2}+\dfrac{\pi}{12}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(\dfrac{5x}{2}+\dfrac{\pi}{12}\right)=0\\cos\left(\dfrac{x}{2}+\dfrac{\pi}{12}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5x}{2}+\dfrac{\pi}{12}=\dfrac{\pi}{2}+k\pi\\\dfrac{x}{2}+\dfrac{\pi}{12}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

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

NP là giao tuyến của \(\left(MNP\right)\) với \(\left(ABCD\right)\).

NP cắt AB, AD lần lượt tại E, F.

ME cắt SB tại H \(\Rightarrow MH\) là giao tuyến của \(\left(MNP\right)\) với \(\left(SAB\right)\).

Kẻ HN \(\Rightarrow HN\) là giao tuyến của \(\left(MNP\right)\) với \(\left(SBC\right)\).

MF cắt SD tại T \(\Rightarrow MT\) là giao tuyến của \(\left(MNP\right)\) với \(\left(SAD\right)\).

Kẻ PT \(\Rightarrow PT\) là giao tuyến của \(\left(MNP\right)\) với \(\left(SDC\right)\)