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 :)

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.

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

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

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

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

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

8.

\(sin4x+\sqrt{3}cos4x=2sinx\)

\(\Leftrightarrow\dfrac{1}{2}sin4x+\dfrac{\sqrt{3}}{2}cos4x=sinx\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{9}+\dfrac{k2\pi}{3}\\x=\dfrac{2\pi}{15}+\dfrac{k2\pi}{5}\end{matrix}\right.\)

17. coming 18. iron___ doing 19. wakes 20. buying

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17. to come    18.iron , doing.  19.wake.  

Đề ko rõ ràng \(\sqrt{x^2}+x+\dfrac{1}{4}\) hay \(\sqrt{x^2+x+\dfrac{1}{4}}\)??

m??

Có 2 nghiệm 

Đặt B=0

=>x^2-9=0

=>x^2=9

=>x=3 hoặc x=-3

`B=x^2-9=0`

`-> x^2=0+9`

`-> x^2=9`

`-> x^2=(+-3)^2`

`-> x=+-3`

Vậy, đa thức `B` có `2` nghiệm là `x={3 ; -3}`.

có [x-y]2=1

suy ra [x-y]mũ 2= 1 mũ 2

suy ra x-1=1

x=1+1

x=2

x = 2 nha bạn

đáp án :
(x-1)²= 1
(x-1)² = 1²

x-1=1
x= 1+1
x=2 
vậy x = 2