4: BH*AC=AB*CK

=>BH/CK=AB/AC=5/6

\(a>0\Leftrightarrow a< a+1\Leftrightarrow\dfrac{1}{a}>\dfrac{1}{1+a}\)

Chọn A

8: D

11.

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

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\\sinx-3cosx=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=cosx\\sinx=3cosx\end{matrix}\right.\)

Với \(cosx=0\Rightarrow\) pt vô nghiệm

Với \(cosx\ne0\)

\(pt\Leftrightarrow\left[{}\begin{matrix}tanx=0\\tanx=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=arctan3+k\pi\end{matrix}\right.\)

12.

\(pt\Leftrightarrow\sqrt{3}tanx+1=0\)

\(\Leftrightarrow tanx=-\dfrac{\sqrt{3}}{3}\)

\(\Leftrightarrow x=-\dfrac{\pi}{6}+k\pi\)

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

17.

\(2tan^2x+5tanx+3=0\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=-\dfrac{3}{2}\end{matrix}\right.\)

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

Nghiệm âm lớn nhất là \(x=-\dfrac{\pi}{4}\)

18.

Pt vô nghiệm khi:

\(m^2+4^2< 6^2\)

\(\Leftrightarrow m^2< 20\)

\(\Rightarrow-2\sqrt{5}< m< 2\sqrt{5}\)

\(ab=20\)

19.

Pt có nghiệm khi:

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

\(\Leftrightarrow3m^2-4m-3\le0\)

Theo Viet: \(\left\{{}\begin{matrix}a+b=\dfrac{4}{3}\\ab=-1\end{matrix}\right.\)

\(\Rightarrow a^2+b^2=\left(a+b\right)^2-2ab=\dfrac{34}{9}\)

20.

\(cos\left(2x-60^0\right)=sin\left(x+60^0\right)=cos\left(30^0-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-60^0=30^0-x+k360^0\\2x-60^0=x-30^0+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=30^0+k120^0\\x=30^0+k360^0\end{matrix}\right.\) \(\Leftrightarrow x=30^0+k120^0\)

21.

Đáp án C, kiểm tra điều kiện có nghiệm \(a^2+b^2\ge c^2\) thì chỉ có \(1^2+\left(-1\right)^2>1^2\) thỏa mãn

22.

\(cos^2x-cosx=0\Rightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)

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

\(\Rightarrow x=\dfrac{\pi}{2}\)

23.

\(2sin^2x+5sinx-3=0\Rightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx=-3\left(loại\right)\end{matrix}\right.\)

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

Nghiệm dương bé nhất là \(x=\dfrac{\pi}{6}\)

24.

\(1-cos^2x-3cosx-4=0\)

\(\Leftrightarrow cos^2x+3cosx+3=0\)

Pt bậc 2 nói trên vô nghiệm nên pt đã cho vô nghiệm

25.

\(\Leftrightarrow\left(tanx+1\right)^2=0\)

\(\Leftrightarrow tanx=-1\)

\(\Rightarrow x=-\dfrac{\pi}{4}+k\pi\)

26.

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

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

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

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

28.

\(\Leftrightarrow sinx-\left(1-2sin^2x\right)-2=0\)

\(\Leftrightarrow2sin^2x+sinx-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\dfrac{3}{2}\left(loại\right)\end{matrix}\right.\)

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