Câu 21: Ý A

Pt \(\Leftrightarrow tanx=\sqrt{3}\Leftrightarrow x=\dfrac{\pi}{3}+k\pi\left(k\in Z\right)\)

Ý C

\(T=\dfrac{2\pi}{\left|2\right|}=\pi\)

Câu 19:  Ý A

Pt \(\Leftrightarrow sinx=\dfrac{\sqrt{3}}{2}\)

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

Các nghiệm thuộc \(\left[0;2\pi\right]\) là: \(\dfrac{\pi}{3};\dfrac{2\pi}{3}\)

Tổng các nghiệm là:\(\dfrac{\pi}{3}+\dfrac{2\pi}{3}=\pi\)

1.a) Để căn thức có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x^2}{2x-1}\ge0\\2x-1\ne0\end{matrix}\right.\)

\(\Leftrightarrow2x-1>0\Leftrightarrow x>\dfrac{1}{2}\)

Vậy...

b, \(\dfrac{\sqrt[3]{625}}{\sqrt[3]{5}}-\sqrt[3]{-216}.\sqrt[3]{\dfrac{1}{27}}=\sqrt[3]{\dfrac{625}{5}}-\sqrt[3]{-\dfrac{216}{27}}=\sqrt[3]{125}-\sqrt[3]{-8}=5-\left(-2\right)=7\)

\(\sqrt{36x^4\left(b-2\right)^2}=6x^2\left|b-2\right|=6x^2\left(2-b\right)\) (vì b<2 nên b-2<0)

a)Pt\(\Leftrightarrow\left|x+1\right|=3\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

b)Đk:\(x\ge-1\)

Sửa đề: \(3\sqrt{4x+4}-\sqrt{9x+9}-8\sqrt{\dfrac{x+1}{16}}=5\)

Pt \(\Leftrightarrow6\sqrt{x+1}-3\sqrt{x+1}-2\sqrt{x+1}=5\)

\(\Leftrightarrow\sqrt{x+1}=5\)

\(\Leftrightarrow x=24\left(tm\right)\)

Câu 18: Ý B

Để pt vô nghiệm \(\Leftrightarrow m^2+\left(\sqrt{3}\right)^2< \left(m+1\right)^2\)

\(\Leftrightarrow3< 2m+1\Leftrightarrow m>1\)

Câu 16: Ý D

Pt\(\Leftrightarrow2sin\left(x+\dfrac{\pi}{6}\right)=0\)

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

Tại \(k=0\Rightarrow x=-\dfrac{\pi}{6}\)

Tại \(k=-1\Rightarrow x=-\dfrac{7\pi}{6}\)

Câu 15: Ý B

\(cos^2x-cosx=0\)\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=k2\pi\end{matrix}\right.\)(\(k\in Z\))

Do \(0< x< \pi\Rightarrow\left[{}\begin{matrix}0< \dfrac{\pi}{2}+k\pi< \pi\\0< k2\pi< \pi\end{matrix}\right.\)(\(k\in Z\))\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{1}{2}< k< \dfrac{1}{2}\\0< k< \dfrac{1}{2}\end{matrix}\right.\)(\(k\in Z\))

\(\Rightarrow\left[{}\begin{matrix}k=0\\k\in\varnothing\end{matrix}\right.\) \(\Rightarrow x=\dfrac{\pi}{2}+0.\pi=\dfrac{\pi}{2}\)