ĐKXĐ:

a. \(cos\left(x-\dfrac{2\pi}{3}\right)\ne0\Rightarrow x-\dfrac{2\pi}{3}\ne\dfrac{\pi}{2}+k\pi\Rightarrow x\ne\dfrac{\pi}{6}+k\pi\)

b. \(sin\left(x+\dfrac{\pi}{6}\right)\ne0\Rightarrow x+\dfrac{\pi}{6}\ne k\pi\Rightarrow x\ne-\dfrac{\pi}{6}+k\pi\)

c. \(\dfrac{1+x}{2-x}\ge0\Rightarrow-1\le x< 2\)

a) \(y=\dfrac{4}{sin^22x-1}\)

Xác định khi và chỉ khi

\(sin^22x-1\ne0\)

\(\Leftrightarrow sin^22x\ne1\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x\ne1\\sin2x\ne-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x\ne sin\dfrac{\pi}{2}\\sin2x\ne sin\dfrac{3\pi}{2}\end{matrix}\right.\)

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

Vậy tập xác định là \(D=R\)\\(\left\{\pm\dfrac{\pi}{4}+k\pi\right\}\)

2:

a: \(y=4+\left(cos^2x-sin^2x\right)+\left(cos^2x+sin^2x\right)\)

\(=4+1+cos2x=cos2x+5\)

-1<=cos2x<=1

=>-1+5<=cos2x+5<=1+5

=>4<=cos2x+5<=6

TGT là T=[4;6]

b: \(y=5-\dfrac{3}{2}\cdot2sinx\cdot cosx=-\dfrac{3}{2}sin2x+5\)

-1<=sin 2x<=1

=>-3/2<=-3/2sin2x<=3/2

=>-3/2+5<=y<=3/2+5

=>7/2<=y<=13/2

=>TGT là T=[7/2;13/2]

c: -1<=sin x<=1

=>-2<=-2sin x<=2

=>3<=-2sinx+5<=7

=>\(\dfrac{4}{3}>=\dfrac{4}{-2sinx+5}>=\dfrac{4}{7}\)

TGT là T=[4/7;4/3]

Đặt \(sinx=t\Rightarrow t\in\left[-\dfrac{1}{2};1\right]\)

\(y=f\left(t\right)=2t^2+t+4\)

Xét hàm \(f\left(t\right)=2t^2+t+4\) trên \(\left[-\dfrac{1}{2};1\right]\)

\(-\dfrac{b}{2a}=-\dfrac{1}{4}\in\left[-\dfrac{1}{2};1\right]\)

\(f\left(-\dfrac{1}{2}\right)=4\) ; \(f\left(-\dfrac{1}{4}\right)=\dfrac{31}{8}\); \(f\left(1\right)=7\)

\(y_{max}=7\) khi \(t=1\) hay \(x=\dfrac{\pi}{2}\)

\(y_{min}=\dfrac{31}{8}\) khi \(sinx=-\dfrac{1}{4}\)

Hàm số y 1 = sin π 2 − x  có chu kì  T 1 = 2 π − 1 = 2 π

Hàm số y 2 = cot x 3  có chu kì  T 2 = π 1 3 = 3 π

Suy ra hàm số đã cho y = y 1 + y 2  có chu kì T = B C N N 2 , 3 π = 6 π .

Vậy đáp án là D.

\(sin\left(\dfrac{\pi}{3}+x\right)\in\left[-1;1\right]\)

\(\Rightarrow y=\dfrac{3}{2}+sin\left(\dfrac{\pi}{3}+x\right)\in\left[\dfrac{1}{2};\dfrac{5}{2}\right]\)

\(\Rightarrow\left\{{}\begin{matrix}y_{min}=\dfrac{1}{2}\\y_{max}=\dfrac{5}{2}\end{matrix}\right.\)