1. Không dịch được đề

2.

\(-1\le cos2x\le1\Rightarrow1\le y\le3\)

3.

a. \(-2\le2sinx\le2\Rightarrow-1\le y\le3\)

\(y_{min}=-1\) khi \(sinx=-1\Rightarrow x=-\dfrac{\pi}{2}+k2\pi\)

\(y_{max}=3\) khi \(sinx=1\Rightarrow x=\dfrac{\pi}{2}+k2\pi\)

b.

\(0\le cos^2x\le1\Rightarrow-1\le y\le2\)

\(y_{min}=-1\) khi \(cos^2x=1\Rightarrow x=k\pi\)

\(y_{max}=2\) khi \(cosx=0\Rightarrow x=\dfrac{\pi}{2}+k\pi\)

4.

\(y=\left(tanx-1\right)^2+2\ge2\)

\(y_{min}=2\) khi \(tanx=1\Rightarrow x=\dfrac{\pi}{4}+k\pi\)

ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\sinx+tanx\ne0\end{matrix}\right.\) 

\(\Leftrightarrow\left\{{}\begin{matrix}cosx\ne0\\\dfrac{sinx\left(1+cosx\right)}{cosx}\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}cosx\ne0\\sinx\ne0\end{matrix}\right.\)

\(\Leftrightarrow sin2x\ne0\)

\(\Rightarrow2x\ne k\pi\Rightarrow x\ne\dfrac{k\pi}{2}\)

1.

Hàm số xác định khi \(\left\{{}\begin{matrix}\dfrac{1+x}{1-x}\ge0\\1-x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-1\le x< 1\\x\ne1\end{matrix}\right.\Leftrightarrow-1\le x< 1\)

2.

Hàm số xác định khi \(cosx+1\ne0\Leftrightarrow cosx\ne-1\Leftrightarrow x\ne-\pi+k2\pi\)

3.

Hàm số xác định khi \(cosx-cos3x\ne0\Leftrightarrow sin2x.sinx\ne0\Leftrightarrow\left[{}\begin{matrix}x\ne k\pi\\x\ne\dfrac{k\pi}{2}\end{matrix}\right.\)

\(\sqrt{3}sinx+cosx\ne0\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx\ne0\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{6}\right)\ne0\)

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

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

Hàm số \(y=3-\dfrac{cos2x}{tan2x}\) xác định khi \(\left\{{}\begin{matrix}sin2x\ne0\\cos2x\ne0\end{matrix}\right.\Leftrightarrow x\ne\dfrac{k\pi}{4}\).

\(Vì-1\le\cos2x\le1\)

\(\Rightarrow2\le3+\cos2x\le4\)

\(\Rightarrow\sqrt{2}\le\sqrt{3+\cos2x}\le\sqrt{4}\)

\(\Rightarrow\sqrt{2}\le\sqrt{3+\cos2x}\le2\)

\(\Rightarrow\sqrt{2}\le y\le2\)

\(Vậy\) \(y_{max}=2\)

       \(y_{min}=\sqrt{2}\)

\(y=sin^3x+2sin^2x+sinx-2\)

đặt \(t=sinx\) với \(t\in\left[-1;1\right]\)

 pt \(\Leftrightarrow\)\(y=t^3+2t^2+t-2\)

\(y'=3t^2+4t+1\)

\(y'=0\Leftrightarrow\left[{}\begin{matrix}t=-1\\t=-\dfrac{1}{3}\end{matrix}\right.\)

vậy GTLN của y = 2 với t=1 \(\Leftrightarrow sinx=1\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)

GTNN của y=-58/27  với \(t=-\dfrac{1}{3}\Leftrightarrow sinx=-\dfrac{1}{3}\Leftrightarrow x=sin^{-1}\left(-\dfrac{1}{3}\right)\)

ĐKXĐ:

a. \(x-1\ge0\Rightarrow x\ge1\)

b. \(\left\{{}\begin{matrix}cosx\ne0\\cos2x+1\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}cosx\ne0\\cos2x\ne-1\end{matrix}\right.\)

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

c.

\(cosx\ge0\Rightarrow-\dfrac{\pi}{2}+k2\pi\le x\le\dfrac{\pi}{2}+k2\pi\)

d, Hàm số xác định khi:

\(\left\{{}\begin{matrix}cos\left(x+\dfrac{\pi}{4}\right)\ne0\\sinx.cosx+cos2x-3\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{\pi}{4}\ne\dfrac{\pi}{2}+k\pi\\\dfrac{1}{2}sin2x+cos2x\ne3\end{matrix}\right.\)

\(\Leftrightarrow x\ne\dfrac{\pi}{4}+k\pi\)

e, Hàm số xác định khi:

\(\left\{{}\begin{matrix}cosx\ne0\\cos2x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+k\pi\\x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)