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

a: ĐKXĐ: \(\sqrt3\cdot\sin x+cosx<>0\)

=>\(\frac{\sqrt3}{2}\cdot\sin x+\frac12\cdot cosx<>0\)

=>\(\sin\left(x+\frac{\pi}{6}\right)<>0\)

=>\(x+\frac{\pi}{6}<>k\pi\)

=>\(x<>-\frac{\pi}{6}+k\pi\)

\(\frac{2\cdot cos2x+1}{\sqrt3\cdot\sin x+cosx}=2\cdot cosx-1\)

=>\(\frac{2\cdot\left(2\cdot cos^2x-1\right)+1}{2\cdot\sin\left(x+\frac{\pi}{6}\right)}=2\cdot cosx-1\)

=>\(\frac{4\cdot cos^2x-1}{2\cdot\sin\left(x+\frac{\pi}{6}\right)}-\left(2\cdot cosx-1\right)=0\)

=>\(\left(2\cdot cosx-1\right)\left\lbrack\frac{2\cdot cosx+1}{2\cdot\sin\left(x+\frac{\pi}{6}\right)}-1\right\rbrack=0\)

TH1: \(\frac{2\cdot cosx+1}{\sqrt3\cdot\sin x+cosx}-1=0\)

=>\(\frac{2\cdot cosx+1}{\sqrt3\cdot\sin x+cosx}=1\)

=>\(\sqrt3\cdot\sin x+cosx=2\cdot cosx+1\)

=>\(\sqrt3\cdot\sin x-cosx=1\)

=>\(\frac{\sqrt3}{2}\cdot\sin x-\frac12\cdot cosx=\frac12\)

=>\(\sin\left(x-\frac{\pi}{6}\right)=\frac12\)

=>\(\left[\begin{array}{l}x-\frac{\pi}{6}=\frac{\pi}{6}+k2\pi\\ x-\frac{\pi}{6}=\pi-\frac{\pi}{6}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{3}+k2\pi\\ x=\pi+k2\pi\end{array}\right.\)

TH2: \(2\cdot cosx-1=0\)

=>\(cosx=\frac12\)

=>\(\left[\begin{array}{l}x=\frac{\pi}{3}+k2\pi\\ x=-\frac{\pi}{3}+k2\pi\end{array}\right.\)

a: ĐKXĐ: 2*sin x+1<>0

=>sin x<>-1/2

=>x<>-pi/6+k2pi và x<>7/6pi+k2pi

b: ĐKXĐ: \(\dfrac{1+cosx}{2-cosx}>=0\)

mà 1+cosx>=0

nên 2-cosx>=0

=>cosx<=2(luôn đúng)

c ĐKXĐ: tan x>0

=>kpi<x<pi/2+kpi

d: ĐKXĐ: \(2\cdot cos\left(x-\dfrac{pi}{4}\right)-1< >0\)

=>cos(x-pi/4)<>1/2

=>x-pi/4<>pi/3+k2pi và x-pi/4<>-pi/3+k2pi

=>x<>7/12pi+k2pi và x<>-pi/12+k2pi

e: ĐKXĐ: x-pi/3<>pi/2+kpi và x+pi/4<>kpi

=>x<>5/6pi+kpi và x<>kpi-pi/4

f: ĐKXĐ: cos^2x-sin^2x<>0

=>cos2x<>0

=>2x<>pi/2+kpi

=>x<>pi/4+kpi/2

ĐKXĐ: (tất cả \(k\in Z\))

a. \(sinx-1\ge0\Leftrightarrow sinx\ge1\)

\(\Leftrightarrow sinx=1\Rightarrow x=\dfrac{\pi}{2}+k2\pi\)

b. \(\left\{{}\begin{matrix}\dfrac{1-sinx}{1+sinx}\ge0\left(luôn-đúng\right)\\1+sinx\ne0\end{matrix}\right.\) \(\Leftrightarrow sinx\ne-1\)

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

c. \(sinx\ne0\Leftrightarrow x\ne k\pi\)

a: ĐKXĐ: \(cosx-1\ne0\)

=>\(cosx\ne1\)

=>\(x\ne k2\Omega\)

b: ĐKXĐ: sin x-1>=0

=>sin x>=1

mà \(-1< =sinx< =1\)

nên sin x=1

=>\(x=\dfrac{\Omega}{2}+k2\Omega\)

c:

-1<=sin x<=1

=>-1+1<=sin x+1<=1+1

=>0<=sin x+1<=2

ĐKXĐ: \(\dfrac{1+sinx}{1-cosx}>=0\)

mà \(1+sinx>=0\)(cmt)

nên \(1-cosx>0\)

=>\(cosx< 1\)

mà -1<=cosx<=1

nên \(cosx\ne1\)

=>\(x\ne k2\Omega\)

1: ĐKXĐ: 3-cosx>0

=>cosx<3(luôn đúng)

2: ĐKXĐ: 1-sin 3x>=0

=>sin 3x<=1(luôn đúng)

3: ĐKXĐ: sin x<>0 và 2x<>pi/2+kpi

=>x<>kpi và x<>pi/4+kpi/2

4: ĐKXĐ: 2x-1>=0

=>x>=1/2

ĐKXĐ: \(sinx+cosx>0\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)>0\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)>0\)

\(\Leftrightarrow k2\pi< x+\dfrac{\pi}{4}< \pi+k2\pi\)

\(\Leftrightarrow-\dfrac{\pi}{4}+k2\pi< x< \dfrac{3\pi}{4}+k2\pi\)

sao ra đc dòng t2 vậy ạ

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

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

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

\(\Leftrightarrow sin2x\ne0\)

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