Đây nè bn:

ơ bạn :\(\dfrac{cos\left(x+y\right)+cosx}{cos\left(x+y\right)-cosx}=\dfrac{2cos\left(\dfrac{2x+y}{2}\right).cos\left(\dfrac{y}{2}\right)}{-2sin\left(\dfrac{2x+y}{2}\right).sin\left(\dfrac{y}{2}\right)}=-2.cot\left(\dfrac{2x+y}{2}\right).cot\left(\dfrac{y}{2}\right)\) L không thể bẳng 0 được

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

1: ĐKXĐ: 2x-1<>0

=>2x<>1

=>x<>1/2

=>TXĐ là D=R\{1/2}

2: ĐKXĐ: \(3x+\frac25\pi<>\frac{\pi}{2}+k\pi\)

=>\(3x<>\frac{\pi}{2}-\frac25\pi+k\pi=\frac{1}{10}\pi+k\pi\)

=>\(x<>\frac{1}{30}\pi+\frac{k\pi}{3}\)

=>TXĐ là D=R\{\(\frac{\pi}{30}+\frac{k\pi}{3}\) }

3: ĐKXĐ: \(2x-\frac13<>k\pi\)

=>\(2x<>\frac13+k\pi\)

=>\(x<>\frac16+\frac{k\pi}{2}\)

=>TXĐ là D=R\{\(\frac{k\pi}{2}+\frac16\) }

4: ĐKXĐ: sin x-cosx<>0

=>\(\sqrt2\cdot\sin\left(x-\frac{\pi}{4}\right)<>0\)

=>\(\sin\left(x-\frac{\pi}{4}\right)<>0\)

=>\(x-\frac{\pi}{4}<>k\pi\)

=>\(x<>\frac{\pi}{4}+k\pi\)

=>TXĐ là D=R\{\(\frac{\pi}{4}+k\pi\) }

5: ĐKXĐ: \(\begin{cases}\sin x<>0\\ cosx<>0\end{cases}\Rightarrow\begin{cases}x<>k\pi\\ x<>\frac{\pi}{2}+k\pi\end{cases}\Rightarrow x<>\frac{k\pi}{2}\)

=>TXĐ là D=R\{\(\frac{k\pi}{2}\) }

6: ĐKXĐ: \(\begin{cases}1-\sin x\ge0\\ cosx<>0\end{cases}\Rightarrow\begin{cases}\sin x<=1\\ x<>\frac{\pi}{2}+k\pi\end{cases}\)

=>\(x<>\frac{\pi}{2}+k\pi\)

=>TXĐ là D=R\{\(\frac{\pi}{2}+k\pi\) }

7: ĐKXĐ: \(\sin^2x-cos^2x<>0\)

=>\(cos^2x-\sin^2x<>0\)

=>cos2x<>0

=>\(2x<>\frac{\pi}{2}+k\pi\)

=>\(x<>\frac{\pi}{4}+\frac{k\pi}{2}\)

=>TXĐ là D=R\{\(\frac{\pi}{4}+\frac{k\pi}{2}\) }

8: ĐKXĐ: \(\begin{cases}x<>\frac{\pi}{2}+k\pi\\ \sin x<>-1\end{cases}\Rightarrow\begin{cases}x<>\frac{\pi}{2}+k\pi\\ x<>-\frac{\pi}{2}+k2\pi\end{cases}\)

=>\(x<>\frac{\pi}{2}+k\pi\)

=>TXĐ là D=R\{\(\frac{\pi}{2}+k\pi\) }

Giả sử các biểu thức đã cho đều xác định

a/ \(\dfrac{1+sin^2x}{1-sin^2x}=\dfrac{1+sin^2x}{cos^2x}=\dfrac{1}{cos^2x}+\dfrac{sin^2x}{cos^2x}+1+tan^2x+tan^2x=1+2tan^2x\)

b/ \(\dfrac{sinx}{1+cosx}+\dfrac{1+cosx}{sinx}=\dfrac{sin^2x+\left(1+cosx\right)^2}{\left(1+cosx\right)sinx}=\dfrac{sin^2x+cos^2x+2cosx+1}{\left(1+cosx\right)sinx}\)

\(=\dfrac{1+2cosx+1}{\left(1+cosx\right)sinx}=\dfrac{2+2cosx}{\left(1+cosx\right)sinx}=\dfrac{2\left(1+cosx\right)}{\left(1+cosx\right)sinx}=\dfrac{2}{sinx}\)

c/ \(\dfrac{1-sinx}{cosx}=\dfrac{\left(1-sinx\right)cosx}{cos^2x}=\dfrac{\left(1-sinx\right)cosx}{1-sin^2x}\)

\(\dfrac{\left(1-sinx\right)cosx}{\left(1-sinx\right)\left(1+sinx\right)}=\dfrac{cosx}{1+sinx}\)

d/ \(\left(1-cosx\right)\left(1+cot^2x\right)=\left(1-cosx\right).\dfrac{1}{sin^2x}\)

\(=\dfrac{1-cosx}{1-cos^2x}=\dfrac{1-cosx}{\left(1-cosx\right)\left(1+cosx\right)}=\dfrac{1}{1+cosx}\)

e/ \(1-\dfrac{sin^2x}{1+cotx}-\dfrac{cos^2x}{1+tanx}=1-\dfrac{sin^3x}{sinx\left(1+\dfrac{cosx}{sinx}\right)}-\dfrac{cos^3x}{cosx\left(1+\dfrac{sinx}{cosx}\right)}\)

\(=1-\left(\dfrac{sin^3x}{sinx+cosx}+\dfrac{cos^3x}{sinx+cosx}\right)=1-\left(\dfrac{sin^3x+cos^3x}{sinx+cosx}\right)\)

\(=1-\left(\dfrac{\left(sinx+cosx\right)\left(sin^2x-sinx.cosx+cos^2x\right)}{sinx+cosx}\right)\)

\(=1-\left(1-sinx.cosx\right)=sinx.cosx\)

f/ Bạn ghi đề sai à?

a: \(\frac{1}{\sin x}+\frac{1}{cosx}=4\cdot\sin\left(x+\frac{\pi}{4}\right)\)

=>\(\frac{\sin x+cosx}{\sin x\cdot cosx}=4\cdot\frac{\sqrt2}{2}\cdot\left(\sin x+cosx\right)\)

=>\(\left(\sin x+cosx\right)\left(\frac{1}{\sin x\cdot cosx}-2\sqrt2\right)=0\)

TH1: \(\frac{1}{\sin x\cdot cosx}-2\sqrt2=0\)

=>\(\frac{1}{\sin x\cdot cosx}=2\sqrt2\)

=>\(sinx\cdot cosx=\frac{1}{2\sqrt2}\)

=>\(2\cdot\sin x\cdot cosx=\frac{1}{\sqrt2}\)

=>\(\sin2x=\frac{1}{\sqrt2}\)

=>\(\left[\begin{array}{l}2x=\frac{\pi}{4}+k2\pi\\ 2x=-\frac{\pi}{4}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{8}+k\pi\\ x=-\frac{\pi}{8}+k\pi\end{array}\right.\)

TH2: sin x+cosx=0

=>\(\sqrt2\cdot\sin\left(x+\frac{\pi}{4}\right)=0\)

=>\(\sin\left(x+\frac{\pi}{4}\right)=0\)

=>\(x+\frac{\pi}{4}=k\pi\)

=>\(x=-\frac{\pi}{4}+k\pi\)