a/ ĐKXĐ:

\(sin\left(\frac{\pi}{2}.sinx\right)\ne0\Rightarrow\frac{\pi}{2}.sinx\ne k\pi\)

\(\Rightarrow sinx\ne2k\)

Mà \(-1\le sinx\le1\Rightarrow sinx\ne0\Rightarrow x\ne k\pi\)

b/

\(sinx-1\ge0\Leftrightarrow sinx\ge1\Rightarrow sinx=1\)

\(\Rightarrow x=\frac{\pi}{2}+k2\pi\)

c/

\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\\cos2x\ne0\end{matrix}\right.\) \(\Rightarrow sin4x\ne0\)

\(\Rightarrow x\ne\frac{k\pi}{4}\)

d/

\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\\sinx+cotx\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}sin2x\ne0\\sin^2x+cosx\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x\ne k\pi\\-cos^2x+cosx+1\ne0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x\ne\frac{k\pi}{2}\\cosx\ne\frac{1-\sqrt{5}}{2}\\\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne\frac{k\pi}{2}\\x\ne\pm arccos\left(\frac{1-\sqrt{5}}{2}\right)+k2\pi\end{matrix}\right.\)

e/

\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne1\end{matrix}\right.\) \(\Leftrightarrow sinx\ne0\Rightarrow x\ne k\pi\)

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

Chọn C

ĐKXĐ:

29.

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

\(\Leftrightarrow sin2x\ne0\Leftrightarrow x\ne\frac{k\pi}{2}\)

30.

\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\end{matrix}\right.\) \(\Leftrightarrow x\ne\frac{k\pi}{2}\) (như câu trên)

31.

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

32.

\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\\sin2x\ne1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}sin2x\ne0\\sin2x\ne1\end{matrix}\right.\)

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

33.

\(\left\{{}\begin{matrix}cosx\ne0\\cos\frac{x}{2}\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{2}+k\pi\\x\ne\pi+k2\pi\end{matrix}\right.\)

34.

\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\\cotx\ne1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}sin2x\ne0\\cotx\ne1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{k\pi}{2}\\x\ne\frac{\pi}{4}+k\pi\end{matrix}\right.\)

35.

\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne1\end{matrix}\right.\) \(\Leftrightarrow sinx\ne0\)

\(\Leftrightarrow x\ne k\pi\)

36.

\(sin^2x-cos^2x\ne0\Leftrightarrow cos2x\ne0\)

\(\Leftrightarrow x\ne\frac{\pi}{4}+\frac{k\pi}{2}\)

37.

\(cos3x\ne cosx\Leftrightarrow\left\{{}\begin{matrix}3x\ne x+k2\pi\\3x\ne-x+k2\pi\end{matrix}\right.\)

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

38.

\(\left\{{}\begin{matrix}x\ge0\\sin\pi x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\pi x\ne k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne k\end{matrix}\right.\)

39.

\(\left\{{}\begin{matrix}cos\left(x-\frac{\pi}{3}\right)\ne0\\tan\left(x-\frac{\pi}{3}\right)\ne-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-\frac{\pi}{3}\ne\frac{\pi}{2}+k\pi\\x-\frac{\pi}{3}\ne-\frac{\pi}{4}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{5\pi}{6}+k\pi\\x\ne-\frac{\pi}{12}+k\pi\end{matrix}\right.\)

1.

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

\(\Leftrightarrow\left\{{}\begin{matrix}cosx\ne0\\sinx\ne0\\cosx\ne1\end{matrix}\right.\) \(\Leftrightarrow sin2x\ne0\Leftrightarrow x\ne\dfrac{k\pi}{2}\)

2.

ĐKXĐ: \(sin2x\ne0\Leftrightarrow x\ne\dfrac{k\pi}{2}\)

3. 

ĐKXĐ: \(\left\{{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)\ne0\\cos\left(x-\dfrac{\pi}{4}\right)\ne0\end{matrix}\right.\)

\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{2}\right)\ne0\Leftrightarrow cos2x\ne0\)

\(\Leftrightarrow x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

câu 2 ..... \(\dfrac{cos^22x}{sin^22x}=cot^22x\) nên suy ra sin2x khác 0 đúng hơm

còn câu 3, tui ko hiểu chỗ sin(2x-pi/4).. sao ở đây rớt xuống dợ

đề bài đầy đủ: rút gọn các biểu thức lượng giác sau trên điều kiện xác định của chúng:

\(\frac{sin^2x}{cosx+cosx.\frac{sinx}{cosx}}-\frac{cos^2x}{sinx+sinx.\frac{cosx}{sinx}}=\frac{sin^2x}{sinx+cosx}-\frac{cos^2x}{sinx+cosx}=\frac{sin^2x-cos^2x}{sinx+cosx}\)

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

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

\(=\left(\frac{sinx+1}{cosx\left(1+sinx\right)}\right)\left(\frac{cosx+1}{sinx\left(1+cosx\right)}\right)=\frac{1}{sinx.cosx}\)

\(y=sinx\Rightarrow y'=cosx;y''=-sinx;y'''=-cosx\)

Bằng quy nạp toán học ; ta c/m được : \(y^{\left(n\right)}sinx=sin\left(x+n\dfrac{\pi}{2}\right)\)

\(a,\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4sin^2x.cos^2x}=-1\)

\(VT=\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4.sin^2x.cos^2x}=\left(\frac{1}{tan2x}\right)^2-\frac{1}{sin^22x}=\left(\frac{cos2x}{sin2x}\right)^2-\frac{1}{sin^22x}=\frac{cos^22x-1}{sin^22x}=\frac{-sin^22x}{sin^22x}=-1=VP\)

b, \(VT=\frac{cos^2x-sin^2x}{sin^4x+cos^4x-sin^2x}=\frac{cos2x}{\left(sin^2x+cos^2x\right)^2-sin^2x-2.sin^2x.cos^2x}=\frac{cos2x}{1-sin^2x-2.sin^2x.cos^2x}=\frac{cos2x}{cos^2x-2.sin^2x.cos^2x}\)

=\(\frac{cos2x}{cos^2x.\left(1-2.sin^2x\right)}=\frac{cos2x}{cos^2x.cos2x}=\frac{1}{cos^2x}=1+tan^2x=VP\)

d, \(VT=\left(\frac{cosx}{1+sinx}+tanx\right).\left(\frac{sinx}{1+cosx}+cotx\right)=\left(\frac{cosx}{1+sinx}+\frac{sinx}{cosx}\right).\left(\frac{sinx}{1+cosx}+\frac{cosx}{sinx}\right)\)

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

=\(\frac{1}{cosx.sinx}=VP\)

e, \(VT=cos^2x.\left(cos^2x+2sin^2x+sin^2x.tan^2x\right)=cos^2x.\left(1+sin^2x.\left(1+tan^2x\right)\right)=cos^2x.\left(1+tan^2x\right)=cos^2x.\frac{1}{cos^2x}=1=VP\)

c, \(VT=\frac{sin^2x}{cosx.\left(1+tanx\right)}-\frac{cos^2x}{sinx.\left(1+cosx\right)}=\frac{sin^3x.\left(1+cosx\right)-cos^3x.\left(1+tanx\right)}{sinx.cosx.\left(1+tanx\right).\left(1+cosx\right)}\)

=\(\frac{sin^3x+sin^3x.cotx-cos^3x-cos^3.tanx}{\left(sinx+cosx\right)^2}=\frac{sin^3x+sin^2xcosx-cos^3x-cos^2sinx}{\left(sinx+cosx\right)^2}=\frac{sin^2x.\left(sinx+cosx\right)-cos^2x.\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}\)

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

Đây nha bạn