Tìm TXĐ( giúp mình vs ạ :(( )
a) y=sin 1\x
b) y=tanx+cotx-4
c) y= tanx\cosx-1
d) y=sin(x+1\x-1)
e) y=tan(3x+pi\4)
f) y= tan2(x+pi\3)\cosx+1
g) y=cot(x-3 pi)
Tìm TXĐ của các hàm số sau
\(a,\dfrac{1-cosx}{2sinx+1}\)
\(b,y=\sqrt{\dfrac{1+cosx}{2-cosx}}\)
\(c,\sqrt{tanx}\)
\(d,\dfrac{2}{2cos\left(x-\dfrac{\Pi}{4}\right)-1}\)
\(e,tan\left(x-\dfrac{\Pi}{3}\right)+cot\left(x+\dfrac{\Pi}{4}\right)\)
\(f,y=\dfrac{sinx}{cos^2x-sin^2x}\)
\(g,y=\dfrac{2}{cosx+cos2x}\)
\(h,y=\dfrac{1+cos2x}{1-cos4x}\)
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
Xét tính chẵn, lẻ của các hàm số
1,\(y=cosx+sin^2x\)
2,\(y=sinx+cosx\)
3,\(y=tanx+2sinx\)
4,\(y=tan2x-sin3x\)
5,\(sin2x+cosx\)
6,\(y=cosx.sin^2x-tan^2x\)
7,\(y=cos\left(x-\dfrac{\pi}{4}\right)+cos\left(x+\dfrac{\pi}{4}\right)\)
8,\(y=\dfrac{2+cosx}{1+sin^2x}\)
9,\(y=\left|2+sinx\right|+\left|2-sinx\right|\)
Tìm tập xác đinh của các hàm số sau
29 , \(y=\frac{tanx+cosx}{sinx}\)
30 , \(y=\frac{1}{sinx}-\frac{1}{cosx}\)
31 , \(y=\frac{cosx+cotx}{sinx}\)
32 , \(y=\frac{tanx+cotx}{1-sin2x}\)
33 , \(y=tanx+\frac{1}{cos\frac{x}{2}}\)
34 , \(y=\frac{1-tanx}{1-cotx}\)
35 , \(y=\frac{cotx}{cosx-1}\)
36 , \(y=\frac{3}{sin^2x-cos^2x}\)
37 , \(y=\frac{2}{cosx-cos3x}\)
38 , \(y=\frac{\sqrt{x}}{sin\pi x}\)
39 , \(y=\frac{2-cosx}{1+tan\left(x-\frac{\pi}{3}\right)}\)
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.
\(\left\{{}\begin{matrix}cos\left(x-\frac{\pi}{3}\right)\ne0\\tan\left(x-\frac{\pi}{3}\right)\ne-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-\frac{\pi}{3}\ne\frac{\pi}{2}+k\pi\\x-\frac{\pi}{3}\ne-\frac{\pi}{4}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{5\pi}{6}+k\pi\\x\ne-\frac{\pi}{12}+k\pi\end{matrix}\right.\)
Tìm TXĐ của hàm số: y=7sin(x-pi/5)/cos(x-pi) y=3-cot3x/ sin2x+1 y=cot(2x-pi/8)/(tanx-1)sin^2x
1/ mình giải ở bài kia rồi
Bạn viết đề rõ ràng hơn đi
VD: 1/y=\(\dfrac{7sin\left(x-\dfrac{\Pi}{5}\right)}{cos\left(x-\Pi\right)}\)
1) Tìm TXĐ của các hàm số sau:
a) y= tan ( x - \(\frac{\Pi}{4}\) ) + cos2x
b) y= \(cos^3\frac{x}{x^2-1}\)
c) y= \(\frac{cosx+1}{x^2+1}\)
d) y= \(\frac{tanx}{x^2-x+2}\)
2) Xét tính chẵn lẻ của các hàm số sau:
a) f(x) = \(\frac{x+tanx}{x^2+1}\)
b) f(x) = \(\frac{5x.cos5x}{sin^2x+2}\)
c) f(x) = (2x-3). sin4x
d) f(x)= \(sin^42x+cos^4\left(2x-\frac{\Pi}{6}\right)\)
1. ĐKXĐ:
a.
\(cos\left(x-\frac{\pi}{4}\right)\ne0\)
\(\Leftrightarrow x-\frac{\pi}{4}\ne\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow x\ne\frac{3\pi}{4}+k\pi\)
b.
\(x^2-1\ne0\Leftrightarrow x\ne\pm1\)
c.
Hàm xác định trên R
d.
\(cosx\ne0\Leftrightarrow x\ne\frac{\pi}{2}+k\pi\)
2.
a. ĐKXĐ: \(x\ne\frac{\pi}{2}+k\pi\)
Miền xác định đối xứng
\(f\left(-x\right)=\frac{-x+tan\left(-x\right)}{\left(-x\right)^2+1}=\frac{-x-tanx}{x^2+1}=-\frac{x+tanx}{x^2+1}=-f\left(x\right)\)
Hàm lẻ
b. \(f\left(-x\right)=\frac{5\left(-x\right).cos\left(-5x\right)}{sin^2\left(-x\right)+2}=\frac{-5x.cos5x}{sin^2x+2}=-f\left(x\right)\)
Hàm lẻ
c. \(f\left(-x\right)=\left(-2x-3\right)sin\left(-4x\right)=\left(2x+3\right)sin4x\)
Hàm không chẵn không lẻ
d. \(f\left(-x\right)=sin^4\left(-2x\right)+cos^4\left(-2x-\frac{\pi}{6}\right)\)
\(=sin^42x+cos^4\left(2x+\frac{\pi}{6}\right)\)
Hàm ko chẵn ko lẻ
Tìm đạo hàm các hàm số:
1, \(y=\tan(3x-\dfrac{\pi}{4})+\cot(2x-\dfrac{\pi}{3})+\cos(x+\dfrac{\pi}{6})\)
2, \(y=\dfrac{\sqrt{\sin x+2}}{2x+1}\)
3, \(y=\cos(3x+\dfrac{\pi}{3})-\sin(2x+\dfrac{\pi}{6})+\cot(x+\dfrac{\pi}{4})\)
a.
\(y'=\dfrac{3}{cos^2\left(3x-\dfrac{\pi}{4}\right)}-\dfrac{2}{sin^2\left(2x-\dfrac{\pi}{3}\right)}-sin\left(x+\dfrac{\pi}{6}\right)\)
b.
\(y'=\dfrac{\dfrac{\left(2x+1\right)cosx}{2\sqrt{sinx+2}}-2\sqrt{sinx+2}}{\left(2x+1\right)^2}=\dfrac{\left(2x+1\right)cosx-4\left(sinx+2\right)}{\left(2x+1\right)^2}\)
c.
\(y'=-3sin\left(3x+\dfrac{\pi}{3}\right)-2cos\left(2x+\dfrac{\pi}{6}\right)-\dfrac{1}{sin^2\left(x+\dfrac{\pi}{4}\right)}\)
tìm tập xác định của hàm số lượng giác sau
a)\(y=\dfrac{tan\left(2x-\dfrac{\pi}{4}\right)}{\sqrt{1-sin\left(x-\dfrac{\pi}{8}\right)}}\)
b)\(y=\dfrac{tan\left(x-\dfrac{\pi}{4}\right)}{1-cos\left(x+\dfrac{\pi}{3}\right)}\)
c)\(y=\dfrac{3}{cosx-cos3x}\)
d)\(y=\dfrac{4}{sin^2x-cos^2x}\)
e)\(y=\dfrac{1+cot\left(\dfrac{\pi}{3}+x\right)}{tan^2\left(3x-\dfrac{\pi}{4}\right)}\)
Tìm tập xác định của các hàm số sau:
1) a) y=tanx+3
b) y=3-4cotx
c) y=tan2x+1
d) y=4-5cot3x
e) \(y=tan\left(x+\dfrac{\pi}{3}\right)\)-3
f) \(y=4-2cot\left(x-\dfrac{\pi}{6}\right)\)
2) a) y=3sinx-4cosx+5
b) y=3cos2x-4sin2x+1
c) \(y=\dfrac{3}{1-cosx}+5\)
d) \(\dfrac{1}{1+cosx}+2\)
e) \(y=\dfrac{sinx+2}{cosx+3}\)
f) \(y=1-\dfrac{2}{sinx-1}\)
g) \(y=2x+\dfrac{3}{1+sinx}\)
h) \(y=x^2-x+\dfrac{1}{sin^2x-sinx}\)
j) y=2tanx-3cotx+5
h) \(y=\sqrt{\dfrac{1-sin^2x}{1+cos^2x}}\)
1:
a: ĐKXĐ: \(x< >\dfrac{\Omega}{2}+k\Omega\)
=>TXĐ: \(D=R\backslash\left\{\dfrac{\Omega}{2}+k\Omega\right\}\)
b: ĐKXĐ: \(x< >k\Omega\)
=>TXĐ: \(D=R\backslash\left\{k\Omega\right\}\)
c: ĐKXĐ: \(2x< >\dfrac{\Omega}{2}+k\Omega\)
=>\(x< >\dfrac{\Omega}{4}+\dfrac{k\Omega}{2}\)
TXĐ: \(D=R\backslash\left\{\dfrac{\Omega}{4}+\dfrac{k\Omega}{2}\right\}\)
d: ĐKXĐ: \(3x< >\Omega\cdot k\)
=>\(x< >\dfrac{k\Omega}{3}\)
TXĐ: \(D=R\backslash\left\{\dfrac{k\Omega}{3}\right\}\)
e: ĐKXĐ: \(x+\dfrac{\Omega}{3}< >\dfrac{\Omega}{2}+k\Omega\)
=>\(x< >\dfrac{\Omega}{6}+k\Omega\)
TXĐ: \(D=R\backslash\left\{\dfrac{\Omega}{6}+k\Omega\right\}\)
f: ĐKXĐ: \(x-\dfrac{\Omega}{6}< >\Omega\cdot k\)
=>\(x< >k\Omega+\dfrac{\Omega}{6}\)
TXĐ: \(D=R\backslash\left\{k\Omega+\dfrac{\Omega}{6}\right\}\)
Tìm TXĐ
1. y=\(\frac{cotx}{1-sinx}\)
2.y=\(\frac{1+tan\left(2x+\frac{\pi}{3}\right)}{cot^{2^{ }}x+1}\)
3.y=\(\sqrt{\frac{5-3cos2x}{1+sin\left(2x-\frac{\pi}{2}\right)}}\)
4.y=\(\frac{1+cot\left(x+\frac{\pi}{3}\right)}{tan^2\left(3x-\frac{\pi}{4}\right)}\)
\(\text{1) Đ}K:\left\{{}\begin{matrix}sinx\ne0\\1-sinx\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne m\pi\\x\ne\frac{\pi}{2}+n2\pi\end{matrix}\right.\)
\(2\text{) }ĐK:\left\{{}\begin{matrix}cos\left(2x+\frac{\pi}{3}\right)\ne0\\sinx\ne0\end{matrix}\right.\Leftrightarrow\\ \left\{{}\begin{matrix}2x+\frac{\pi}{3}\ne\frac{\pi}{2}+m\pi\\x\ne n\pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{12}+\frac{m\pi}{2}\\x\ne n\pi\end{matrix}\right.\)
\(3\text{) }ĐK:\left\{{}\begin{matrix}\frac{5-3cos2x}{1+sin\left(2x-\frac{\pi}{2}\right)}\ge0\\1+sin\left(2x-\frac{\pi}{2}\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5-3cos2x\ge0\\sin\left(2x-\frac{\pi}{2}\right)\ne-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}cos2x\le\frac{5}{3}\left(T/m\right)\\2x-\frac{\pi}{2}\ne\frac{3\pi}{2}+k2\pi\end{matrix}\right.\Leftrightarrow x\ne\pi+k\pi\)
\(4\text{) }ĐK:\left\{{}\begin{matrix}sin\left(x+\frac{\pi}{3}\right)\ne0\\cos\left(3x-\frac{\pi}{4}\right)\ne0\\tan\left(3x-\frac{\pi}{4}\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+\frac{\pi}{3}\ne a\pi\\3x-\frac{\pi}{4}\ne\frac{\pi}{2}+b\pi\\3x-\frac{\pi}{4}\ne c\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\frac{\pi}{3}+a\pi\\x\ne\frac{\pi}{4}+\frac{b\pi}{3}\\x\ne\frac{\pi}{12}+\frac{c\pi}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-\frac{\pi}{3}+a\pi\\x\ne\frac{\pi}{12}+\frac{k\pi}{6}\end{matrix}\right.\)