CMR: \(\frac{Sinx+Sin\frac{x}{2}}{1+Cosx+Cos\frac{x}{2}}=Tan\frac{x}{2}\)
Những câu hỏi liên quan
Chứng minh đẳng thức sau :
a, left(frac{tan^2x-1}{2tanx}right)^2 - frac{1}{4sin^2x.cos^2x} -1
b, frac{cos^2x-sin^2x}{sin^4x+cos^4x-sin^2x} 1 + tan2x
c, frac{sin^2x}{cosx.left(1+tanxright)}-frac{cos^2x}{sinx.left(1+cotxright)}sinx-cosx
d, left(frac{cosx}{1+sinx}+tanxright).left(frac{sinx}{1+cosx}+cotxright)frac{1}{sinx.cosx}
e, cos2x.(cos2x + 2sin2x + sin2x.tan2x) 1
Đọc tiếp
Chứng minh đẳng thức sau :
a, \(\left(\frac{tan^2x-1}{2tanx}\right)^2\) - \(\frac{1}{4sin^2x.cos^2x}\) = -1
b, \(\frac{cos^2x-sin^2x}{sin^4x+cos^4x-sin^2x}\) = 1 + tan2x
c, \(\frac{sin^2x}{cosx.\left(1+tanx\right)}-\frac{cos^2x}{sinx.\left(1+cotx\right)}=sinx-cosx\)
d, \(\left(\frac{cosx}{1+sinx}+tanx\right).\left(\frac{sinx}{1+cosx}+cotx\right)=\frac{1}{sinx.cosx}\)
e, cos2x.(cos2x + 2sin2x + sin2x.tan2x) = 1
\(a,\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4sin^2x.cos^2x}=-1\)
\(VT=\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4.sin^2x.cos^2x}=\left(\frac{1}{tan2x}\right)^2-\frac{1}{sin^22x}=\left(\frac{cos2x}{sin2x}\right)^2-\frac{1}{sin^22x}=\frac{cos^22x-1}{sin^22x}=\frac{-sin^22x}{sin^22x}=-1=VP\)
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
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CMR:
a, \(\frac{\cot^2x-\sin^2x}{\cot^2x-tan^2x}=sin^2x.\cos^2x\)
b, \(\frac{\tan x}{1-\tan^2x}.\frac{\cot^2-1}{\cot x}=1\)
c, \(\frac{1+\sin x.\cos x}{\sin^2x-\cos^2x}=\frac{\tan x+1}{\cot x+1}\)
d, \(\frac{\sin x+\cos x-1}{\sin x-cosx+1}=\frac{\cos x}{1+sinx}\)
Chứng minh|
a) \(\frac{1+sin2x}{sinx+cosx}-\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}=sinx\)
b) \(sin^4x+cos^4\left(x+\frac{\pi}{4}\right)=\frac{3}{4}-\frac{\sqrt{2}}{2}sin\left(2x+\frac{\pi}{4}\right)\)
\(\frac{sin^2x+cos^2x+2sinx.cosx}{sinx+cosx}-\left(1-tan^2\frac{x}{2}\right).cos^2\frac{x}{2}\)
\(=\frac{\left(sinx+cosx\right)^2}{sinx+cosx}-\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)\)
\(=sinx+cosx-cosx=sinx\)
\(sin^4x+cos^4\left(x+\frac{\pi}{4}\right)=\left(\frac{1}{2}-\frac{1}{2}cos2x\right)^2+\left(\frac{1}{2}+\frac{1}{2}cos\left(2x+\frac{\pi}{2}\right)\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\left(\frac{1}{2}-\frac{1}{2}sin2x\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\frac{1}{4}-\frac{1}{2}sin2x+\frac{1}{4}sin^22x\)
\(=\frac{1}{4}-\frac{1}{2}\left(cos2x+sin2x\right)+\frac{1}{4}\left(cos^22x+sin^22x\right)\)
\(=\frac{3}{4}-\frac{\sqrt{2}}{2}sin\left(2x+\frac{\pi}{4}\right)\)
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14 tháng 6 2021 lúc 9:28
Chứng minh:
a) \(tan(\frac\pi4+\frac{x}2).\frac{1+cos(\frac\pi2+x)}{sin(\frac\pi2+x)}=1\)
b) \(tan(\frac\pi4+x)=\frac{1+sin2x}{cos2x}\)
c) \(\frac{cosx}{1-sinx}=cot(\frac\pi4-\frac{x}{2})\)
d) \(tanx.tan3x=\frac{tan^22x-tan^2x}{1-tan^2x.tan^22x}\)
a) nếu sin x = 3.cos x Tính sin x. cos x = .....?
b) cho sin x\(=\frac{2}{3}\). tính \(5cos^2x+2sin^2x=...?\)
c) sin x + cos x = \(\frac{5}{7}\). tính tan x = ...?
d) tan x = \(\frac{1}{2}\) tính \(\frac{sinx+cosx}{cosx-sinx}=...?\)
m.n giúp mk nha, ai nhanh mk tick cho nha
a) 1-cot^4xfrac{2}{sin^2x}-frac{1}{sin^4x}b)frac{1-2sinx.cosx}{cos^2-sin^2}frac{1-tanx}{1+tanx}c)frac{sin^2x}{sinx-cosx}+frac{sinx+cosx}{1-tanx}sinx+cosxd)sqrt{frac{1+cosx}{1-cosx}}-sqrt{frac{1-cosx}{1+cosx}}frac{2.cosx}{|sin|}e)tan^3x+tan^2x+tanx+1frac{sinx+cosx}{cos^3x}
Đọc tiếp
a) \(1-cot^4x=\frac{2}{sin^2x}-\frac{1}{sin^4x}\)
b)\(\frac{1-2sinx.cosx}{cos^2-sin^2}\)\(=\frac{1-tanx}{1+tanx}\)\(\)
c)\(\frac{sin^2x}{sinx-cosx}+\frac{sinx+cosx}{1-tanx}=sinx+cosx\)
d)\(\sqrt{\frac{1+cosx}{1-cosx}}-\sqrt{\frac{1-cosx}{1+cosx}}=\frac{2.cosx}{|sin|}\)
e)\(tan^3x+tan^2x+tanx+1=\frac{sinx+cosx}{cos^3x}\)
24 tháng 7 2020 lúc 18:32
giải các pt
a) sinleft(frac{3pi}{10}-frac{x}{2}right)frac{1}{2}sinleft(frac{pi}{10}+frac{3x}{2}right)
b) 4left(sin^2x+frac{1}{sin^2x}right)+4left(sinx+frac{1}{sinx}right)7
c) 9left(frac{2}{cosx}+cosxright)+2left(cos^2x+frac{4}{cos^2x}right)1
d) 2left(cos^2x+frac{4}{cos^2x}right)+9left(frac{2}{cosx}-cosxright)1
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giải các pt
a) \(sin\left(\frac{3\pi}{10}-\frac{x}{2}\right)=\frac{1}{2}sin\left(\frac{\pi}{10}+\frac{3x}{2}\right)\)
b) \(4\left(sin^2x+\frac{1}{sin^2x}\right)+4\left(sinx+\frac{1}{sinx}\right)=7\)
c) \(9\left(\frac{2}{cosx}+cosx\right)+2\left(cos^2x+\frac{4}{cos^2x}\right)=1\)
d) \(2\left(cos^2x+\frac{4}{cos^2x}\right)+9\left(\frac{2}{cosx}-cosx\right)=1\)
a/
\(\Leftrightarrow cos\left(\frac{x}{2}+\frac{\pi}{5}\right)=\frac{1}{2}sin\left(\frac{3x}{2}+\frac{\pi}{10}\right)\)
Đặt \(\frac{x}{2}+\frac{\pi}{5}=a\Rightarrow\frac{x}{2}=a-\frac{\pi}{5}\Rightarrow\frac{3x}{2}=3a-\frac{3\pi}{5}\)
Pt trở thành:
\(cosa=\frac{1}{2}sin\left(3a-\frac{3\pi}{5}+\frac{\pi}{10}\right)\)
\(\Leftrightarrow cosa=\frac{1}{2}sin\left(3a-\frac{\pi}{2}\right)\)
\(\Leftrightarrow cosa=-\frac{1}{2}sin\left(\frac{\pi}{2}-3a\right)=-\frac{1}{2}cos3a\)
\(\Leftrightarrow cos3a+2cosa=0\)
\(\Leftrightarrow4cos^3a-3cosa+2cosa=0\)
\(\Leftrightarrow4cos^3a-cosa=0\)
\(\Leftrightarrow cosa\left(4cos^2a-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}cosa=0\\cosa=\frac{1}{2}\\cosa=-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}cos\left(\frac{x}{2}+\frac{\pi}{5}\right)=0\\cos\left(\frac{x}{2}+\frac{\pi}{5}\right)=\frac{1}{2}\\cos\left(\frac{x}{2}+\frac{\pi}{5}\right)=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{2}+\frac{\pi}{5}=\frac{\pi}{2}+k\pi\\\frac{x}{2}+\frac{\pi}{5}=\pm\frac{\pi}{3}+k2\pi\\\frac{x}{2}+\frac{\pi}{5}=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\) \(\Rightarrow x=...\) (5 nghiệm bạn tự biến đổi)
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b/
ĐKXĐ: ...
Đặt \(sinx+\frac{1}{sinx}=a\Rightarrow sin^2x+\frac{1}{sin^2x}=a^2-2\)
Pt trở thành:
\(4\left(a^2-2\right)+4a=7\)
\(\Leftrightarrow4a^2+4a-15=0\Rightarrow\left[{}\begin{matrix}a=\frac{3}{2}\\a=-\frac{5}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sinx+\frac{1}{sinx}=\frac{3}{2}\\sinx+\frac{1}{sinx}=-\frac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin^2x-\frac{3}{2}sinx+1=0\left(vn\right)\\sin^2x+\frac{5}{2}sinx+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sinx=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
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c/
ĐKXĐ: ...
Đặt \(cosx+\frac{2}{cosx}=a\Rightarrow cos^2x+\frac{4}{cos^2x}=a^2-4\)
Pt trở thành:
\(9a+2\left(a^2-4\right)=1\)
\(\Leftrightarrow2a^2+9a-9=0\)
Pt này nghiệm xấu quá bạn :(
d/ĐKXĐ: ...
Đặt \(\frac{2}{cosx}-cosx=a\Rightarrow cos^2x+\frac{4}{cos^2x}=a^2+4\)
Pt trở thành:
\(2\left(a^2+4\right)+9a-1=0\)
\(\Leftrightarrow2a^2+9a+7=0\Rightarrow\left[{}\begin{matrix}a=-1\\a=-\frac{7}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{2}{cosx}-cosx=-1\\\frac{2}{cosx}-cosx=-\frac{7}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-cos^2x+cosx+2=0\\-cos^2x+\frac{7}{2}cosx+2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}cosx=-1\\cosx=2\left(l\right)\\cosx=4\left(l\right)\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pi+k2\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
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Mọi người giúp em giải bài này ạ, em cảm ơn
Bài 1: Rút gọn biểu thức:
Afrac{sin2x+sin x}{1+cos2x+cos x}
Bcotaleft(frac{1+sin^2a}{cos a}-cosaright)
Cfrac{1+cos x+cos2x+cos3x}{2cos^2x+cos x-1}
Dfrac{2cosleft(frac{pi}{2}-xright)cdotsinleft(frac{pi}{2}+xright)cdottanleft(pi-xright)}{cotleft(frac{pi}{2}+xright)cdotsinleft(pi-xright)}-2cos x
Ecos^2xcdotcot^2x+3cos^2x-cot^2x+2sin^2x
Ffrac{sin^2x+sin^2xtan^2x}{cos^2x+cos^2xtan^2x}
Gfrac{1+cos2a-cosa}{2sina-sina}
Hsin^{^{ }4}left(frac{pi}{2}+alp...
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Mọi người giúp em giải bài này ạ, em cảm ơn
Bài 1: Rút gọn biểu thức:
A=\(\frac{\sin2x+\sin x}{1+\cos2x+\cos x}\)
B=\(cota\left(\frac{1+\sin^2a}{\cos a}-cosa\right)\)
C=\(\frac{1+\cos x+\cos2x+\cos3x}{2\cos^2x+\cos x-1}\)
D=\(\frac{2\cos\left(\frac{\pi}{2}-x\right)\cdot\sin\left(\frac{\pi}{2}+x\right)\cdot\tan\left(\pi-x\right)}{\cot\left(\frac{\pi}{2}+x\right)\cdot\sin\left(\pi-x\right)}-2\cos x\)
E=\(\cos^2x\cdot\cot^2x+3\cos^2x-\cot^2x+2\sin^2x\)
\(F=\frac{\sin^2x+\sin^2x\tan^2x}{\cos^2x+\cos^2x\tan^2x}\)
\(G=\frac{1+cos2a-cosa}{2sina-sina}\)
H=\(sin^{^{ }4}\left(\frac{\pi}{2}+\alpha\right)-cos^4\left(\frac{3\pi}{2}-\alpha\right)+1\)
Bài 2: chứng minh
a) cho \(\Delta ABCchứngminhsin\frac{A+B}{2}=cos\frac{C}{2}\)
b) chứng minh biểu thức sau độc lập với biến x:
A=\(cosx+cos\left(x+\frac{2\pi}{3}\right)+cos\left(x+\frac{4\pi}{3}\right)\)
c)cho \(\Delta\) ABC chứng minh : sin A+sin B+ sin C= \(4cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}\)
d)CMR: \(\frac{cos2a}{1+sin2a}=\frac{cosa-sina}{cosa+sina}\)
e) CMR:\(E=\frac{sin\alpha+cos\alpha}{cos^3\alpha}=1+tan\alpha+tan^2\alpha+tan^3\alpha\)
f) CMR \(\Delta\)ABC cân khi và chỉ khi \(sinB=2cosAsinC\)
g) CM: \(\frac{1-cosx+cos2x}{sin2x-sinx}=cotx\)
h)CM: \(\left(cos3x-cosx\right)^2+\left(sin3x-sinx\right)^2=4sin^2x\)
k) CMR trong tam giac ABC ta có: \(sin2A+sin2B+sin2C=4sinA\cdot sinB\cdot sinC\)
Bài 3: giải bất phương trình:
a)\(\frac{\left(1-3x\right)\left(2x^2+1\right)}{-2x^2-3x+5}>0\)
b)\(\frac{2x+1}{\left(x-1\right)\left(x+2\right)}\ge0\)
c)\(\frac{\left(3x-2\right)\left(x^2-9\right)}{x^2-4x+4}\le0\)
d)\(\frac{\left(2x^2+3x\right)\left(3-2x\right)}{1-x^2}\ge0\)
e)\(\frac{\left(x^2+2x+1\right)\left(x-1\right)}{3-x^2}\)
f)\(\frac{2x+1}{-x^2+x+6}\ge0\)
\(A=\frac{2sinx.cosx+sinx}{1+2cos^2x-1+cosx}=\frac{sinx\left(2cosx+1\right)}{cosx\left(2cosx+1\right)}=\frac{sinx}{cosx}=tanx\)
\(B=\frac{cosa}{sina}\left(\frac{1+sin^2a}{cosa}-cosa\right)=\frac{cosa}{sina}\left(\frac{1+sin^2a-cos^2a}{cosa}\right)=\frac{cosa}{sina}.\frac{2sin^2a}{cosa}=2sina\)
\(C=\frac{1+cos2x+cosx+cos3x}{2cos^2x-1+cosx}=\frac{1+2cos^2x-1+2cos2x.cosx}{cos2x+cosx}=\frac{2cosx\left(cosx+cos2x\right)}{cos2x+cosx}=2cosx\)
\(D=\frac{2sinx.cosx.\left(-tanx\right)}{-tanx.sinx}-2cosx=2cosx-2cosx=0\)
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\(E=cos^2x.cot^2x-cot^2x+cos^2x+2cos^2x+2sin^2x\)
\(E=cot^2x\left(cos^2x-1\right)+cos^2x+2=\frac{cos^2x}{sin^2x}\left(-sin^2x\right)+cos^2x+2=2\)
\(F=\frac{sin^2x\left(1+tan^2x\right)}{cos^2x\left(1+tan^2x\right)}=\frac{sin^2x}{cos^2x}=tan^2x\)
Câu G mẫu số có gì đó sai sai, sao lại là \(2sina-sina?\)
\(H=sin^4\left(\frac{\pi}{2}+a\right)-cos^4\left(\frac{3\pi}{2}-a\right)+1=cos^4a-sin^4a+1\)
\(=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+1=cos^2a-\left(1-cos^2a\right)+1=2cos^2a\)
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Bài 2:
\(sin\frac{A+B}{2}=sin\left(\frac{180^0-C}{2}\right)=sin\left(90^0-\frac{C}{2}\right)=cos\frac{C}{2}\)
b/
\(A=cosx+cos\left(x+\frac{2\pi}{3}\right)+cos\left(x+\frac{4\pi}{3}\right)=cosx+2cos\left(x+\pi\right).cos\frac{\pi}{3}\)
\(=cosx-2cosx.\frac{1}{2}=0\)
c/
\(sinA+sinB+sinC=2sin\frac{A+B}{2}cos\frac{A-B}{2}+2sin\frac{C}{2}cos\frac{C}{2}=2cos\frac{C}{2}.cos\frac{A-B}{2}+2sin\frac{C}{2}cos\frac{C}{2}\)
\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+sin\frac{C}{2}\right)=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+cos\frac{A+B}{2}\right)=4cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}\)
d/ \(\frac{cos2a}{1+sin2a}=\frac{cos^2a-sin^2a}{cos^2a+sin^2a+2sina.cosa}=\frac{\left(cosa-sina\right)\left(cosa+sina\right)}{\left(cosa+sina\right)^2}=\frac{cosa-sina}{cosa+sina}\)
e/
\(E=\frac{sina+cosa}{cos^3a}=\frac{1}{cos^2a}\left(tana+1\right)=\left(1+tan^2a\right)\left(tana+1\right)\)
\(E=tan^3a+tan^2a+tana+1\)
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Tìm tập xác đinh của các hàm số sau
29 , yfrac{tanx+cosx}{sinx}
30 , yfrac{1}{sinx}-frac{1}{cosx}
31 , yfrac{cosx+cotx}{sinx}
32 , yfrac{tanx+cotx}{1-sin2x}
33 , ytanx+frac{1}{cosfrac{x}{2}}
34 , yfrac{1-tanx}{1-cotx}
35 , yfrac{cotx}{cosx-1}
36 , yfrac{3}{sin^2x-cos^2x}
37 , yfrac{2}{cosx-cos3x}
38 , yfrac{sqrt{x}}{sinpi x}
39 , yfrac{2-cosx}{1+tanleft(x-frac{pi}{3}right)}
Đọc tiếp
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.
\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\\sin2x\ne1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}sin2x\ne0\\sin2x\ne1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{k\pi}{2}\\x\ne\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
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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.
\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne1\end{matrix}\right.\) \(\Leftrightarrow sinx\ne0\)
\(\Leftrightarrow x\ne k\pi\)
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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.
\(\left\{{}\begin{matrix}cos\left(x-\frac{\pi}{3}\right)\ne0\\tan\left(x-\frac{\pi}{3}\right)\ne-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-\frac{\pi}{3}\ne\frac{\pi}{2}+k\pi\\x-\frac{\pi}{3}\ne-\frac{\pi}{4}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{5\pi}{6}+k\pi\\x\ne-\frac{\pi}{12}+k\pi\end{matrix}\right.\)
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