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Bài 1.6
a) \(\cos14^0=\sin76^0\)
\(\cos87^0=\sin3^0\)
Do đó: \(\cos87^0< \sin47^0< \cos14^0< \sin78^0\)
b) \(\cot25^0=\tan65^0\)
\(\cot38^0=\tan52^0\)
Do đó: \(\cot38^0< \tan62^0< \cot25^0< \tan73^0\)
1.1
a) \(\sin40^012'\simeq0,65\)
b) \(\cos52^054'\simeq0,603\)
c) \(\tan63^036'\simeq2,014\)
d) \(\cot25^018'\simeq2,116\)
e) \(\sin39^013'\simeq0,632\)
f) \(\cos52^018'\simeq0,612\)
g) \(\tan13^020'\simeq0,237\)
h) \(\cot10^017'\simeq5,511\)
i) \(\sin70^013'\simeq0,941\)
j) \(\cos25^032'\simeq0,902\)
k) \(\tan43^010'\simeq0,938\)
l) \(\cot32^015'\simeq1,585\)
A = 2014 . (2015+1) = 2014 . 2015 + 2014
B= 2015^2 = 2015(2014 + 1) = 2014 . 2015 +2015
Vì 2014<2015 => 2014.2015 + 2014 < 2014.2015 +2015
=> A< B
Vậy A<B
A = 2014 . (2015+1) = 2014 . 2015 + 2014
B= 2015^2 = 2015(2014 + 1) = 2014 . 2015 +2015
Vì 2014<2015 => 2014.2015 + 2014 < 2014.2015 +2015
=> A< B
Vậy A<B
\(2\sqrt{3+\sqrt{5}}=\sqrt{2}\cdot\sqrt{6+2\sqrt{5}}\)
\(=\sqrt{2}\cdot\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{2}\cdot\left(\sqrt{5}+1\right)\)
\(=\sqrt{10}+\sqrt{2}>\sqrt{10}+1\)
Vậy ....
1.3
a) \(\sin20^0< \sin70^0\)
b) \(\cos25^0>\cos63^015'\)
c) \(\tan73^020'>\tan45^0\)
d) \(\cot2^0>\cot37^040'\)
e) \(\tan45^0>\cos45^0\)
f) \(\cot32^0>\cos32^0\)
g) \(\tan25^0>\sin25^0\)
h) \(\cot60^0>\sin30^0\)
1.4
a) \(\dfrac{\sin25^0}{\cos65^0}=\dfrac{\sin25^0}{\sin25^0}=1\)
b) \(\tan58^0-\cot32^0=\cot32^0-\cot32^0=0\)