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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) \(\tan58^0-\cot32^0=\cot32^0-\cot32^0=0\)
\(2\sqrt{3+\sqrt{5}}=\sqrt{2}\cdot\sqrt{6+2\sqrt{5}}\)
\(=\sqrt{2}\cdot\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{2}\cdot\left(\sqrt{5}+1\right)\)
\(=\sqrt{10}+\sqrt{2}>\sqrt{10}+1\)
Vậy ....
Dễ c/m đẳng thức: \(\left(n-1\right)\left(n+1\right)=n^2-1\)
Lúc đó: \(A=2014^2-1+2015^2-1=2014^2+2015^2-2=B\)
Vậy A = B
\(A=2013.2015+2014.2016\)
\(=\left(2015-2\right).2015+2014\left(2014+2\right)\)
\(=(2015^2-4030)+(2014^2+4028)\)
\(=\left(2015^2+2014^2\right)-\left(4030-4028\right)\)
\(=2014^2+2015^2-2\)
\(\Rightarrow A=B\)
a) A = 20152
B = 2014.2016 = ( 2015 - 1 ) . ( 2015 + 1 ) = 20152 - 1
Vì 20152 > 20152 - 1
=> A > B
b) C = 316 - 1
D = 8. ( 32 + 1 ).( 34 + 1 ). ( 38 + 1 )
= ( 32 - 1 ).( 32 + 1 ).( 34 + 1 ). ( 38 + 1 )
= ( 34 - 1 ).( 34 + 1 ). ( 38 + 1 )
= ( 38 - 1 ) . ( 38 + 1 )
= 316 - 1
Vì 316 - 1 = 316 - 1
=> C = D
A = 262 - 242 = (26 - 24)(26 + 24) = 2.50 = 100
B = 272 - 252 = (27 - 25)(27 + 25) = 2.52 = 104
\(A=26^2-24^2\)
\(A=\left(26-24\right)\left(26+24\right)\)
\(A=2.50\)
\(B=27^2-25^2\)
\(B=\left(27-25\right)\left(27+25\right)\)
\(B=2.52\)
\(2.50< 2.52\)
\(< =>A< B\)
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
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