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a) (-21) + |-50| + (-29) – |-2016|
= (-21) + 50 + (-29) – 2016
= [(-21) + (-29) + 50] – 2016 = ( -50 +50 ) – 2016
= 0 – 2016 = - 2016 .
b) 36 : 3 2 + 3 2 . 2 3 - 15 0
= 36 : 9 + 9 . 8 – 1 = 4 + 72 – 1 = 76 – 1 = 75
c) ( 5 103 – 5 102 – 5 101 ) : ( 5 99 . 26 – 5 99 )
= ( 5 103 – 5 102 – 5 101 ) : 5 99 . ( 26 – 1 )
= ( 5 103 – 5 102 – 5 101 ) : ( 5 99 . 25 )
= 5 101 ( 5 2 – 5 1 – 5 0 ) : ( 5 99 . 5 2 )
= ( 5 101 . 19 ) : 5 101 = 19
a) (-21) + | -50 | + (-29) – | -2016 |
= -21 + 50 - 29 -2016
= (-21 - 29 +50 ) -2016
= -2016
b) 36 : 32 + 32 . 23 -150 ( xem lại đề nhé )
c) ( 5103 – 5102 – 5101) : ( 599 . 26 – 599)
= ( 1 - 5101 ) : ( 599 . ( 26 - 1 ) )
= - 5100 : (599 . 25)
= \(-\frac{204}{599}\)
c) ( 5 103 - 5 102 - 5 101 ) : ( 5 99 . 26 - 5 99 )
= ( 5 103 - 5 102 - 5 101 ) : [ 5 99 . ( 26 – 1)]
= ( 5 103 - 5 102 - 5 101 ) : ( 5 99 . 25 )
= 5 101 ( 52 – 51 – 50) : ( 5 99 . 52 )
= ( 5 101 . 19 ) : 5 101 = 19
\(A=5^2+5^4+5^6+...+5^{100}+5^{102}\\ =5^2.\left(1+5^2+5^4+...+5^{98}+5^{100}\right)\\ =25.\left(1+5^2+5^4+...+5^{98}+5^{100}\right)⋮25\)
\(4+4\cdot5+4\cdot5^2+...+4\cdot5^{100}\\ =4\left(1+5+5^2+...+5^{100}\right)\left(1\right)\)
Đặt \(A=1+5+5^2+...+5^{100}\)
\(\Leftrightarrow5A=5+5^2+...+5^{101}\\ \Leftrightarrow4A=5^{101}-1\\ \Leftrightarrow A=\dfrac{5^{101}-1}{4}\)
Thay vào (1)
\(\left(1\right)=4\cdot\dfrac{5^{101}-1}{4}=5^{101}-1:5^{101}-1=1\)
Vậy \(4+4\cdot5+4\cdot5^2+...+4\cdot5^{100}:5^{101}-1=1\)
\(F=5+5^3+5^5+...+5^{101}\)
=>\(25F=5^3+5^5+...+5^{103}\)
=>\(25F-F=5^3+5^5+...+5^{103}-5-5^3-...-5^{101}\)
=>\(24F=5^{103}-5\)
=>\(F=\dfrac{5^{103}-5}{24}\)
a, 4 5 : 4 = 4 4
b, 2 10 : 2 3 = 2 7
c, x 9 : x 3 = x 6 x ≠ 0
d, 5 103 : 5 3 = 5 100
\(B=1+22+24+26+28+...+2200\)
\(=1+\dfrac{\left(2200+22\right).\left[\left(2200-22\right):2+1\right]}{2}\)
\(=1+\dfrac{2222.1090}{2}\)
\(=1+1210990\)
\(=1210991\)
\(C=5+53+55+57+...+5101\)
\(=5+\dfrac{\left(5101+53\right).\left[\left(5101-53\right):2+1\right]}{2}\)
\(=5+\dfrac{5154.2525}{2}\)
\(=5+6506925\)
\(=6506930\)