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A = \(\dfrac{1}{2021.2022}\) + \(\dfrac{1}{2022.2023}\) + \(\dfrac{1}{2023.2024}\) + \(\dfrac{1}{2024.2025}\) - \(\dfrac{4}{2021.2025}\)
A = \(\dfrac{1}{2021}\) - \(\dfrac{1}{2022}\) + \(\dfrac{1}{2022}\) - \(\dfrac{1}{2023}\) + \(\dfrac{1}{2023}\) - \(\dfrac{1}{2024}\) + \(\dfrac{1}{2024}\) - \(\dfrac{1}{2025}\) - \(\dfrac{1}{2021}\) + \(\dfrac{1}{2025}\)
A = (\(\dfrac{1}{2021}\) - \(\dfrac{1}{2021}\)) + (\(\dfrac{1}{2022}\) - \(\dfrac{1}{2022}\)) + (\(\dfrac{1}{2023}\) - \(\dfrac{1}{2023}\)) + (\(\dfrac{1}{2024}\) - \(\dfrac{1}{2024}\)) + (\(\dfrac{1}{2025}\) - \(\dfrac{1}{2025}\))
A = 0 + 0 +0 + 0+ ... + 0
A = 0
a, 1 - 2 - 3 + 4 + 5 + 6 - 7 + ... + 1996 + 1997 - 1998 - 1999 + 2000 + 2001
= ( 1 - 2 - 3 + 4 ) + ( 5 - 6 - 7 + 8 ) + ... +( 1993 - 1994 - 1995 + 1996 ) + ( 1997 - 1998 - 1999 + 2000 ) + 2001
= 0 + 0 + ... + 0 + 0 + 2001
= 2001
b, 1 - 3 + 5 -7 + ...+ 2001 - 2003 + 2005
= ( 1 -3 ) + ( 5 - 7 ) + ... + ( 2001 - 2003 ) + 2005
= -2 + (-2 )+ ... + (-2) + 2005
Có 501 số ( - 2 )
= ( - 2 ) . 501 + 2005
= -1002 + 2005
= 1003
P/s : Tham khảo ( Chép đầu bài thôi cũng sai )
e) \(1997^{1^{9^{9^7}}}\)
= \(1997^{1^{9^{...1}}}\)
=\(1997^{1^{...9}}\)
=\(1997^1\)
= 1997
f) ( 3333)33
= 331089
= ( 33 4 ) 272 . 33
= ( ...1 ) 33
= ...3
g) 375735
= ...5
( 144 ) 68
= ...6
Học tốt!!!!
ta có
\(S_2=\left(1-3\right)+\left(5-7\right)+..+\left(1997-1999\right)+2001\)
ha y \(S_2=-2-2-2..+2001=-2.500+2001=1001\)
\(S_3=\left(1-2-3+4\right)+\left(5-6-7+8\right)+..+\left(1997-1998-1999+2002\right)\)
hay \(S_3=0+0+..+0=0\)
\(S_2=\left(1-3\right)+\left(5-7\right)+...+\left(1997-1999\right)+2001\)
\(=\left(-2\right)+\left(-2\right)+....+\left(-2\right)+2001=\left(-2\right).500+2001=-1000+2001=1001\)
\(S_3=\left(0+1-2-3\right)+\left(4+5-6-7\right)+...+\left(1996+1997-1998-1999\right)+2000\)
\(=-4+\left(-4\right)+...+\left(-4\right)+2000=\left(-4\right).500+2000=0\)
sách Nâng cao phát triển lớp 6 tập 1