so sánh các số
a. A=1+2+22 + ......+ 250; B= 251
b. 2300 và 3200
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Ta có: A=212121/202020=21/20
Ta thấy: A=21/20=1+1/20
B=22/21=1+1/21
Vì 1/20>1/21 nên 21/20>22/21
Vậy A>B
a: 3/5>-19/5
b: 8/7<8/3
c: 3/4=15/20
2/5=8/20
mà 15>8
nên 3/4>2/5
d: -3/5=-18/30
-4/6=-20/30
mà -18>-20
nên -3/5>-4/6
a: 3/5>-19/5
b: 8/7<8/3
c: 3/4=15/20
2/5=8/20
mà 15>8
nên 3/4>2/5
d: -3/5=-18/30
-4/6=-20/30
mà -18>-20
nên -3/5>-4/6
a.2021/2023 < 2017/2019
b.2005/2007 > 2009/2011
Giải thích : So sánh mẫu số, phân số có mẫu số bé hơn thì nó lớn hơn
2:
=1-1+1-1=0
3:
a: =>34*(100+1)/2:a=17
=>a=101
b: =>5/3(x-1/2)=5/4
=>x-1/2=5/4:5/3=3/4
=>x=5/4
1a, \(\dfrac{2005}{2001}\) = 1+\(\dfrac{4}{2001}\); \(\dfrac{2009}{2005}\)=1+\(\dfrac{4}{2005}\)vì\(\dfrac{4}{2001}\)>\(\dfrac{4}{2005}\)nên\(\dfrac{2005}{2001}\)>\(\dfrac{2009}{2005}\)
1b,\(\dfrac{1313}{1515}\)=\(\dfrac{1313:101}{1515:101}\)= \(\dfrac{13}{15}\); \(\dfrac{131313}{151515}\)=\(\dfrac{131313:10101}{151515:10101}\)=\(\dfrac{13}{15}\)
Vậy \(\dfrac{13}{15}\)=\(\dfrac{1313}{1515}\)=\(\dfrac{131313}{151515}\)
Ta có \(A=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\)
\(2A=1+\dfrac{2}{2}+\dfrac{3}{2^2}+...+\dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\)
\(2A-A=\left(1+\dfrac{2}{2}+\dfrac{3}{2^2}+...+\dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\right)-\left(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\right)\)\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\) - \(\dfrac{2023}{2^{2023}}\)
Đặt B = \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\)
2B = \(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\)
2B - B = \(\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\right)\)B = 2 - \(\dfrac{1}{2^{2022}}\)
Suy ra A = 2 - \(\dfrac{1}{2^{2022}}\) - \(\dfrac{2023}{2^{2023}}\) < 2
Vậy A < 2
\(A=\dfrac{1}{2}+\dfrac{2}{2^{2}}+\dfrac{3}{2^{3}}+...+\dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\)
\(2A=1+\dfrac22+\dfrac3{2^2}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\\2A-A=\left(1+\dfrac22+\dfrac3{2^2}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\right)-\left(\dfrac12+\dfrac2{2^2}+\dfrac3{2^3}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\right)\\A=1+\dfrac12+\dfrac1{2^3}\ +\,.\!.\!.+\ \dfrac1{2^{2021}}+\dfrac1{2^{2022}}-\dfrac{2023}{2^{2023}}\\2\left(A+\dfrac{2023}{2^{2023}}\right)=2+1+\dfrac12+\dfrac1{2^2}\ +\,.\!.\!.+\ \dfrac1{2^{2020}}+\dfrac1{2^{2021}}\\A+\dfrac{2023}{2^{2023}}=2-\dfrac1{2^{2022}}\\A=2-\dfrac1{2^{2022}}+\dfrac{2023}{2^{2023}}<2\)
Bài 1. Rút gọn các phân số:
24/36 =2/3
12/81=4/27
125/250 =1/2
131313/151515=13/15
20052005/20062006= 2005/2006
Bài 2: So sánh các phân số sau;
3/8 và 1/3
Ta có: 3/8 = 9/24 ; 1/3 = 8/24
=> 3/8 > 1/3
2/5 và 4/7
Ta có: 2/5 = 14/35 ; 4/7 = 20/35
=> 2/5 < 4/7
2004/2005 và 2005/2006
Ta có: 2004/2005 = 1 - 2004/2005 = 1/2005
2005/2006 = 1 - 2005/2006 = 1/2006
=> 2004/2005 < 2005/2006
Bài 4 Tim y biết:
y : 2 + y + y : 3 + y : 4 = 25
=> y : 2 + y : 1 + y : 3 + y : 4 = 25
=> y : (2+1+3+4)=25
=> y : 10 = 25
=> y = 25 x 10
=> y = 250
Vậy y=250
b . <
a thì vẫn chưa ra
a) Lấy 2A - A ,được 2^51 - 1 < 2^51
=> A < B
b) 2^300 = (2^3)^100 = 8^100
3^200 = (3^2)^100 = 9^100
=> 2^300 < 3^200