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đặt 22018 = a ; 32019 = b ; 52020 = c
Ta có : \(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)
\(B=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(2B=\frac{2}{1.2}+\frac{2}{3.4}+...+\frac{2}{2019.2020}\)
\(< 1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}+\frac{1}{2019.2020}\)
\(2B< 1+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2019-2018}{2018.2019}+\frac{2020-2019}{2019.2020}\)
\(2B< 1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}=1+\frac{1}{2}-\frac{1}{2020}< 1+\frac{1}{2}\)
\(B< \frac{3}{4}\)
\(\Rightarrow A>1>\frac{3}{4}>B\)
Mình chỉ biết cách tính B thôi, đây nhé:
B= \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2019.2020}\)
B=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(B=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-2\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)
\(B=\frac{1}{1011}+\frac{1}{1012}+....+\frac{1}{2019}+\frac{1}{2020}\)
\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1.\)
Với : \(a=2^{2018};.b=3^{2019};,c=5^{2020}.\)
Và : \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\Leftrightarrow\)
\(B=1-\frac{1}{2020}< 1< A\)
B= 1/1.2+1/2.3+...+1/2019.2020
B=1/1-1/2+1/2-1/3+...+1/2019-1/2020
B=1-1/2020=2020/2020-1/2020=2019/2020
\(A=\frac{2019^{2020}+1}{2019^{2021}+1}\)và \(B=\frac{2019^{2018}+1}{2019^{2019}+1}\)
Xét \(A=\frac{2019^{2020}+1}{2019^{2021}+1}\Rightarrow2019A=\frac{2019^{2021}+2019}{2019^{2021}+1}=1+\frac{2019}{2019^{2021}+1}\)
Xét \(B=\frac{2019^{2018}+1}{2019^{2019}+1}\Rightarrow2019B=\frac{2019^{2019}+2019}{2019^{2019}+1}=1+\frac{2018}{2019^{2019}+1}\)
Vì \(1+\frac{2018}{2019^{2021}+1}< 1+\frac{2018}{2019^{2019}+1}\Rightarrow\frac{2019^{2020}+1}{2019^{2021}+1}< \frac{2018^{2019}+1}{2019^{2019}+1}\)
\(\Rightarrow A< B\)
Ta có:
\(A=\frac{2019^{2020}+1}{2019^{2021}+1}\)
\(\Rightarrow2019A=\frac{2019^{2021}+2019}{2019^{2021}+1}\)
\(\Rightarrow2019A=1+\frac{2019}{2019^{2021}+1}\)
\(\Rightarrow A=1+\frac{2019}{2019^{2021}+1}:2019\)
Ta lại có:
\(B=\frac{2019^{2018}+1}{2019^{2019}+1}\)
\(\Rightarrow2019B=\frac{2019^{2019}+2019}{2019^{2019}+1}\)
\(\Rightarrow2019B=1+\frac{2019}{2019^{2019}+1}\)
\(\Rightarrow B=1+\frac{2019}{2019^{2019}+1}:2019\)
Do \(2019^{2021}+1>2019^{2019}+1\)
\(\Rightarrow\frac{2019}{2019^{2021}+1}< \frac{2019}{2019^{2019}+1}\)
\(\Rightarrow1+\frac{2019}{2019^{2021}+1}:2019< 1+\frac{2019}{2019^{2019}+1}:2019\)
\(\Rightarrow A< B\)
Vậy \(A< B.\)
a) Ta có :
N = 2018 + 2019/2019 + 2020
= 2018/2019 + 2020 + 2019/2019 + 2020
Ta thấy : 2018/2019 + 2020 < 2018/2019 ( Vì 2019 + 2020 > 2019 )
2019/2019 + 2020 < 2019/2020 ( Vì 2019 + 2020 > 2020 )
=> 2018/2019 + 2020 + 2019/2019 + 2020 < 2018/2019 + 2019/2020
=> M > N
b) Mk ko bt làm !!
c) Ta có :
19/31 > 1/2
17/35 < 1/2
=> 19/31 > 17/35
d) Ta có :
3535/3434 = 1 + 1/3534
2323/2322 = 1 + 1/2322
Ta thấy :
1/3534 < 1/2322 ( Vì 3534 > 2322 )
=> 1 + 1/3534 < 1 + 1/2322
=> 3535/3534 < 2323/2322
Hok tốt !
Lời giải:
\(B=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{2019.2020}\)
\(\Rightarrow 2B=\frac{2}{1.2}+\frac{2}{3.4}+\frac{2}{5.6}+....+\frac{2}{2019.2020}\)
\(< 1+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{2018.2019}+\frac{1}{2019.2020}\)
\(2B< 1+\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+....+\frac{2019-2018}{2018.2019}+\frac{2020-2019}{2019.2020}\)
\(2B< 1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\( 2B< 1+\frac{1}{2}-\frac{1}{2020}< 1+\frac{1}{2}\)
\(B< \frac{3}{4}\)
---------------------
Đặt \(2^{2018}=a; 3^{2019}=b; 5^{2020}=c(a,b,c>0)\)
\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}> \frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)
\(\Rightarrow A>1> \frac{3}{4}> B\)
\(A=\frac{2017^{2018}+1}{2017^{2018}-3}\)\(=\frac{2017^{2018}-3+4}{2017^{2018}-3}\)\(=1+\frac{4}{2017^{2018}-3}\)
\(B=\frac{2017^{2018}-1}{2017^{2018}-5}=\frac{2017^{2018}-5+4}{2017^{2018}-5}\)\(=1+\frac{4}{2017^{2018}-5}\)
Vì \(2017^{2018}-3>2017^{2018}-5\)(vì cái nào trừ đi ít thì còn nhiều,cái nào trừ đi nhiều thì còn ít)
\(\Rightarrow1+\frac{4}{2017^{2018}-3}< 1+\frac{4}{2017^{2018}-5}\)(vì trong 2 phân số cùng tử, phân số nào có mẫu nhỏ hơn thì lớn hơn)
\(\Rightarrow A< B\)
Mình sửa lại đề bài nha!Đề của mình mới đúng!CHÚC BẠN HỌC TỐT!
Ta có :
A = \(\frac{2017^{2018}}{2017^{2018}}+\frac{1}{-3}\)= 1 + \(\frac{1}{-3}\)
B = \(\frac{2017^{2018}-1}{2017^{2018}-5}\)= \(\frac{2017^{2018}-5}{2018^{2018}-5}+\frac{4}{2017^{2018}-5}\)= 1 + \(\frac{4}{2017^{2018}-5}\)
Mà 1 + \(\frac{4}{2017^{2018}-5}\)> 1 + \(\frac{1}{-3}\)Do đó A < B
Vậy A < B
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