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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
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=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(B=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-2\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)
\(B=\frac{1}{1011}+\frac{1}{1012}+....+\frac{1}{2019}+\frac{1}{2020}\)
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 \(2^{2018}=a; 3^{2019}=b; 5^{2020}=c(a,b,c>0)\)
\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}> \frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)
\(\Rightarrow A>1> \frac{3}{4}> B\)
\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1.\)
Với : \(a=2^{2018};.b=3^{2019};,c=5^{2020}.\)
Và : \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\Leftrightarrow\)
\(B=1-\frac{1}{2020}< 1< A\)
HD: Vũ Phương Vy em chỉ cần đặt ts c rồi rút gọn
ko chép lại đề nha
=\(A=\frac{2\left(1-\frac{2}{19}+\frac{2}{23}\right)-\frac{1}{1010}}{3\left(1-\frac{1}{19}+\frac{1}{23}-\frac{1}{2020}\right)}\)\(.\frac{4\left(1-\frac{1}{29}+\frac{1}{41}\right)-\frac{1}{505}}{5\left(1-\frac{1}{29}+\frac{1}{41}\right)-\frac{1}{404}}\)
rồi em chỉ cần rút gọn tiếp
p/s đến đây thấy đề kì kì sao đó
em chek lại đề đc k
1 3/7-4/5=10/7-4/5=50/35-28/35=22/35
6/13x-3/10+2/5x4/13
=-9/65+8/65
=-1/65
câu đầu mik tính ra số to mà cx ko chắc là đúng nên mik ko viết
*chúc bn học tốt đạt nhiều điểm cao*
1/2.x-3/5=-4/5
1/2.x=-4/5+3/5
1/2.x=-1/5
x=-1/5:1/2
x=-2/5
kl:.....
câu đầu mik tính ra sốn to lắm
câu cuối mik tính ko chia hết nên chỉ làm đc câu giữa
Mk sửa đề nha :
20202020 x ( 710 : 78 - 3 x 24 - 22020 : 22020 )
= 20202020 x ( 72 - 48 - 20 )
= 20202020 x ( 49 - 48 - 1 )
= 20202020 x 0
= 0
Study well ! >_<
\(A=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^5}+...+\frac{2020}{5^{2020}}\)
\(\Rightarrow5A=1+\frac{2}{5}+\frac{3}{5^2}+\frac{4}{5^3}+...+\frac{2020}{5^{2019}}\)
\(\Rightarrow5A-A=4A=1+\left(\frac{2}{5}-\frac{1}{5}\right)+\left(\frac{3}{5^2}-\frac{2}{5^2}\right)+...+\left(\frac{2020}{5^{2019}}-\frac{2019}{5^{2019}}\right)-\frac{2020}{5^{2020}}\)
\(\Leftrightarrow4A=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2019}}-\frac{2020}{5^{2020}}\)
\(B=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2019}}\)
\(\Rightarrow5B=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2019}}\)
\(\Rightarrow4B=1-\frac{1}{5^{2019}}\)
\(\Rightarrow B=\frac{1}{4}-\frac{1}{4.5^{2019}}\)
\(\Rightarrow4A=1+B-\frac{2020}{5^{2020}}\)
\(\Rightarrow A=\frac{5}{16}-\frac{1}{5^{2019}}\left(\frac{1}{4}+\frac{2020}{5}\right)=\frac{5}{16}-\frac{1617}{4.5^{2019}}\)
\(16>\frac{1617}{4.5^{2019}}\Rightarrow A=\frac{1}{4}+\left(\frac{1}{16}-\frac{1617}{4.5^{2019}}\right)>\frac{1}{4}\)
\(\frac{5}{16}< \frac{1}{3}\Rightarrow A< \frac{1}{3}\)
\(\Rightarrow\frac{1}{4}< A< \frac{1}{3}\left(Đpcm\right)\)