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khỏi ghi lại đề nha
A=1-1/2+1/2-1/3+1/3-1/4+......+1/49-1/50
A=1-1/50
A=49/50
Bài 15 :
a) Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2020}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(A=1-\frac{1}{2020}=\frac{2019}{2020}< \frac{2020}{2020}=1\)
b) Ta có : \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}\)
\(2A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1001}}\)
\(2A-A=\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1001}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{1000}}\right)\)
\(A=\frac{1}{2^{1001}}-\frac{1}{2}\)
Tới đây là so sánh đi nhé
Cái này mình làm hôm qua rồi mà '-'
a) Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2020}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(A=\frac{1}{1}-\frac{1}{2020}=\frac{2019}{2020}\)
\(\Rightarrow A< 1\)
b) \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}\)
\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}\right)\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{999}}\)
\(2A-A=A\)
\(=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{999}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{1000}}\right)\)
\(=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^{999}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{1000}}\)
\(=1-\frac{1}{2^{1000}}\)
\(\Rightarrow A=1-\frac{1}{2^{1000}}< 1\left(đpcm\right)\)
\(\frac{-7}{11}.\frac{11}{19}+\frac{-7}{11}.\frac{8}{19}+\frac{-4}{11}\)
\(=\frac{-7}{11}.\left(\frac{11}{19}+\frac{8}{19}\right)+\frac{-4}{11}\)
\(=\frac{-7}{11}.1+\frac{-4}{11}\)
\(=\frac{-7}{11}+\frac{-4}{11}=\frac{-11}{11}=-1\)
~ Hok tốt ~
Đặt \(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)
\(\Rightarrow B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(\Rightarrow B=1-\frac{1}{2019}\)
\(\Rightarrow B=\frac{2018}{2019}\)
=1/1-1/2+1/2-1/3+1/3-1/4+.........+1/1999-1/2000
=1/1-1/2000
=1999/2000<3/4
a. \(\frac{1}{1.2}+...+\frac{1}{x.\left(x+1\right)}=99\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+...+\frac{1}{x}-\frac{1}{x+1}=99\)
\(\Rightarrow1-\frac{1}{x+1}=99\)
\(\Rightarrow\frac{1}{x+1}=1-99=-98\)
\(\Rightarrow x=\frac{1}{-98}-1\)
\(\Rightarrow x=-\frac{99}{98}\)
P/s : Bạn ơi đề sai, x sai hay mk sai ạ???
A, 1/(1.2)+...+1/[x(x+1)]=99
=>1-1/2+...+1/x+1/(x-1)=99
=>1-1/(x-1)=99
=>1/(x-1)=-98
=>1/(x-1)=-98/1
=>1.(-98)=(x-1).1(tích chéo)
=>x-1=-98
=>x=-97
Câu 3 : \(2+4+6+.........+2n=156\)
\(\Leftrightarrow2\left(1+2+3+.....+n\right)=156\)
\(\Leftrightarrow1+2+3+.........+n=78\)
\(\Leftrightarrow\frac{n\left(n+1\right)}{2}=78\)\(\Leftrightarrow n\left(n+1\right)=156=12.13\)\(\Leftrightarrow n=12\)
Vậy \(n=12\)
2) hc sinh giỏi lớp 6B là
35.40%=14(hs)
Số hc sinh khá lớp 6B là
14.\(\frac{9}{7}\)=17(hs)
Số hc sinh trung bình lớp 6B là
35-(14+17)=4(hs)
kl...
3)
Số hs trung bình là
1200.\(\frac{5}{8}\)=750 (hs)
Số hc sinh khá là
1200.\(\frac{1}{3}\)=400(hs)
Số hc sinh giỏi là
1200-750-400=50(hs)
kl....
a) 1/1.2 + 1/2.3 + ... + 1/2019.2020
= 1 - 1/2 + 1/2 - 1/3 + ... + 1/2019 - 1/2020
= 1 - 1/2020
= 2019/2020
b) 1/1.4 + 1/4.7 + ... + 1/100.103
= 1/3.(1 - 1/4 + 1/4 - 1/7 + ... + 1/100 - 1/103)
= 1/3.(1 - 1/103)
= 1/3.102/103
= 34/103
\(a,\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{2019.2020}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2019}-\frac{1}{2020}\)
\(=1-\frac{1}{2020}=\frac{2019}{2020}\)
\(b,\frac{1}{1.4}+\frac{1}{4.7}+....+\frac{1}{100.103}\)
\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{1}{4.7}+....+\frac{1}{100.103}\right)\)
\(=\frac{1}{3}.\left(1-\frac{1}{103}\right)=\frac{1}{3}.\frac{102}{103}=\frac{34}{103}\)