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1/2 + 1/6 + 1/12 + 1/20 + ... + 1/9900
= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/99.100
= 1 - 1/2 + 1/2 - 1/2 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
Mk nhanh nhất đó
Đúng 100%
Tk mk mk tk lại
Cảm ơn bạn nhiều
Thank you very much
( ^ _ ^ )
99/100
Buổi chiều hôm nay cô giáo mới dạy cho mình mà nên mình chắc chắn 100%
T = \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}=\)
T = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}=\)
T = \(\frac{1}{1}-\frac{1}{100}=\)
T = \(\frac{99}{100}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}\)
\(=1-\frac{1}{5}\)
\(=\frac{4}{5}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Ta có \(\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{9900}\right):x=\frac{1}{5}\)
\(\left(\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}+...+\frac{1}{99\times100}\right):x=\frac{1}{5}\)
\(\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right):x=\frac{1}{5}\)
\(\left(\frac{1}{3}-\frac{1}{100}\right):x=\frac{1}{5}\)
\(\frac{97}{300}:x=\frac{1}{5}\)
\(x=\frac{97}{300}:\frac{1}{5}\)
\(x=\frac{97}{60}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
1/2 + 1/6 + 1/12 + ... + 1/9900
1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +... + 1/99 - 1/100
1/1 - 1/100
= 99/100
A = \(\dfrac{1}{12}\)+ \(\dfrac{1}{20}\)+ \(\dfrac{1}{30}\)+...+\(\dfrac{1}{9900}\)
A = \(\dfrac{1}{3\times4}\)+ \(\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{99\times100}\)
A = \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
A = \(\dfrac{1}{3}\) - \(\dfrac{1}{100}\)
A = \(\dfrac{97}{300}\)
Lời giải:
Gọi tổng trên là $A$
$A=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{99.100}$
$=\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}$
$=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}$
$=\frac{1}{3}-\frac{1}{100}=\frac{97}{300}$
A) bạn xem lại đề ạ
B) 1/2 + 1/6 + 1/ 12 + 1/ 20 + ...+ 1/ 9900
=1/2+1/6+1/12+...+1/9900
=1/1.2+1/2.3+1/3.4+...+1/99.100
=1/1-1/2+1/2-1/3+...+1/99-1/100
=1/1-1/100
=99/100
C) Biến đổi tử số và mẫu số ta có
- Tử số: 20,2 x 5,1 - 30,3 x 3,4 + 14,58
= 103,02 - 103,02 + 14,58
= 14,58
- Mẫu số: 14,58 x 460 + 7,29 x 540 x 2
= 14,58 x 460 + 14,58 x 540
= 14,58 x (460 + 540)
= 14,58 x 1000
= 14580
Thay vào ta có: = 14,58 : 14580
= 0,001
Vậy 20.2*5.1-30.3*3.4+14.56/ 14.58*460+7.29 *540*2 = 0,001.
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\frac{1}{54}+\frac{1}{162}\)
\(\Rightarrow A=\frac{1}{2}\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\right)\)
Gọi \(B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
\(\Rightarrow\frac{1}{3}B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow B-\frac{1}{3}B=1-\frac{1}{243}\)
\(\Rightarrow\frac{2}{3}B=\frac{242}{243}\)
\(\Rightarrow B=\frac{121}{81}\)
Suy ra \(A=\frac{1}{2}B=\frac{1}{2}.\frac{121}{81}=\frac{121}{162}\)
bài toán này làm = cách đứt ruột đó
thầy mk vừa dạy mk xong
Ta có: \(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)