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DV
1
Q
9 tháng 4 2017
a) \(P=\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+...\dfrac{10}{46.56}\)
\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...\dfrac{1}{46}-\dfrac{1}{56}\)
\(P=1-\dfrac{1}{56}\)
\(P=\dfrac{55}{56}\)
b) \(A=\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+...+\dfrac{3}{99.100}\)
\(A=3\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)
\(A=3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=3\left(1-\dfrac{1}{100}\right)\)
\(A=3.\dfrac{99}{100}\)
\(A=\dfrac{297}{100}\)
c) \(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\)
\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(B=1-\dfrac{1}{103}\)
\(B=\dfrac{102}{103}\)
d) \(C=\dfrac{5}{1.4}+\dfrac{5}{4.7}+\dfrac{5}{7.10}+...+\dfrac{5}{100.103}\)
\(C=\dfrac{5}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\right)\)
\(C=\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(C=\dfrac{5}{3}\left(1-\dfrac{1}{103}\right)\)
\(C=\dfrac{5}{3}.\dfrac{102}{103}\)
\(C=\dfrac{170}{103}\)
e) \(D=\dfrac{7}{1.5}+\dfrac{7}{5.9}+\dfrac{7}{9.13}+...+\dfrac{7}{101.105}\)
\(D=\dfrac{7}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{101.105}\right)\)
\(D=\dfrac{7}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{101}-\dfrac{1}{105}\right)\)
\(D=\dfrac{7}{4}\left(1-\dfrac{1}{105}\right)\)
\(D=\dfrac{7}{4}.\dfrac{104}{105}\)
\(D=\dfrac{26}{15}\)
D
9 tháng 4 2017
a)\(P=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{46}-\frac{1}{56}\)
=\(1-\frac{1}{56}=\frac{55}{56}\)
b)\(A.\frac{1}{3}=\frac{1}{3}.\left(\frac{3}{1.2}+\frac{3}{2.3}+....+\frac{3}{99.100}\right)\)
= \(\frac{1}{3}A=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{3}{99.100}\)
=> \(\frac{1}{3}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
=> \(\frac{1}{3}A=1-\frac{1}{100}=\frac{99}{100}\)
=> \(A=\frac{99}{100}.3=\frac{297}{100}\)
c)\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)
=\(1-\frac{1}{103}=\frac{102}{103}\)
d) \(\frac{3}{5}C=\frac{3}{5}.\left(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\right)\)
=\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\)
=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{103}\)
=\(1-\frac{1}{103}=\frac{102}{103}\)
=>\(C=\frac{102}{103}.\frac{5}{3}=\frac{170}{103}\)
e) \(\frac{4}{7}D=\frac{4}{7}.\left(\frac{7}{1.5}+\frac{7}{5.9}+...+\frac{7}{101.105}\right)\)
=\(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{101.105}\)
=\(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{105}\)
=\(1-\frac{1}{105}=\frac{104}{105}\)
=< D=\(\frac{104}{105}.\frac{7}{4}=\frac{26}{15}\)
20 tháng 4 2015
S=\(\frac{1}{1}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{10}\)+...+\(\frac{1}{100}\)-\(\frac{1}{103}\)+\(\frac{1}{103}\)-\(\frac{1}{104}\)+\(\frac{1}{104}\)-\(\frac{1}{105}\)+\(\frac{1}{105}\)-\(\frac{1}{106}\)+\(\frac{1}{106}\)-\(\frac{1}{107}\)
S=1-\(\frac{1}{107}\)
S=\(\frac{106}{107}\)
(Ở đề bài, ở phân số cuối cùng 1/106+107 nên sửa lại thành 1/106.107 sẽ được kết quả như trên)
NV
20 tháng 4 2015
Ta có: \(S=\frac{1}{1}-\frac{1}{103}+\frac{1}{103}-\frac{1}{107}\)
\(S=1-\frac{1}{107}=\frac{106}{107}\)
18 tháng 4 2015
Trong trường hợp bn viết nhầm 1/106.107 chứ ko phải 1/106+107
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}+\frac{1}{103.104}+\frac{1}{104.105}+\frac{1}{105.106}+\frac{1}{106.107}\)
\(S=\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)+\left(\frac{1}{103}-\frac{1}{104}+\frac{1}{104}-\frac{1}{105}+\frac{1}{105}-\frac{1}{106}+\frac{1}{106}-\frac{1}{107}\right)\)
\(S=\left(1-\frac{1}{103}\right)+\left(\frac{1}{103}-\frac{1}{107}\right)\)
\(S=\frac{102}{103}+\frac{4}{11021}\)
\(S=\frac{106}{107}\)
18 tháng 4 2015
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}+\frac{1}{103.104}+\frac{1}{104.105}+\frac{1}{105.106}+\frac{1}{106+107}\)
\(S=\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)+\left(\frac{1}{103}-\frac{1}{104}+\frac{1}{104}-\frac{1}{105}+\frac{1}{105}-\frac{1}{106}\right)+\frac{1}{106+107}\)
\(S=\left(1-\frac{1}{103}\right)+\left(\frac{1}{103}-\frac{1}{106}\right)+\frac{1}{106+107}\)
\(S=\frac{102}{103}+\frac{3}{10918}+\frac{11343}{106}\)
\(S=108\)
LT
0
NT
2
29 tháng 4 2023
\(\dfrac{3}{2}\)B= \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\)
\(\dfrac{3}{2}\)B= \(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\) \(\dfrac{3}{2}\)B= \(\dfrac{102}{103}\) \(\)B= \(\dfrac{102}{103}:\dfrac{3}{2}\) B=\(\dfrac{68}{103}\)
TH
4
6 tháng 4 2017
\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{100.103}\)
\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(=\frac{5}{3}\left(1-\frac{1}{103}\right)\)
\(=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)
6 tháng 4 2017
\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{100.103}=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 6 2019
\(A=\frac{2}{4.9}+\frac{2}{9.14}+\frac{2}{14.19}+...+\frac{2}{504.509}\)
\(A=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{504}-\frac{1}{509}\right)\)
\(A=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{509}\right)=...\)
\(B=\frac{1.4+1}{1.4}+\frac{4.7+1}{4.7}+\frac{7.10+1}{7.10}+...+\frac{100.103+1}{100.103}\)
\(B=1+\frac{1}{1.4}+1+\frac{1}{4.7}+...+1+\frac{1}{100.103}\)
\(B=34+\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(B=34+\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(B=34+\frac{1}{3}\left(1-\frac{1}{103}\right)=...\)
Ta có :
\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{100.103}\)
\(=\)\(\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)
\(=\)\(\frac{2}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(=\)\(\frac{2}{3}\left(1-\frac{1}{103}\right)\)
\(=\)\(\frac{2}{3}.\frac{102}{103}\)
\(=\)\(\frac{68}{103}\)
Chúc bạn học tốt ~