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22 tháng 3 2018
Ta có : \(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{2}+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{4}+\frac{1}{4}\right)+...+\left(-\frac{1}{99}+\frac{1}{99}\right)-\frac{1}{100}\)
\(A=\frac{1}{2}+0+0+..+0-\frac{1}{100}\)
\(A=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(B=\frac{5}{1.4}+\frac{5}{4.7}+..+\frac{5}{100.103}\)
\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)
\(B=1+\left(-\frac{1}{4}+\frac{1}{4}\right)+\left(-\frac{1}{7}+\frac{1}{7}\right)+...+\left(-\frac{1}{100}+\frac{1}{100}\right)-\frac{1}{103}\)
\(B=1+0+0+...+0-\frac{1}{103}\)
\(B=1-\frac{1}{103}=\frac{102}{103}\)
So sánh : A < B vì 49/100 < 102/103 (49.103 < 102 . 100)
8 tháng 5 2015
\(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(2A=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)
\(2A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(2A=\frac{1}{1}-\frac{1}{100}\)
\(2A=\frac{99}{100}\Rightarrow A=\frac{99}{100}:2\Rightarrow A=\frac{99}{200}\)
Câu B và C làm tương tự.
8 tháng 5 2015
bạn Nhi làm sai rồi
\(\frac{2}{2\cdot3}\) sao có thể bằng \(\frac{1}{2}-\frac{1}{3}\) được
\(\frac{1}{2\cdot3}\) mới bằng \(\frac{1}{2}-\frac{1}{3}\)
kết quả là : \(\frac{49}{100}\)
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=\(\frac{104}{105}.\frac{7}{4}=\frac{26}{15}\)
PV
3
30 tháng 4 2017
a) \(A=\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\)
\(\Leftrightarrow A=\dfrac{5}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\right)\)
\(\Leftrightarrow\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(\Leftrightarrow\dfrac{5}{3}\left(1-\dfrac{1}{103}\right)\)
\(\Leftrightarrow\dfrac{5}{3}.\dfrac{102}{103}\)
\(\Leftrightarrow\) \(A=\dfrac{170}{103}\)
b) \(B=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)
\(B=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)
\(B=\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{51}\right)\)
\(B=\dfrac{1}{2}.\dfrac{16}{51}\)
\(B=\dfrac{8}{51}\)
2 tháng 6 2017
A = \(\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\)
A = \(\dfrac{5}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\right)\)
A = \(\dfrac{5}{3}.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-...-\dfrac{1}{100}+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
A = \(\dfrac{5}{3}.\left[\dfrac{1}{1}-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\left(\dfrac{1}{100}-\dfrac{1}{100}\right)-\dfrac{1}{103}\right]\)
A = \(\dfrac{5}{3}.\left[\dfrac{1}{1}-0-0-...-0-\dfrac{1}{103}\right]\)
A = \(\dfrac{5}{3}.\left[\dfrac{1}{1}-\dfrac{1}{103}\right]\)
A = \(\dfrac{5}{3}.\left[\dfrac{103}{103}-\dfrac{1}{103}\right]\)
A = \(\dfrac{5}{3}.\dfrac{102}{103}\)
A = \(\dfrac{170}{103}\)
B = \(\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)
B = \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)
B = \(\dfrac{1}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\)
B = \(\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...-\dfrac{1}{49}+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
B = \(\dfrac{1}{2}.\left[\dfrac{1}{3}-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\left(\dfrac{1}{49}-\dfrac{1}{49}\right)-\dfrac{1}{51}\right]\)
B = \(\dfrac{1}{2}.\left[\dfrac{1}{3}-0-0-...-0-\dfrac{1}{51}\right]\)
B = \(\dfrac{1}{2}.\left[\dfrac{1}{3}-\dfrac{1}{51}\right]\)
B = \(\dfrac{1}{2}.\left[\dfrac{17}{51}-\dfrac{1}{51}\right]\)
B = \(\dfrac{1}{2}.\dfrac{16}{51}\)
B = \(\dfrac{8}{51}\)
8 tháng 8 2018
Làm từng phần nha bạn
\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{298\cdot301}+x=\frac{299}{301}\)
Đặt \(A+x=\frac{299}{301}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{298}-\frac{1}{301}\)
\(A=1-\frac{1}{301}\)
\(A=\frac{300}{301}\)
=> \(\frac{300}{301}+x=\frac{299}{301}\)
\(x=\frac{299-300}{301}\)
\(x=-\frac{1}{301}\)
8 tháng 8 2018
\(A=5\cdot\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{301\cdot304}\right)\)
\(\frac{3A}{5}=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{301\cdot304}\)
\(\frac{3}{5}\cdot A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{301}-\frac{1}{304}\)
\(\frac{3}{5}\cdot A=1-\frac{1}{304}\)
\(\frac{3}{5}\cdot A=\frac{303}{304}\)
\(A=\frac{505}{304}\)
6 tháng 5 2018
A=2.(1/1.3 + 1/3.5 + 1/5.7 +.......+1/99.101)
=2.(1/1 + 1/3 + 1/5 + 1/5 + 1/7 +...+1/99 + 1/101)
=2.(1-1/101)
=2.(101/101-1/101)
=2.100/101
200/101
6 tháng 5 2018
B=2.(1/1.3+1/3.5+1/3.1+....+1/99.101)
=2.(1/1+1/3+1/3+1/5+1/3+1/7+....+1/99+1/101)
=2.(1/1+1/101)
=2.(101/101+1/101)
=2.102/101
=204/101
Q
22 tháng 4 2017
\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
\(3B=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{100.103}\right)\)
\(3B=5\left(1-\frac{1}{103}\right)\)
\(3B=5.\frac{102}{103}\)
\(3B=\frac{510}{103}\)
\(\Rightarrow B=\frac{170}{103}\)
22 tháng 4 2017
Ta có:
B=\(\frac{5}{1.4}\)+\(\frac{5}{4.7}+.....+\frac{5}{100.103}\)
B=\(\frac{5}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{100.103}\right)\)
B=\(\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{103}\right)\)
B=\(\frac{5}{3}\left(1-\frac{1}{103}\right)\)
B=\(\frac{5}{3}.\frac{102}{103}\)
B=\(\frac{170}{103}\)
Vậy B=\(\frac{170}{103}\)
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