\(B=\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)=\dfrac{5}{3}\left(\dfrac{102}{103}\right)=\dfrac{170}{103}\)

\(\frac{5}{1.4}+\frac{5}{4.7}+................+\frac{5}{100.103}\)

\(\frac{1}{3}.\left(5-\frac{5}{4}+\frac{5}{4}-\frac{5}{7}+..............+\frac{5}{100}-\frac{5}{103}\right)\)

\(\frac{1}{3}.\left(5-\frac{5}{103}\right)\)

\(\frac{1}{3}.\left(\frac{510}{103}\right)=\frac{170}{103}\)

\(=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

\(=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{....}{....}\)

xin lỗi mình ko có máy tính nên ko tính đc phép tính cuối

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) \(D=\dfrac{7}{1.5}+\dfrac{7}{5.9}+\dfrac{7}{9.13}+...+\dfrac{7}{101.105}\)

\(D=\dfrac{7}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{101.105}\right)\)

\(D=\dfrac{7}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{101}-\dfrac{1}{105}\right)\)

\(D=\dfrac{7}{4}\left(1-\dfrac{1}{105}\right)\)

\(D=\dfrac{7}{4}.\dfrac{104}{105}\)

\(D=\dfrac{26}{15}\)

=5/3.(1/1-1/4+1/4-1/7+...+1/100-1/103)

=5/3.(1/1-1/103)

=5/3.102/103

=170/103

=5/3.(1/1-1/4+1/4-1/7+...+1/100-1/103)

=5/3.(1/1-1/103)

=5/3.102/103

=170/103

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)

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=\(\frac{104}{105}.\frac{7}{4}=\frac{26}{15}\)

S = \(5-\frac{5}{4}+\frac{5}{4}-\frac{5}{7}+.......+\frac{5}{100}-\frac{5}{103}\)

S = \(5-\frac{5}{103}\)

S = \(\frac{510}{103}\)

B= 5/3 .(1-1/4+1/4-1/7+........+1/100-1/103 )

B= 5/3 . (1-1/103 )

B=5/3 .102 /103 =170 /103

 Đ lm tương tự nha.......

\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)

\(3B=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{100.103}\right)\)

\(3B=5\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(3B=5\left(1-\frac{1}{103}\right)\)

\(3B=5.\frac{102}{103}\)

\(3B=\frac{510}{103}\)

\(\Rightarrow B=\frac{170}{103}\)

\(D=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{36.41}\)

\(\Rightarrow D=5\left(\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{36.41}\right)\)

\(D=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{36}-\frac{1}{41}\right)\)

\(D=5\left(1-\frac{1}{41}\right)\)

\(D=5.\frac{40}{41}\)

\(D=\frac{200}{41}\)

\(B=\dfrac{5}{1.4}+\dfrac{5}{4.7}+.....+\dfrac{5}{100.103}\)
\(B=\dfrac{5}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\right)\)
\(B=\dfrac{5}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(B=\dfrac{5}{3}.\left(1-\dfrac{1}{103}\right)\)
\(B=\dfrac{5}{3}.\dfrac{102}{103}\)
\(B=1\dfrac{67}{103}\)

Ta có: \(B=\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\)

\(=\dfrac{5}{3}\left(1-\dfrac{1}{4}\right)+\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}\right)+...+\dfrac{5}{3}\left(\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{5}{3}\left(1-\dfrac{1}{103}\right)\)

\(=\dfrac{5}{3}.\dfrac{102}{103}=\dfrac{170}{103}\)

Vậy \(B=\dfrac{170}{103}\).

Ta có : B = \(\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\)

\(\Rightarrow B=\dfrac{5}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\right)\)

\(\Rightarrow B=\dfrac{5}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(\Rightarrow B=\dfrac{5}{3}.\left(1-\dfrac{1}{103}\right)\)

\(\Rightarrow B=\dfrac{5}{3}.\dfrac{102}{103}\)

\(\Rightarrow B=\dfrac{170}{103}\)

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