B = 21450 nhé đúng 100% luôn

=21450

Hoàng chó

a, A = 1.2 + 2.3 + 3.4 + 4.5 + 5.6 + 6.7 + 7.8 + 8.9 + 9.10 + 10.11

= (1.2 + 2.3) + (3.4 + 4.5) + (5.6 + 6.7) + (7.8 + 8.9) + (9.10 + 10.11)

= 2( 1 + 3) + 4( 3 + 5) + 6( 5 + 7) + 8 ( 7 + 9) + 10( 9 + 11)

= 2.4 + 4.8 + 6.12 + 8.16 + 10.20 = 2.2.2 + 2.4.4 + 2.6.6 + 2.8.8 + 2.10.1

b,tương tự nhé

B=1/1.2+1/3.4+1/5.6+...+1/99.100

=1-1/2+1/3-1/4+1/5-1/6+...+1/99-1/100

=(1+1/3+1/5+...+1/99)-(1/2+1/4+1/6+...+1/100)

=(1+1/2+1/3+1/4+1/5+1/6+...+1/99+1/100)-2(1/2+1/4+1/6+...+1/100)

=(1+1/2+1/3+1/4+...+1/100)-(1+1/2+1/3+..+1/50)

=1/51+1/52+1/53+..+1/100 (1)

A=1/51+1/52+1/53+..+1/100 (2)

(1),(2)=> A/B=1

\(\frac{A}{B}=1\)

\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+\dfrac{8-7}{7.8}+\dfrac{9-8}{8.9}+\dfrac{10-9}{9.10}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\\ =1-\dfrac{1}{10}\\ =\dfrac{10-1}{10}=\dfrac{9}{10}\)

1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10

=2-1/1.2+3-2/2.3+4-3/3.4+...+10-9/9.10

=1-1/2+1/2-1/3+1/3-1/4+....+1/9-1/10

=1-1/10

=9/10

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}\)

\(=\dfrac{9}{10}\)

\(\frac{3}{2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{17.18}\)

\(=\frac{3.1}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{17.18}\)

\(=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{17.18}\right)\)

\(=3.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{17}-\frac{1}{18}\right)\)

\(=3.\left(1-\frac{1}{18}\right)\)

\(=3.\frac{17}{18}\)

\(=\frac{17}{6}\)

cùng hình nè

đặt A =\(\frac{3}{2}\)+\(\frac{3}{3x4}\)+...........+\(\frac{3}{17x18}\)

nhân cả 2 vế với \(\frac{1}{3}\)

A x \(\frac{1}{3}\)= \(\frac{1}{2}\)+\(\frac{1}{3x4}\)+..........+\(\frac{1}{17x18}\)

A x \(\frac{1}{3}\) =\(\frac{1}{2}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...........+\(\frac{1}{17}\)-\(\frac{1}{18}\)

A x \(\frac{1}{3}\)=\(\frac{1}{2}\)+\(\frac{1}{3}\)-\(\frac{1}{18}\)

A x \(\frac{1}{3}\)=\(\frac{7}{9}\)

A = \(\frac{7}{9}\):\(\frac{1}{3}\)

A= \(\frac{7}{3}\)

\(S_1=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{48\cdot49}+\frac{1}{49\cdot50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}=\frac{49}{50}\)

\(S_2=\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+....+\frac{1}{94\cdot97}+\frac{1}{97\cdot100}\)

\(3S_2=\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+....+\frac{3}{94\cdot97}+\frac{3}{97\cdot100}\)

\(=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+....+\frac{1}{97}-\frac{1}{100}\)

\(=\frac{1}{4}-\frac{1}{100}=\frac{6}{25}\)

=> \(S_2=\frac{6}{25}:3=\frac{2}{25}\)