Lời giải:

$A=12.13+13.14+14.15+...+99.100$

$3A=12.13(14-11)+13.14(15-12)+14.15(16-13)+...+99.100(101-98)$

$=(12.13.14+13.14.15+14.15.16+...+99.100.101) - (11.12.13+12.13.14+13.14.15+...+98.99.100)$
$=99.100.101-11.12.13$

$\Rightarrow A=\frac{99.100.101-11.12.13}{3}$

12.13+14.15+16.17+18.19

\(\dfrac{3}{10.11}\) + \(\dfrac{3}{11.12}\) + \(\dfrac{3}{12.13}\) + \(\dfrac{3}{13.14}\) + \(\dfrac{3}{14.15}\)

= \(\dfrac{3}{10}\) - \(\dfrac{3}{11}\) + \(\dfrac{3}{11}\) - \(\dfrac{3}{12}\) + \(\dfrac{3}{12}\) - \(\dfrac{3}{13}\) + \(\dfrac{3}{13}\) - \(\dfrac{3}{14}\) + \(\dfrac{3}{14}\) - \(\dfrac{3}{15}\)

= \(\dfrac{3}{10}\) - \(\dfrac{3}{15}\) = \(\dfrac{1}{10}\)

\(\dfrac{3}{10.11}+\dfrac{3}{11.12}+\dfrac{3}{12.13}+\dfrac{3}{13.14}+\dfrac{3}{14.15}\)

\(=\dfrac{3}{1}.\left(\dfrac{3}{10}-\dfrac{3}{11}+\dfrac{3}{11}-\dfrac{3}{12}+\dfrac{3}{12}-\dfrac{3}{13}+\dfrac{3}{13}-\dfrac{3}{14}+\dfrac{3}{14}-\dfrac{3}{15}\right)\)

\(=\dfrac{3}{1}.\left(\dfrac{3}{10}-\dfrac{3}{15}\right)\)

\(=\dfrac{3}{1}.\left(\dfrac{9}{30}-\dfrac{6}{30}\right)\)

\(=\dfrac{3}{1}.\dfrac{1}{10}\)

\(=\dfrac{3}{10}\)

A= 5/10.11+5/11.12+5/12.13+5/13.14+5/14.15

A= (1/10.11+1/11.12+1/12.13+1/13.14+1/14.15) :5

A= [(1/10-1/11)+(1/11-1/12)+(1/12-1/13)+(1/13-1/14)+(1/14-1/15)] :5

A= (1/10-1/15):5

A= 1/30:5

A= 1/150

Ta có : (93.2727 - 9393.27).(11.12 + 12.13 + 13.14 + 14.15)

=         (93.101.27 - 9393.27).(11.12 + 12.13 + 13.14 + 14.15)

=         (9393.27 - 9393.27).(11.12 + 12.13 + 13.14 + 14.15)

=          0.(11.12 + 12.13 + 13.14 + 14.15)

=          0

Vậy ...

               Chúc các bn học giỏi ! ❤️ 

   Kb và Tk nhaaa  ♥♥!♥♥

5/10.11+5/11.12+5/12.13+5/13.14+5/14.15

=1/10-1/11+1/11-1/12+.....+1/14-1/15

=1/10-1/15

=1/30

\(=5\left(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{14.15}\right)\)

\(5\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-...+\frac{1}{14}-\frac{1}{15}\right)\)

\(5\left(\frac{1}{10}-\frac{1}{15}\right)\)

\(=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)

bài này là tính hả bạn ??

\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{13\cdot14}+\frac{1}{14\cdot15}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{14}-\frac{1}{15}\)

\(=\frac{1}{2}-\frac{1}{15}\)

\(=\frac{13}{30}\)

Sửa đề :\(\frac{5}{11.12}+\frac{5}{12.13}+\frac{5}{13.14}+\frac{5}{14.15}\)

\(=5\left(\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}\right)\)

\(=5\left(\frac{1}{11}-\frac{1}{15}\right)\)

\(=\frac{5}{11}-\frac{1}{3}\)

\(=\frac{15-11}{33}=\frac{14}{33}\)

\(A = {5\over10}-{5\over11}+{5\over11}-{5\over12}+...+{5\over14}-{5\over15} = {5\over10}-{5\over15}={1\over6} \)Ban chep de thieu nha^^

\(\frac{5}{10.11}+\frac{5}{12.13}+\frac{5}{13.14}+\frac{5}{14.15}\)

\(=5.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}\right)\)

\(=5\left(\frac{1}{10}-\frac{1}{15}\right)\)

\(=5.\frac{5}{150}\)

\(=\frac{25}{150}=\frac{1}{6}\)

Barack mới tự học toán 6 hơn 1 tuần, có cách giải này

= 10² + 10 + 11² + 11 + 12² + 12 + 13² + 13 + 14² + 14 + 15² + 15 +16² +16 +17² +17 + 18² + 18 + 19² +19 

= 10² + 11² + 13² + 14² + 15² + 16² + 17² + 18² +19² + 10 + 11 +12 +13 + 14 + 15 + 16 +17 + 18 + 19

Lớp 6 gì xương thế