SSH:(20152-12):10+1=2015

(12-22)+(32-42)+(52-62)+...+(20132-20142)+20152

-10+(-10)+(-10)+...+(-10)+20152

-10x(2015-1):2+20152=12

=> C=12

\(\dfrac{2^2}{1\times3}\times\dfrac{3^2}{2.4}\times\dfrac{4^2}{3.5}\times\dfrac{5^2}{4.6}=\dfrac{2^2.3^2.4^2.5^2}{1.3.2.4.3.5.4.6}=\dfrac{2^2.3^2.4^2.5^2}{1.2.3.3.4.4.5.2.3}=\dfrac{2^2.3^2.4^2.5^2}{3^3.2^2.4^2.5.1}=\dfrac{5}{3.1}=\dfrac{5}{3}\)

\(\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot\dfrac{5^2}{4.6}\\ =\dfrac{2^2\cdot3^2\cdot4^2\cdot5^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot4\cdot6}\\ =\dfrac{2^2\cdot3^2\cdot4^2\cdot5^2}{1\cdot2\cdot4^2\cdot4^2\cdot5\cdot6}\\ =\dfrac{2\cdot5}{6}=\dfrac{5}{3}\)

= 50

Nho t ick nha

     \(\left(102+82+62+42+22\right)-\left(12+32+53+72+92\right)\)

\(=102+82+62+42+22-12-32-52-72-92\)

\(=\left(102-92\right)+\left(82-72\right)+\left(62-52\right)+\left(42-32\right)+\left(22-12\right)\)

\(=10+10+10+10+10\)

\(=10.5\)

\(=50\)

12 + 22 + 32 + 42 + 52 + 62 = 222

bạn k mình, mình k lại

12 + 22 + 32 + 42 + 52 + 62 = 222

k mk , mk k lai

12 + 22 + 32 + 42 + 52 + 62 =222

k nha

https://olm.vn/cau-hoi/a-cho-a12211216211002-ctr-a12-b-cho-p122132142120232-ctr-p-khong-la-so-tu-nhien-c-cho-c132152172120211.8293222842881

Cô làm rồi em nhá

Câu a, xem lại đề bài

Câu b: 

    P =  \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + ...+ \(\dfrac{1}{2023^2}\)

   Vì  \(\dfrac{1}{2^2}\) < \(\dfrac{1}{1.2}\)                =  \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)

         \(\dfrac{1}{3^2}\) < \(\dfrac{1}{2.3}\)                = \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)

         \(\dfrac{1}{4^2}\)  < \(\dfrac{1}{3.4}\)               = \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) 

     ........................

        \(\dfrac{1}{2023^2}\) < \(\dfrac{1}{2022.2023}\) = \(\dfrac{1}{2022}\) - \(\dfrac{1}{2023}\)

Cộng vế với vế ta có:  

0< P < 1 - \(\dfrac{1}{2023}\) < 1

Vậy 0 < P < 1 nên P không phải là số tự nhiên vì không tồn tại số tự nhiên giữa hai số tự nhiên liên tiếp

Câu c:  

C = \(\dfrac{1}{3^2}\) + \(\dfrac{1}{5^2}\) + \(\dfrac{1}{7^2}\) + ....+ \(\dfrac{1}{2021^2}\) + \(\dfrac{1}{2023^2}\) = C 

B =  \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\)+.......+ \(\dfrac{1}{2020^2}\) + \(\dfrac{1}{2023^2}\) > 0 

Cộng vế với vế ta có: 

C+B =  \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{5^2}\)+ \(\dfrac{1}{6^2}\)+...+ \(\dfrac{1}{2023^2}\) > C + 0 = C > 0

             Mặt khác ta có: 

1 > \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\)+...+ \(\dfrac{1}{2023^2}\) (cm ở ý b)

Vậy 1 > C > 0 hay C không phải là số tự nhiên (đpcm)

\(A=\dfrac{2}{1x3}+\dfrac{2}{3x5}+\dfrac{2}{5x7}+...+\dfrac{2}{21x23}\)

\(A=2x\left(\dfrac{1}{1x3}+\dfrac{1}{3x5}+\dfrac{1}{5x7}+...+\dfrac{1}{21x23}\right)\)

\(A=2x\dfrac{1}{2}x\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{21}-\dfrac{1}{23}\right)\)

\(A=1-\dfrac{1}{23}\)

\(A=\dfrac{22}{23}\)

\(B=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)

\(B=\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+\dfrac{1}{5x6}+\dfrac{1}{6x7}+\dfrac{1}{7x8}+\dfrac{1}{8x9}+\dfrac{1}{9x10}\)

\(B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(B=\dfrac{1}{2}-\dfrac{1}{10}\)

\(B=\dfrac{5}{10}-\dfrac{1}{10}\)

\(B=\dfrac{4}{10}\)

\(B=\dfrac{2}{5}\)

\(C=\dfrac{1}{28}+\dfrac{1}{70}+\dfrac{1}{130}+\dfrac{1}{208}\)

\(C=\dfrac{1}{4x7}+\dfrac{1}{7x10}+\dfrac{1}{10x13}+\dfrac{1}{13x16}\)

\(C=\dfrac{1}{3}x\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right)\)

\(C=\dfrac{1}{3}x\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)

\(C=\dfrac{1}{3}x\left(\dfrac{4}{16}-\dfrac{1}{16}\right)\)

\(C=\dfrac{1}{3}x\dfrac{3}{16}\)

\(C=\dfrac{1}{16}\)