Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có :
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(A=\frac{49}{50}\)
Vậy \(A=\frac{49}{50}\)
Chúc bạn học tốt ~
A= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)
A= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)
A= \(1-\frac{1}{50}\)
A= \(\frac{49}{50}\)
Ta có: \(\frac{-3}{1.2.3}+\frac{-3}{2.3.4}+\frac{-3}{3.4.5}+...+\frac{-3}{18.19.20}\)
\(=\frac{-3}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{18.19.20}\right)\)
\(=\frac{-3}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{-3}{2}\left(\frac{1}{2}-\frac{1}{19.20}\right)=\frac{-3}{2}.\frac{189}{380}=\frac{-567}{760}\)
\(A=2.3+3.4+4.5+...+49.50\)
\(3A=2.3.3+3.4.3+4.5.3+...+49.50.3\)
\(3A=2.3.\left(4-1\right)+3.4.\left(5-2\right)+4.5.\left(6-3\right)+...+49.50.\left(51-48\right)\)
\(3A=2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+49.50.51-48.49.50\)
\(3A=-1.2.3+49.50.51\)
\(3A=-6+48450\)
\(3A=48444\)
\(A=\frac{48444}{3}\)
\(A=16148\)
Chúc bạn học tốt ~
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{25.26.27}\)
\(2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{25.26.27}\)
\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{25.26}-\frac{1}{26.27}\)
\(2B=\frac{1}{1.2}-\frac{1}{26.27}\)
\(2B=\frac{1}{2}-\frac{1}{702}\)
\(2B=\frac{175}{351}\)
\(B=\frac{175}{251}:2\)
\(B=\frac{175}{502}\)
Chúc bạn học tốt ~
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}\)
\(A=\frac{1}{4.6}+\frac{1}{10.12}+\frac{1}{18.20}+...+\frac{1}{810.812}\)
.......
~ Chúc học tốt ~
Ai ngang qua xin để lại 1 L - I - K - E
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+.....+\frac{1}{27.28.29.30}\)
\(3A=3.\left(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+......+\frac{1}{27.28.29.30}\right)\)
\(3A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+..........+\frac{3}{27.28.29.30}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+........+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{28.29.30}\)
\(3A=\frac{1}{6}-\frac{1}{24360}\)
\(3A=\frac{1353}{8120}\)
\(A=\frac{1353}{8120}:3\)
\(A=\frac{451}{8120}\)
\(B1\)
\(=\frac{1}{1}-\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{37}-\frac{1}{38}-\frac{1}{39}\)
\(=1-\frac{1}{39}\)
\(=\frac{38}{39}\)
\(B2\)
\(=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+.....+\frac{1}{99\cdot100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+......+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}\)
\(=\frac{25}{100}-\frac{1}{100}\)
\(=\frac{24}{100}\)
\(=\frac{6}{25}\)
Bài 1 :
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)
\(=\frac{1}{1.2}-\frac{1}{38.39}\)
\(=\frac{370}{741}\)
\(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}\)
\(A=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=2.\left(1-\frac{1}{7}\right)\)
\(A=2.\frac{6}{7}\)
\(A=\frac{12}{7}\)
\(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}\)
\(A=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=2.\left(1-\frac{1}{7}\right)\)
\(A=2.\left(\frac{7}{7}-\frac{1}{7}\right)\)
\(A=2.\frac{6}{7}\)
\(A=\frac{12}{7}\)
Chúc bạn học tốt !!!
\(\frac{27^3.4^5}{6^8}:\left(\frac{5^5.2^4}{10^4}:\frac{6^4}{2^6.3^4}\right)\)
\(=\frac{3^9.2^{10}}{6^8}:\left(5:\frac{1}{2^2}\right)\)\(=3.2^2:20=\frac{12}{20}=\frac{3}{5}\)
\(\frac{27^3.4^5}{6^8}:\left(\frac{5^5.2^4}{10^4}:\frac{6^4}{2^6.3^4}\right)\)
\(=\frac{\left(3^3\right)^3.\left(2^2\right)^5}{\left(3.2\right)^8}:\left(\frac{5^5.2^4}{\left(5.2\right)^4}:\frac{\left(2.3\right)^4}{2^6.3^4}\right)\)
\(=\frac{3^9.2^{10}}{3^8.2^8}:\left(\frac{5^5.2^4}{5^4.2^4}:\frac{2^4.3^4}{2^6.3^4}\right)\)
\(=3.2^2:\left(5:\frac{1}{2^2}\right)\)
\(=3.4:\left(5.4\right)\)
\(=12:20\)
\(=\frac{12}{20}=\frac{3}{5}\)
\(\frac{3^4.4-3^6}{3^5.5+10.3^2.4}=\frac{3^4\left(4-3^2\right)}{3^5.5+2.5.3^2.2^2}=\frac{3^4\left(-5\right)}{3^2.5\left(3^3+2^3\right)}=\frac{3^2.3^2.5.\left(-1\right)}{3^2.5.35}=\frac{3^2\left(-1\right)}{35}=\frac{-9}{35}\)