Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bạn chuyển thành dạng các phân số có tử số bằng 3 bằng cách nhân mỗi phân số với 3 rồi cả tổng tất cả nhân với 1/3. Sau đó làm như bình thường
Ta có:
A\(=\frac{1}{2x5}+\frac{1}{5x3}+\frac{1}{3x7}+\frac{1}{7x4}+...+\frac{1}{14x29}+\frac{1}{29x15}\)
\(=\frac{2}{2x\left(2x5\right)}+\frac{2}{\left(5x3\right)x2}+\frac{2}{2x\left(3x7\right)}+\frac{2}{\left(7x4\right)x2}+...+\frac{2}{2x\left(14x29\right)}+\frac{2}{\left(29x15\right)x2}\)
\(=\frac{2}{4x5}+\frac{2}{5x6}+\frac{2}{6x7}+\frac{2}{7x8}+...+\frac{2}{28x29}+\frac{2}{29x30}\)
\(=2x\left(\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{7x8}+...+\frac{1}{28x29}+\frac{1}{29x30}\right)\)
\(=2x\left(\frac{5-4}{4x5}+\frac{6-5}{5x6}+\frac{7-6}{6x7}+\frac{8-7}{7x8}+...+\frac{29-28}{28x29}+\frac{30-29}{29x30}\right)\)
\(=2x\left(\frac{5}{4x5}-\frac{4}{4x5}+\frac{6}{5x6}-\frac{5}{5x6}+\frac{7}{6x7}-\frac{6}{6x7}+\frac{8}{7x8}-\frac{7}{7x8}+...+\frac{29}{28x29}-\frac{28}{28x29}+\frac{30}{29x30}-\frac{29}{29x30}\right)\)
\(=2x\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{28}-\frac{1}{29}+\frac{1}{29}-\frac{1}{30}\right)\)
\(=2x\left(\frac{1}{4}-\left(\frac{1}{5}-\frac{1}{5}\right)-\left(\frac{1}{6}-\frac{1}{6}\right)-...-\left(\frac{1}{28}-\frac{1}{28}\right)-\left(\frac{1}{29}-\frac{1}{29}\right)-\frac{1}{30}\right)\)
\(=2x\left(\frac{1}{4}-0-0-...-0-0-\frac{1}{30}\right)\)
\(=2x\left(\frac{1}{4}-\frac{1}{30}\right)\)
\(=2x\frac{1}{4}-2x\frac{1}{30}\)
\(=\frac{1}{2}-\frac{1}{15}\)
=15/30-2/30=13/30
\(\frac{3}{2\times5}+\frac{2}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{602\times605}\)
\(=\frac{5-2}{2\times5}+\frac{8-5}{5\times8}+\frac{11-8}{8\times11}+...+\frac{605-602}{602\times605}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{602}-\frac{1}{605}\)
\(=\frac{1}{2}-\frac{1}{605}=\frac{603}{1210}\)
\(\frac{4}{3\times7}+\frac{5}{7\times12}+\frac{1}{12\times13}+\frac{2}{13\times15}\)
\(=\frac{7-4}{3\times7}+\frac{12-7}{7\times12}+\frac{13-12}{12\times13}+\frac{15-13}{13\times15}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)
a) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{x\times\left(x+3\right)}=\dfrac{99}{200}\)
Ta có: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=1-\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=\dfrac{1}{100}\)
\(\Rightarrow x+1=100\)
\(x=100-1\)
\(x=99\)
=(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9) chia 2
=(1-1/9)chia 2
=8/9 chia 2
=4/9
\(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+...+\dfrac{1}{121\times123}\)
\(=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{121}-\dfrac{1}{123}\right)\)
\(=\dfrac{1}{2}.\left(1-\dfrac{1}{123}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{123}{123}-\dfrac{1}{123}\right)\)
\(=\dfrac{1}{2}.\dfrac{122}{123}\)
\(=\dfrac{61}{123}\)
\(#NqHahh\)
\(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{31.35}+\frac{1}{35.39}\) ( . là nhân nhé )
\(=\frac{1}{4}.\left[4.\left(\frac{1}{3.7}+\frac{1}{4.11}+\frac{1}{11.15}+...+\frac{1}{31.35}+\frac{1}{35.39}\right)\right]\)
\(=\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{31.35}+\frac{4}{35.39}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{39}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{39}\right)=\frac{1}{4}.\left(\frac{13-1}{39}\right)=\frac{1}{4}.\frac{12}{39}=\frac{1}{4}.\frac{4}{13}=\frac{1}{13}\)
1/1 x 3 + 1/3 x 5 + 1/5 x 7 + 1/7 x 9 + 1/9 x 11
= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11
= 1 - 1/11
= 10/11
\(\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{9.11}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{1}{3.5}+....+\frac{2}{9.11}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{11}\right)=\frac{1}{2}.\left(1-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\frac{10}{11}=\frac{5}{11}\)
\(\frac{1}{1.3}+\frac{1}{3.2}+\frac{1}{2.5}+...+\frac{1}{99.100}\)
= \(2.\left(\frac{1}{1.3.2}+\frac{1}{3.2.2}+\frac{1}{2.5.2}+...+\frac{1}{99.50.2}\right)\)
= \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)
= \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)
= \(2.\left(\frac{1}{2}-\frac{1}{100}\right)\)
= \(2.\frac{49}{100}\)
= \(\frac{49}{50}\)