\(\frac{3}{2x5}\)+ \(\frac{3}{5x8}\)+\(\frac{3}{8x11}\)+ ... + \(\frac{3}{\left(X-3\right)xX}\)
Giúp mk nhé!!!Thanks <3
\(\frac{3}{2x5}\)+ \(\frac{3}{5x8}\)+ \(\frac{3}{8x11}\)+ ... + \(\frac{3}{\left(X-3\right)xX}\)= \(\frac{1}{6}\)
Giúp mk nhé!!!
Giải:
Ta có:
\(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{\left(x-3\right)\times x}=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x-3}-\frac{1}{x}=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x}=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{x}=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}\Leftrightarrow x=3\)
De thoi ma,minh ko ghi lai de nha
=\(\frac{1}{2}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{8}\)+...+...=\(\frac{1}{6}\)
=\(\frac{1}{2}\)-\(\frac{1}{x}\)=\(\frac{1}{6}\)
Con lai bn tu lam nha . chuc bn hok tot !!!
\(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+....+\frac{3}{\left(x-3\right).x}=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{\left(x-3\right)}-\frac{1}{x}=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x}=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{x}=\frac{1}{2}-\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{x}=\frac{1}{3}\)
\(\Leftrightarrow x=1\div\frac{1}{3}=3\)
\(\frac{1}{5x8}+\frac{1}{8x11}+...+\frac{1}{Xx\left(X+3\right)}=\frac{101}{1540}\)
\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\) (dấu . là nhân nhé)
=> \(\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right):3=\frac{101}{1540}\)
=> \(\left(\frac{1}{5}-\frac{1}{x+3}\right):3=\frac{101}{1540}\)
=> \(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}\cdot3=\frac{303}{1540}\)
=> \(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
=> \(x+3=308\Rightarrow x=308-3=305\)
\(\frac{1}{5x8}+\frac{1}{8x11}+...+\frac{1}{Xx\left(X+3\right)}=\frac{101}{1540}\)
<=>\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{101}{1540}\)
<=>\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}\)
<=>\(\frac{x-2}{5x+15}=\frac{101}{1540}\)
<=>1540x-3080=505x+1515
<=>1035x=4595
<=>x=919/207
Tìm X biết \(\frac{1}{5x8}+\frac{1}{8x11}+\frac{1}{11x14}+...+\frac{1}{Xx\left(x+3\right)}=\frac{101}{1504}\)
Tính B = \(\frac{3}{2x5}+\frac{3}{5x8}+\frac{3}{8x11}+\frac{3}{11x14}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)
\(=\frac{1}{2}-\frac{1}{14}\)
\(=\frac{7}{14}-\frac{1}{14}\)
\(=\frac{6}{14}\)
\(=\frac{3}{7}\)
3/2x5 + 3/5x8 + 3/8x11 + 3/11x14
= 3/2 - 3/5 + 3/5 - 3/8 + 3/8 - 3/11 + 3/11 - 3/14
= 3/2 - 3/14
= 21/14 - 3/14
= 18/14
= 9/5
\(B=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}=\frac{1}{2}-\frac{1}{14}=\frac{3}{7}\)
Tìm x biết: \(\frac{1}{5X8}+\frac{1}{8X11}+\frac{1}{11X14}+.....+\frac{1}{xX\left(x+3\right)}=\frac{101}{1540}\)
Tính tổng
\(\frac{3^2}{2x5}+\frac{3^2}{5x8}\frac{3^2}{8x11}+......+\frac{3^2}{27x30}\)
Câu 1:Tìm a biết :
\(\frac{15,2x0,25-48,51:14,7}{a}\)=\(\frac{\left(\frac{13}{44}-\frac{2}{11}-\frac{5}{66}:2\frac{1}{2}\right)x1\frac{1}{5}}{3,2+0,8x\left(5\frac{1}{2}-3,25\right)}\)
Câu 2 :TÌM n
\(\frac{1}{2x5}\)+\(\frac{1}{5x8}\)+\(\frac{1}{8x11}\)+..........+\(\frac{1}{n\left(n+3\right)}\)=\(\frac{31}{198}\)
\(\frac{6}{2x5}+\frac{6}{5x8}+\frac{6}{8x11}+.......+\frac{6}{98x101}\)
Đặt \(Shin=\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+...+\frac{6}{98.101}\)
\(\Rightarrow3Shin=\frac{6}{2}-\frac{6}{5}+\frac{6}{5}-\frac{6}{8}+\frac{6}{8}-\frac{6}{11}+...+\frac{6}{98}-\frac{6}{101}\)
\(\Leftrightarrow3Shin=\frac{6}{2}-\frac{6}{101}=\frac{297}{101}\)
N/t: . là dấu nhân nha! Cái đó lớp 5 chưa biết đâu! Lên cấp 2 mới học. Trong bài làm bạn cứ ghi là dấu " x" thay cho dấu "." của mình nha!
Đặt A = 6/2.5 + 6/5.8 + ... + 6/98.101
=> A = 6.(1/2.5 + 1/5.8 + ... + 1/98.101)
=> 3A = 6.(3/2.5 + 3/5.8 + ... + 3/98.101)
=> 3A = 6.(1/2 - 1/5 + 1/5 - 1/8 + ... + 1/98 - 1/101)
=> 3A = 6.99/202
=> 3A = 297/101
=> A = 99/101
x là nhân hả
\(\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+...+\frac{6}{98.101}\)
\(=2\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+......+\frac{3}{98.101}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{98}-\frac{1}{101}\right)\)
\(=2\left(\frac{1}{2}+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{8}+\frac{1}{8}\right)+....+\left(\frac{-1}{98}+\frac{1}{98}\right)-\frac{1}{101}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=2.\frac{99}{202}\)
\(=\frac{99}{101}\)
\(\frac{1}{2x5}+\frac{1}{5x8}+\frac{1}{8x11}+\frac{1}{11x14}+\frac{1}{14x17}+\frac{1}{17x20}\)
\(\frac{1}{2\times5}+\frac{1}{5\times8}+\frac{1}{8\times11}+\frac{1}{11\times14}+\frac{1}{14\times17}+\frac{1}{17\times20}\)
\(=\frac{1}{3}\times\left(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+\frac{3}{14\times17}+\frac{3}{17\times20}\right)\)
\(=\frac{1}{3}\times\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)
\(=\frac{1}{3}\times\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(=\frac{1}{3}\times\frac{9}{20}\)
\(=\frac{3}{20}\)
_Chúc bạn học tốt_
Đặt \(A=\frac{1}{2x5}+\frac{1}{5x8}+..+\frac{1}{17x20}\)
\(3xA=3x\left(\frac{1}{2x5}+\frac{1}{5x8}+...+\frac{1}{17x20}\right)\)
\(3xA=\frac{3}{2x5}+\frac{3}{5x8}+....+\frac{3}{17x20}\)
\(3xA=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+..+\frac{1}{17}-\frac{1}{20}\)
\(3xA=\frac{1}{2}-\frac{1}{20}\)
\(3xA=\frac{9}{20}\)
\(\Rightarrow A=\frac{3}{20}\)