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\(\frac{1}{2}-\frac{1}{14}=\frac{3}{7}\). Mà ở đây kết quả là \(\frac{1}{21}\)nên phân số phải nhóm ra ngoài là:
\(\frac{1}{21}:\frac{3}{7}=\frac{1}{9}\). Ta có:
\(\frac{1}{9}.\left(\frac{3x}{2.5}+\frac{3x}{5.8}+\frac{3x}{8.11}+\frac{3x}{11.14}\right)=\frac{1}{21}\). Suy ra 3x=9. Vậy x=3
Nhầm nha: Vì có 3x nên phân số nhóm ra ngoài là \(\frac{1}{3}\). Ta có tương tự. Suy ra 3x=3. Vậy x=1
Bài làm
\(\frac{3x}{2.5}+\frac{3x}{5.8}+\frac{3x}{8.11}+\frac{3x}{11.14}=\frac{1}{21}\)
\(\Leftrightarrow\frac{x}{2}-\frac{x}{5}+\frac{x}{5}-\frac{x}{8}+\frac{x}{8}-\frac{x}{11}+\frac{x}{11}-\frac{x}{14}=\frac{1}{21}\)
\(\Leftrightarrow\frac{x}{2}-\frac{x}{14}=\frac{1}{21}\)
\(\Leftrightarrow\frac{7x}{14}-\frac{x}{14}=\frac{1}{21}\)
\(\Leftrightarrow\frac{6x}{14}=\frac{1}{21}\)
\(\Leftrightarrow126x=14\)
\(\Leftrightarrow x=\frac{1}{9}\)
Học tôt
a, \(3x\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}\right)=\frac{1}{21}\)
\(3x.\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)
\(x.\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{21}\)
\(\frac{3x}{7}=\frac{1}{21}\Rightarrow x=\frac{1}{9}\)
Quên mất câu b.
Gọi số kg của quả dưa là x
=>\(\frac{3}{4}x=3\frac{1}{2}\Rightarrow\frac{3}{4}x=\frac{7}{2}\Rightarrow x=\frac{14}{3}=4\frac{2}{3}kg\)
Vậy....
\(\frac{3}{2.5}\)+\(\frac{3}{5.8}\)+\(\frac{3}{8.11}\)+\(\frac{3}{11.14}\)+\(\frac{3}{14.17}\)
=\(\frac{1}{2}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{8}\)+......+\(\frac{1}{14}\)-\(\frac{1}{17}\)
=\(\frac{1}{2}\)-\(\frac{1}{17}\)
=\(\frac{15}{34}\)
Đặt \(A=\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1504}\)
\(A=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{x}-+\frac{1}{x+3}\right)\)
\(A=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)\)
\(A=\frac{1}{3}\left(\frac{\left(x+3\right)-5}{5\left(x+3\right)}\right)\)
\(A=\frac{\left(x+3\right)-5}{15\left(x+3\right)}\)
1504 không chia hết cho 3;5 nên ta xét tủ :
x + 3 - 5 = 101
x + 3 = 106
x = 103
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(x=308-3=305\)
VẬY x = 305
\(=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+....+\frac{3}{80.83}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{80}-\frac{1}{83}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{83}\right)\)
\(=\frac{54}{83}\)
\(\frac{4}{2\cdot5}+\frac{4}{5\cdot8}+............+\frac{4}{80\cdot83}\)
\(=4\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+............+\frac{4}{80\cdot83}\right)\)
\(=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...........+\frac{1}{80}-\frac{1}{83}\right)\)
\(=\frac{4}{3}\cdot\frac{81}{166}\)
\(=\frac{54}{83}\)
CHúc các bạn học tôt !!!!!!!!!!!!!!!!!!!
\(\frac{3x}{4.7}+\frac{3x}{7.10}+\frac{3x}{10.13}+\frac{3x}{13.16}+...+\frac{3x}{19.22}=\frac{-5}{88}\)
\(\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+...+\frac{3}{19.22}\right)x=\frac{-5}{88}\)
\(\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+...+\frac{1}{19}-\frac{1}{22}\right)x=\frac{-5}{88}\)
\(\left[\frac{1}{4}+\left(\frac{1}{7}-\frac{1}{7}\right)+\left(\frac{1}{10}-\frac{1}{10}\right)+...+\left(\frac{1}{19}-\frac{1}{19}\right)-\frac{1}{22}\right]x=\frac{-5}{88}\)
\(\left[\frac{1}{4}-\frac{1}{22}\right]x=\frac{-5}{88}\)
\(\frac{9}{44}x=\frac{-5}{88}\)
\(x=\frac{-5}{88}:\frac{9}{44}\)
\(x=\frac{-5}{18}\)
~ Hok tốt ~
#)Giải :
Đặt \(A=\frac{3x}{2.7}+\frac{3x}{7.10}+\frac{3x}{10.13}+\frac{3x}{13.16}+...+\frac{3x}{19.22}=-\frac{5}{88}\)
\(A=\frac{3x}{2}+\frac{3x}{7}-\frac{3x}{7}+\frac{3x}{10}-\frac{3x}{10}+\frac{3x}{13}-\frac{3x}{13}+\frac{3x}{16}-...-\frac{3x}{19}+\frac{3x}{22}=-\frac{5}{88}\)
\(A=\frac{3x}{2}+0+0+0+...+0+\frac{3x}{22}=-\frac{5}{88}\)
\(A=\frac{3x}{2}+\frac{3x}{22}=-\frac{5}{88}\)
\(3x:\left(2+22\right)=-\frac{5}{88}\)
\(3x:24=-\frac{5}{88}\)
\(3x=-\frac{5}{88}.24\)
\(3x=-\frac{7}{11}\)
\(x=-\frac{7}{11}:3\)
\(x=-\frac{7}{33}\)
#~Will~be~Pens~#
x. [3/2.5 +3/5.8+ 3/8.11+3/11.14] = 1/21
=> x . [1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14] = 1/21
=> x. [1/2-1/14] = 1/21 => x . 3/7= 1/21
=> x = 1/9
x. [3/2.5 +3/5.8+ 3/8.11+3/11.14] = 1/21
=> x . [1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14] = 1/21
=> x. [1/2-1/14] = 1/21 => x . 3/7= 1/21
=> x = 1/9
mình nhé