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TÌM X
\(\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+...+\frac{1}{X.\left(X+3\right)}=\frac{1}{8}\)
\(\frac{3x}{4.7}+\frac{3x}{7.10}+\frac{3x}{10.13}+\frac{3x}{13.16}+...+\frac{3x}{19.22}=\frac{-5}{88}\)
\(\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+...+\frac{3}{19.22}\right)x=\frac{-5}{88}\)
\(\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+...+\frac{1}{19}-\frac{1}{22}\right)x=\frac{-5}{88}\)
\(\left[\frac{1}{4}+\left(\frac{1}{7}-\frac{1}{7}\right)+\left(\frac{1}{10}-\frac{1}{10}\right)+...+\left(\frac{1}{19}-\frac{1}{19}\right)-\frac{1}{22}\right]x=\frac{-5}{88}\)
\(\left[\frac{1}{4}-\frac{1}{22}\right]x=\frac{-5}{88}\)
\(\frac{9}{44}x=\frac{-5}{88}\)
\(x=\frac{-5}{88}:\frac{9}{44}\)
\(x=\frac{-5}{18}\)
~ 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~#
\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+.....+\frac{3^2}{97.100}\)
\(=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\right)\)
Ta thấy :
\(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)
\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)
\(.........\)
\(\frac{3}{97.100}=\frac{100-97}{97.100}=\frac{1}{97}-\frac{1}{100}\)
\(\Rightarrow A=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{100}\right)\)
\(=3\left(1-\frac{1}{100}\right)=3\cdot\frac{99}{100}=\frac{297}{100}\)
đáp án = \(\frac{297}{100}\)
đúng không?
kết bạn với mh nha
\(\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\frac{3^2}{10.13}+...+\frac{3^2}{97.100}\)
\(=3.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{97.100}\right)\)
\(=3.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=3.\left(1-\frac{1}{100}\right)\)
\(=3.\frac{99}{100}\)
Mk sẽ giải từng câu :)
Bài 1 :
Gọi \(ƯCLN\left(2n+2;6n+5\right)=d\)
\(\Rightarrow\hept{\begin{cases}2n+2⋮d\\6n+5⋮d\end{cases}\Rightarrow\hept{\begin{cases}6\left(2n+2\right)⋮d\\2\left(6n+5\right)⋮d\end{cases}\Rightarrow}\hept{\begin{cases}12n+12⋮d\\12n+10⋮d\end{cases}}}\)
\(\Rightarrow\)\(\left(12n+12\right)-\left(12n+10\right)⋮d\)
\(\Rightarrow\)\(2⋮d\)
\(\Rightarrow\)\(d\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
Mà \(6n+5\) không chia hết cho \(2\) và \(-2\) nên \(ƯCLN\left(2n+2;6n+5\right)=\left\{1;-1\right\}\)
Vậy \(\frac{2n+2}{6n+5}\) là phân số tối giản với mọi n
Chúc bạn học tốt ~
1. Gọi d = ƯCLN (2n+2,6n+5)
=>\(\hept{\begin{cases}2n+2\\6n+5\end{cases}}\)chia hết cho d
=>\(\hept{\begin{cases}3.\left(2n+2\right)\\6n+5\end{cases}}\)chia hết cho d
=>\(\hept{\begin{cases}6n+6^{\left(1\right)}\\6n+5^{\left(2\right)}\end{cases}}\)chia hết cho d
Từ (1) và (2) => (6n+6) - (6n+5) chia hết cho d
=> 6n + 6 - 6n - 5 chia hết cho d
=> 1 chia hết cho d
=> d =1
=> ƯCLN (2n+2,6n+5) = 1
Vậy \(\frac{2n+2}{6n+5}\) là phân số tối giản
2. Ta có:
B = 32. (\(\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}+...+\frac{3}{67.70}\))
B = 32. (\(\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+...+\frac{1}{67}-\frac{1}{70}\))
B = 32. (\(\frac{1}{10}-\frac{1}{70}\))
B = 27/35
Vì \(\frac{27}{35}< 1\)
=> B < 1
3. x + \(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{-37}{45}\)
x + ( \(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{-37}{45}\)
x + (\(\frac{1}{5}-\frac{1}{45}\)) = \(\frac{-37}{45}\)
x + \(\frac{8}{45}=\frac{-37}{45}\)
x = \(\frac{-37}{45}-\frac{8}{45}\)
x = -1
\(A=3.\left(\frac{3}{1.4}+\frac{3}{4.7}+.......+\frac{3}{97.100}\right)\)
\(=3.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.........+\frac{1}{97}-\frac{1}{100}\right)\)
\(=3.\left(1-\frac{1}{100}\right)\)
\(=3.\frac{99}{100}\)
\(=\frac{297}{100}\)
Dễ thôi bạn mẫu cách nhau 3 đơn vị tử xuất hiện 3 chỉ cần rút rọn đi 3 là tử có nhé
Ta có: \(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+....+\frac{3^2}{97.100}\)
\(\frac{1}{3}A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+.......+\frac{3}{97.100}\)
\(\frac{1}{3}A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{100}\)
\(\frac{1}{3}A=1-\frac{1}{100}\)
\(\frac{1}{3}A=\frac{99}{100}\)
\(A=\frac{99}{100}.3=\frac{297}{100}\)
\(\frac{\frac{6}{7}-\frac{6}{9}+\frac{6}{11}-\frac{6}{13}}{\frac{8}{7}-\frac{8}{9}+\frac{8}{11}-\frac{8}{13}}\)
=\(\frac{6.\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{11}-\frac{1}{13}\right)}{8.\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{11}-\frac{1}{13}\right)}\)
=\(\frac{6}{8}\)
=\(\frac{3}{4}\)
\(E=\frac{6\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{11}-\frac{1}{13}\right)}{8\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{11}-\frac{1}{13}\right)}=\frac{6}{8}=\frac{3}{4}\)
\(S = \frac{8}{10.13}+\frac{8}{13.16}+\frac{8}{16.19}+..+\frac{8}{307.310}\)
\(=\frac{8}{3}(\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-....+\frac{1}{307}-\frac{1}{310})\)
\(=\frac{8}{3}(\frac{1}{10}-\frac{1}{310})\)
\(=\frac{8}{3}.\frac{30}{310}\)
\(=\frac{8}{3}.\frac{3}{31}\)
\(=\frac{24}{91}\)