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\(A=\frac{2^2}{1.3}\cdot\frac{3^2}{2.4}....\frac{999^2}{998.1000}\)
\(A=\frac{2^2.3^2....999^2}{1.3.2.4.998.100}=\frac{\left(2.3.....999\right)\left(2.3....999\right)}{\left(1.2....998\right)\left(3.4....1000\right)}\)
\(A=999\cdot\frac{1}{500}=\frac{999}{500}\)( khúc này mk làm tắt, bn bỏ dấu ở trên rồi bỏ từng tử)
=?????????????????????????????????????????????????????????????????????????????????????????????????????????????????
mk ko ghi đb nhé
\(=\frac{1\cdot3+1}{1\cdot3}+\frac{2\cdot4+1}{2\cdot4}+...+\frac{99\cdot101+1}{99\cdot101}.\)
\(=1+\frac{1}{1\cdot3}+1+\frac{1}{2\cdot4}+...+1+\frac{1}{99\cdot101}\)
\(=99+\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{2\cdot4}+...+\frac{2}{99\cdot101}\right)\)
\(=99+\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{100}+\frac{1}{99}-\frac{1}{101}\right)\)
\(=99+\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)
phần còn lại bn tự tính nha
a, Ta có:
\(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{0,6-\frac{3}{9}+\frac{3}{11}}+\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{14}}{-1-\frac{3}{7}+\frac{3}{28}}=\frac{2\left(0,2-\frac{1}{9}+\frac{1}{11}\right)}{3\left(0,2-\frac{1}{9}+\frac{1}{11}\right)}+\frac{2\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{28}\right)}{-3\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{28}\right)}=\frac{2}{3}+\frac{-2}{3}=0\)
k đúng cho mình nha. Thanks!!!
a, bày cho mình cách viết bằng phân số đi , mình trình bày cách làm cho. k đúng cho mình nha.
\(N=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{9^2}\)
\(=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{9.9}\)
\(N\)bé hơn \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{8.9}=N_1\)
\(N_1=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{8.9}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.........-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\) \((1)\)
\(N\)lớn hơn \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}=N_2\)
\(\Rightarrow N_2=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.....-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}\)
\(=\frac{5}{10}-\frac{1}{10}=\frac{2}{5}\) \((2)\)
Từ \((1)\)và \((2)\)suy ra ; \(\frac{2}{5}\)bé hơn N bé hơn \(\frac{8}{9}\)
Học tốt
Nhớ kết bạn với mình
a/ \(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+........+\frac{99}{100!}\)
\(\Leftrightarrow A=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+......+\frac{100-1}{100!}\)
\(\Leftrightarrow A=\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+.....+\frac{100}{100!}-\frac{1}{100!}\)
\(\Leftrightarrow A=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+....+\frac{1}{99!}-\frac{1}{100!}\)
\(\Leftrightarrow A=1-\frac{1}{100!}\)
b/ \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+.....+\frac{1}{98.99.100}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.....+\frac{1}{98.99}-\frac{1}{99.100}\)
\(=\frac{1}{1.2}-\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{9900}\)
\(=1-\left(\frac{2}{1.3}-\frac{2}{3.5}-...-\frac{2}{2005-2007}\right)\)
\(=1-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2005}-\frac{1}{2007}\right)\)
\(=1-\left[1+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{5}+\frac{1}{5}\right)+...+\left(-\frac{1}{2005}+\frac{1}{2005}\right)-\frac{1}{2007}\right]\)
\(=1-\left(1+0+0+...+0-\frac{1}{2007}\right)\)
\(=1-\left(1-\frac{1}{2007}\right)\)
\(=1-1+\frac{1}{2007}\)
\(=0+\frac{1}{2007}\)
\(=\frac{1}{2007}\)
Ai thấy tớ đúng k nha
Đặt A = \(1-\frac{2}{1.3}-\frac{2}{3.5}-.....-\frac{2}{2005.2007}\)
= \(1-\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{2005.2007}\right)\)
=\(1-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{2005}-\frac{1}{2007}\right)\)
= \(1-\left(1-\frac{1}{2017}\right)\)
=\(1-1+\frac{1}{2017}\)
=\(0+\frac{1}{2017}\)
=\(\frac{1}{2017}\)
\(B=\frac{2^2}{1.3}+\frac{2^2}{3.5}+...+\frac{2^2}{195.197}=2\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{195.197}\right)\)
\(=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{195}-\frac{1}{197}\right)=2\left(1-\frac{1}{197}\right)=2.\frac{196}{197}=\frac{392}{197}\)