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=1/1-1/2+1/2-1/3+1/3-1/4+.........+1/1999-1/2000
=1/1-1/2000
=1999/2000<3/4
\(7.\left[\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right):2\right]\)
\(7.\left[\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right):2\right]\)
\(7.\left(\frac{1}{3}-\frac{1}{13}\right):2\)
\(7.\frac{10}{39}:2=\frac{35}{39}\)
\(\frac{7}{15}+\frac{7}{35}+\frac{7}{63}+\frac{7}{99}+\frac{7}{143}\)
\(=\frac{7}{2}\cdot\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)\)
\(=\frac{7}{2}\cdot\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{7}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{7}{2}\cdot\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=\frac{7}{2}\cdot\frac{10}{39}\)
\(=\frac{35}{39}\)
\(\frac{-7}{11}.\frac{11}{19}+\frac{-7}{11}.\frac{8}{19}+\frac{-4}{11}\)
\(=\frac{-7}{11}.\left(\frac{11}{19}+\frac{8}{19}\right)+\frac{-4}{11}\)
\(=\frac{-7}{11}.1+\frac{-4}{11}\)
\(=\frac{-7}{11}+\frac{-4}{11}=\frac{-11}{11}=-1\)
~ Hok tốt ~
Đặt \(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)
\(\Rightarrow B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(\Rightarrow B=1-\frac{1}{2019}\)
\(\Rightarrow B=\frac{2018}{2019}\)
kiểm ''cha'' bạn để làm gì vậy. Kiểm ''cha'' mà kiểm bạn đâu, vội vàng làm gì, cứ suy nghĩ đi đã bạn ạ.
A=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2014}{2015}.\frac{2015}{2016}\)
A=\(\frac{1.2.3.4...2015}{2.3.4...2016}=\frac{1}{2016}\)
Hok tốt
A = \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2015}\right).\left(1-\frac{1}{2016}\right)\)
= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2014}{2015}.\frac{2015}{2016}\)
= \(\frac{1}{2016}\)
Vậy ...
đặt \(S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\)
\(2S=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\)
\(\Leftrightarrow2S-S=\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(\Leftrightarrow S=\frac{1}{101}-1=-\frac{100}{101}\)
=\(\left(4-2+3\right)\cdot\frac{-1}{2}\)
=\(5\cdot\left(\frac{-1}{2}\right)\)
=\(\frac{5\cdot\left(-1\right)}{2}\)
=\(\frac{-6}{2}\)
\(=\left(-3\right)\)
Có 2 trg hợp nhé: Nếu x là dấu nhân thì thực hiện theo phép nhân
Nếu x là ẩn số thì ko làm đc nhé vì ko có kết quả
Nên làm theo trường hợp 1
\(4.\frac{-1}{2}-2.\frac{-1}{2}+3.\frac{-1}{2}\)\(=\)\(\left(\frac{-1}{2}\right).\left(4-2+3\right)=\left(\frac{-1}{2}\right).5=\frac{-1.5}{2}=\frac{-5}{2}\)
\(A=\frac{1}{1.5}+\frac{1}{5.10}+\frac{1}{10.15}+...+\frac{1}{95.100}\)
\(\Rightarrow\)\(5A=1+\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{95.100}\)
\(=1+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{100}\)
\(=1+\frac{1}{5}-\frac{1}{100}=\frac{119}{100}\)
\(\Rightarrow\)\(A=\frac{119}{500}\)
A=1/1.5+1/5.10+....+1/95.100
=(5/1.5+5/5.10+...+5/95.100):5
=(1-1/5+1/5-1/10+...+1/95-1/100):5
=(1-1/100):5
=99/100:5
=99/500
khỏi ghi lại đề nha
A=1-1/2+1/2-1/3+1/3-1/4+......+1/49-1/50
A=1-1/50
A=49/50
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)