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\(\left(2018-\frac{2}{135}+\frac{1}{50}\right)-\left(1-\frac{7}{135}+\frac{4}{50}\right)-\left(5+\frac{5}{135}+\frac{3}{50}\right)\)
\(=2018-\frac{2}{135}+\frac{1}{50}-1+\frac{7}{135}-\frac{4}{50}-5-\frac{5}{135}-\frac{3}{50}\)
\(=2012-\frac{6}{50}\)
Đặt A = \(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2015!}\)
A < \(1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{2014.2015}\)
A < \(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2014}-\frac{1}{2015}\)
A < \(2-\frac{1}{2015}\)< 2 < \(2\left(\frac{135^2+136}{136^2-135}\right)\)
=> A < \(2\left(\frac{135^2+136}{136^2-135}\right)\)(Đpcm)
2) 1\26+1\27+1\28+........+1\50=1+1\2+1\3+......+1\50 -( 1+1\2+1\3+.....+1\25)=1+1\2+1\3+....+1\50-2.(1\2+1\4+1\6+....+1\50)=1-1\2+1\3-1\4+.....+1\49-1\50=vế phải(đpcm)
\(P=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}-1\)
\(=2.\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)
\(=2\left[\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{8}\right)-\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right]\)
8/5:(8/5.5/4) / 16/25-1/25 + 1:4/7 / (50/9-9/4).36/17) + 0,6.0,5:2/5
= 8/5:8/5:5/4 / 3/5 + 7/4 / 50/9.36/17-9/4.36/17 + 0,3.5/2
= 4/5:3/5 + 7/4 / 200/17-81/17 + 3/10.5/2
= 4/5.5/3 + 7/4 / 119/17 + 3/4
= 4/3 + 7/4 : 7 + 3/4
= 4/3 + 4 + 3/4
= 16/12 + 48/12 + 9/12
= 73/12
☆★☆★☆
Ta có: \(\left(2008-\dfrac{2}{135}+\dfrac{1}{50}\right)-\left(1-\dfrac{7}{135}+\dfrac{4}{150}\right)-\left(5+\dfrac{5}{135}+\dfrac{3}{50}\right)\)
= \(2008-\dfrac{2}{135}+\dfrac{1}{50}-1+\dfrac{7}{135}-\dfrac{4}{150}-5-\dfrac{5}{135}-\dfrac{3}{50}\)
= (2008-1-5) + \(\left(\dfrac{1}{50}-\dfrac{3}{50}\right)-\left(\dfrac{2}{135}-\dfrac{7}{135}\right)-\dfrac{4}{150}\)
=2002 \(-\dfrac{1}{25}\)+\(\dfrac{1}{27}\)\(-\dfrac{4}{150}\)
=2001,9(3)
hình như sai sai đó bạn