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Câu 1:
a: \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{2}{5}=\dfrac{20-15+24}{60}=\dfrac{29}{60}\)
b: \(\dfrac{1}{3}-\dfrac{4}{5}+\dfrac{4}{3}=\dfrac{5}{3}-\dfrac{4}{5}=\dfrac{25-12}{15}=\dfrac{13}{15}\)
c: \(=\dfrac{1}{2}+\dfrac{1}{6}-\dfrac{1}{4}+\dfrac{1}{5}\)
\(=\dfrac{30}{60}+\dfrac{10}{60}-\dfrac{15}{60}+\dfrac{12}{60}\)
=37/60
A) 7/38 x 9/11 +7/38 x 4/11 -7/38 x 2/11
=7/38.(9/11+4/11-2/11)
=7/38
B) 5/31 x 21/25 + 5/31 x -7/10 - 5/31 x 9/20
=5/31.(21/25-7/10-9/20)
=5/31.(-31/100)
=-1/20
c: Ta có: \(\dfrac{5}{3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{101\cdot103}\)
\(=\dfrac{5}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{101\cdot103}\right)\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{103}\right)\)
\(=\dfrac{5}{2}\cdot\dfrac{102}{103}\)
\(=\dfrac{255}{103}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(A=1-\frac{1}{2020}\)
\(A=\frac{2019}{2020}\)
\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)
\(2B=\frac{2}{1.3}+\frac{2}{3.5}=\frac{2}{5.7}+...+\frac{2}{2017.2019}\)
\(2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}=\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(2B=1-\frac{1}{2019}\)
\(2B=\frac{2018}{2019}\)
\(B=\frac{2018}{2019}:2=\frac{1009}{2019}\)
Tính nhanh 5/8+5/24+5/48+......+5/9800