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a) 4/7 + 7/20 + 1/4 b) 21,15 + 3/5 + 3/4
= 4/7 + 7/20 + 5/20 = 21,15 + 0,6 + 0,75
= 4/7 + 3/5 = 22,5
= 20/35 + 21/35
= 41/35
a,\(\frac{4}{7}\) \(+\frac{7}{20}\) \(+\frac{1}{4}\)
\(=\frac{4}{7}\) \(+\frac{7}{20}\) \(+\frac{5}{20}\)
\(=\frac{4}{7}\) \(+\frac{3}{5}\)
\(=\frac{20}{35}\)\(+\frac{21}{35}\)
\(=\frac{41}{35}\)
#Hemingson
a) Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2A=\left(\frac{1}{2}\times2\right)+\left(\frac{1}{4}\times2\right)+\left(\frac{1}{8}\times2\right)+\left(\frac{1}{16}\times2\right)+\left(\frac{1}{32}\times2\right)\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
Ta lấy : \(2A-1A=1A\)
\(A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)
\(A=1-\frac{1}{32}\)
\(A=\frac{31}{32}\)
Vậy \(A=\frac{31}{32}\)
b) Đặt \(B=\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{18\times19}+\frac{2}{19\times20}\)
\(B=2\times(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20})\)
\(B=2\times\left(1-\frac{1}{20}\right)\)
\(B=2\times\frac{19}{20}\)
\(B=\frac{19}{10}\)
Vậy \(B=\frac{19}{10}\)
Học tốt # ^-<
Ta có : \(\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+\left(x+7\right)=32\)
\(\Leftrightarrow x+x+x+x+1+3+5+7=32\)
\(\Leftrightarrow4x+16=32\Leftrightarrow4x=16\Leftrightarrow x=4.\)
mk nghĩ câu hỏi này ko xứng tầm lớp 5 đâu!
\(\frac{3}{2.3.4}+\frac{3}{3.4.5}+...+\frac{3}{100.101.102}\)
\(=\frac{3}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{100.101}-\frac{1}{101.102}\right)\)
\(=\frac{3}{2}\left(\frac{1}{6}-\frac{1}{101.102}\right)\)