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\(10101.\left(\frac{5}{10101}-\frac{5}{20202}-\frac{5}{30303}-\frac{5}{40404}\right)=\frac{10101.5}{10101}-\frac{10101.5}{10101.2}-\frac{10101.5}{10101.3}-\frac{10101.5}{10101.4}\)\(=5-\frac{5}{2}-\frac{5}{3}-\frac{5}{4}=-\frac{5}{12}\)(nhưng lớp 4 chưa học phân số âm)
\(\frac{1}{2}\cdot\frac{18}{14}+\frac{1}{2}\cdot\frac{3}{14}-\frac{1}{2}\cdot\frac{7}{14}\)
\(=\frac{1}{2}\left(\frac{18}{14}+\frac{3}{14}-\frac{7}{14}\right)\)
\(=\frac{1}{2}\cdot1=\frac{1}{2}\)
9/14 .<.. 11/14 4/25 .<.. 4/23 5/6 ..<. 7/8 1 ..>. 14/15
8/9 ..=. 24/27 20/19 ..>. 20/27 8/7 .>.. 9/8 1 .<..15/14
\(=\dfrac{15}{14}\times\dfrac{15}{14}\times\dfrac{16}{15}\times\dfrac{17}{16}\times\dfrac{18}{17}=\dfrac{15}{14}\times\dfrac{18}{14}=\dfrac{135}{98}\)
\(\left(1+\dfrac{1}{14}\right)\left(1+\dfrac{1}{14}\right)\times\left(1+\dfrac{1}{15}\right)\times\left(1+\dfrac{1}{16}\right)\times\left(1+\dfrac{1}{17}\right)\)
\(=\dfrac{15}{14}.\dfrac{15}{14}.\dfrac{16}{15}.\dfrac{17}{16}.\dfrac{18}{17}\)
\(=\dfrac{15}{14}.\dfrac{18}{14}\)
\(\dfrac{135}{98}\)
\(a:=\left(\dfrac{1}{4}+\dfrac{3}{4}\right)+\left(\dfrac{1}{5}+\dfrac{2}{5}\right)\\ =1+\dfrac{3}{5}=1\dfrac{3}{5}\\ b:=\left(\dfrac{2}{7}+\dfrac{1}{7}\right)+\left(\dfrac{5}{14}+\dfrac{3}{14}\right)\\ =\dfrac{3}{7}+\dfrac{8}{14}\\ =\dfrac{3}{7}+\dfrac{4}{7}=1\)
\(\dfrac{40404}{70707}-\dfrac{1}{14}\\ =\dfrac{4\cdot10101}{7\cdot10101}-\dfrac{1}{14}\\ =\dfrac{4}{7}-\dfrac{1}{14}\\ =\dfrac{8}{14}-\dfrac{1}{14}\\ =\dfrac{7}{14}\\ =\dfrac{1}{2}\)