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a) \(\frac{121212}{151515}+\frac{1414}{2121}=\frac{12}{15}+\frac{14}{21}=\frac{12}{15}+\frac{2}{3}=\frac{12}{15}+\frac{10}{15}=\frac{22}{15}\)
a,\(\frac{2015.2016+2015-1}{2014+2015.2016}=\frac{2015.2016+2014}{2014+2015.2016}=1\)\(1\)
b,\(=1-\frac{1}{5}+\frac{1}{5}...-\frac{1}{2011}+\frac{1}{2011}-\frac{1}{2015}=1-\frac{1}{2015}=\frac{2014}{2015}\)
c,\(=\frac{12}{35}+\frac{12}{35}+\frac{12}{35}+\frac{12}{35}=\frac{12}{35}.4=\frac{48}{35}\)
\(\frac{121212+12121212-1212}{363636+36363636-3636}\)
\(=\frac{12.10101+12.1010101-12.101}{36.10101+36.1010101-36.101}\)
\(=\frac{12\left(10101+1010101-101\right)}{36\left(10101+1010101-101\right)}\)
\(=\frac{12}{12.3}\)
\(=\frac{1}{3}\)
Ta có: \(\frac{121212+12121212-1212}{363636+36363636-3636}\)
=\(\frac{12\times\left(10101+1010101-101\right)}{36\times\left(10101+1010101-101\right)}\)
=\(\frac{12}{36}\)
=\(\frac{1}{3}\)
\(=\frac{13\times10101}{15\times10101}+\frac{13\times10101}{35\times10101}+\frac{13\times10101}{63\times10101}+\frac{13\times10101}{99\times10101}\)
\(=\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}=\frac{13}{2}\times\left(\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{3}{7\times9}+\frac{2}{9\times11}\right)\)
\(=\frac{13}{2}\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)=\frac{13}{2}\times\left(\frac{1}{3}-\frac{1}{11}\right)=\frac{13}{2}\times\frac{8}{33}=\frac{52}{33}\)
\(\frac{13.10101}{15.10101}\)+\(\frac{13.10101}{15.10101}\)+\(\frac{13.10101}{63.10101}\)+ \(\frac{13.10101}{99.10101}\)= \(\frac{13}{15}\) + \(\frac{13}{15}\) + \(\frac{13}{63}\)+ \(\frac{13}{99}\) =\(2\frac{82}{1155}\)
131313/151515+131313/353535+131313/636363+131313/999999=13/15+13/35+13/63+13/99
=52/33=1/19/33
DUYỆT NHÉ
\(\frac{121212}{151515}+\frac{121212}{353535}+\frac{121212}{636363}+\frac{121212}{999999}\)
= \(\frac{12.10101}{15.10101}+\frac{12.10101}{35.10101}+\frac{12.10101}{63.10101}+\frac{12.10101}{99.10101}\)
= \(\frac{12}{15}+\frac{12}{35}+\frac{12}{63}+\frac{12}{99}\)
= \(\frac{12}{3.5}+\frac{12}{5.7}+\frac{12}{7.9}+\frac{12}{9.11}\)
= \(6.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
= \(6.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
= \(6.\left(\frac{1}{3}-\frac{1}{11}\right)\)
= \(6.\frac{8}{33}\)
= \(\frac{16}{11}\)