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b)
\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)
\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(x-2=8\)
=> x = 10
a)
\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)
\(A=\frac{1}{2016}\)
Ta có :
\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)....\left(1+\frac{1}{2014.2016}\right)\)
\(=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.....\frac{4060225}{2014.2016}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}....\frac{2015.2015}{2014.2016}\)
\(=\frac{2.3.4....2015}{1.2.3....2014}.\frac{2.3.4....2015}{3.4.5....2016}\)
\(=\frac{2015}{1}.\frac{2}{2016}\)
\(=2015.\frac{1}{1008}=\frac{2015}{1008}\)
\(\Rightarrow\frac{2015}{1008}=\frac{x}{1008}\Rightarrow x=2015\)
Vậy \(x=2015\)
Ủng hộ mk nha !!! ^_^
a) \(\left(x-\frac{1}{2}\right).\frac{1}{3}+\frac{5}{7}=9\frac{5}{7}\)
\(\left(x-\frac{1}{2}\right).\frac{1}{3}=9\frac{5}{7}-\frac{5}{7}\)
\(x-\frac{1}{2}=9:\frac{1}{3}\)
\(x=27+\frac{1}{2}\)
\(x=\frac{55}{2}\)
b)\(\frac{1}{2}.x+150\%.x=2014\)
\(\left(\frac{1}{2}+150\%\right)x=2014\)
\(2x=2014\)
\(x=2014:2\)
\(x=1007\)
c) \(2-\left|\frac{3}{4}\right|=\frac{7}{2}\)???????
và bài này nữa
tính nhanh
a, \(5\frac{7}{5}\cdot4\frac{2}{7}+5\frac{5}{7}\cdot4\frac{2}{7}\)
b, \(b,\frac{-7}{11}\cdot\frac{4}{19}+\frac{-7}{19}\cdot\frac{7}{11}+2\frac{7}{19}\)
cảm ơn giúp nhiều
\(\left(1+\frac{1}{11}\right)\cdot\left(1+\frac{1}{10}\right)\cdot\left(1+\frac{1}{9}\right)\cdot.........................................\left(1+\frac{1}{2}\right)\)
\(=\frac{12}{11}.\frac{11}{10}.\frac{10}{9}....\frac{3}{2}\)
\(=\frac{12.11.10....3}{11.10.9....2}\)
\(=\frac{12}{2}=6\)
= \(\frac{12}{11}.\frac{11}{10}.....\frac{3}{2}=\frac{12}{2}=6\)
