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\(\frac{4}{3}:\frac{5}{4}:\frac{6}{5}:\frac{7}{6}:\frac{8}{7}:\frac{9}{8}\)
=\(\frac{4}{3}\times\frac{4}{5}\times\frac{5}{6}\times\frac{6}{7}\times\frac{7}{8}\times\frac{8}{9}\)
=\(\frac{16}{27}\)
\(\dfrac{9}{54}=\dfrac{1}{6}=\dfrac{4}{24}< \dfrac{7}{24}\)
\(\dfrac{7}{24}< \dfrac{12}{24}< \dfrac{16}{24}=\dfrac{2}{3}\)
\(\dfrac{2}{3}< \dfrac{2+1}{3+1}=\dfrac{3}{4}\)
\(\dfrac{3}{4}< \dfrac{3+2}{4+2}=\dfrac{5}{6}\)
\(\dfrac{5}{6}< \dfrac{5+2}{6+2}=\dfrac{7}{8}\)
\(\dfrac{7}{8}< \dfrac{7+9}{8+9}=\dfrac{16}{17}\)
Vậy \(\dfrac{9}{54}< \dfrac{7}{24}< \dfrac{2}{3}< \dfrac{3}{4}< \dfrac{5}{6}< \dfrac{7}{8}< \dfrac{16}{17}\)
\(\frac{5}{6}+\frac{7}{9}:\frac{5}{6}-\frac{1}{6}\)
\(=\left(\frac{5}{6}-\frac{1}{6}\right)+\frac{7}{9}.\frac{6}{5}\)
\(=\frac{2}{3}+\frac{14}{15}\)
\(=\frac{10}{15}+\frac{14}{15}=\frac{24}{15}\)
\(\frac{8}{11}.\frac{7}{8}+\frac{8}{11}:\frac{4}{5}\)
\(=\frac{7}{11}+\frac{8}{11}.\frac{5}{4}\)
\(=\frac{7}{11}+\frac{10}{11}\)
\(=\frac{17}{11}\)
\(\frac{5}{6}+\frac{7}{9}:\frac{5}{6}-\frac{1}{6}\) \(\frac{8}{11}x\frac{7}{8}+\frac{8}{11}:\frac{4}{5}\)
\(=\frac{5}{6}+\frac{14}{15}-\frac{1}{6}\) \(=\frac{8}{11}x\frac{7}{8}+\frac{8}{11}x\frac{5}{4}\)
\(=\left(\frac{5}{6}-\frac{1}{6}\right)+\frac{14}{15}\) \(=\frac{8}{11}x\left(\frac{7}{8}+\frac{5}{4}\right)\)
\(=\frac{2}{3}+\frac{14}{15}\) \(=\frac{8}{11}x\frac{17}{8}\)
\(=\frac{8}{5}\) \(=\frac{17}{11}\)
\(P=\dfrac{3}{119}+\dfrac{6}{119}+\dfrac{9}{119}+...+\dfrac{102}{119}\)
\(=\dfrac{3+6+9+...+102}{119}\)
\(=\dfrac{3\times\left(1+2+3+...+34\right)}{119}\)
Ta đi tính tổng \(S=1+2+3+...+34\)
Số các số hạng là 34, như thế \(S=\dfrac{34\times\left(34+1\right)}{2}=595\)
Do đó \(P=\dfrac{3\times595}{119}=15\)
(156+6)-(5+6)/(9*56)+(584+33336)
= 162 - 11 / 504 + 33920
= 152/34424 = 19/4303 ( cả tử và mẫu rút gọn cho 8)