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= \(x^8.\frac{1}{10}.\frac{2}{9}.\frac{3}{8}.\frac{4}{7}.\frac{5}{6}.\frac{6}{5}.\frac{7}{4}.\frac{8}{3}.\frac{9}{2}\)
= \(x^8.\frac{1}{10}.\left(\frac{2}{9}.\frac{9}{2}\right).\left(\frac{3}{8}.\frac{8}{3}\right).\left(\frac{4}{7}.\frac{7}{4}\right).\left(\frac{5}{6}.\frac{6}{5}\right)\)
= \(x^8.\frac{1}{10}.1.1.1.1\)
= \(x^8.\frac{1}{10}\)
Mk ko pik co dung ko nua
\(a,4\frac{1}{2}< 4\frac{3}{4}\)
\(b,2\frac{4}{5}< 3\frac{1}{4}\)
\(c,7\frac{2}{9}>5\frac{2}{9}\)
\(d,13\frac{5}{6}< 13\frac{6}{7}\)
Nao Tomori
\(a,4\frac{1}{2}....4\frac{3}{4}\Rightarrow4\frac{1}{2}=\frac{13}{2};4\frac{3}{4}=\frac{19}{4}\)
\(=4\frac{1}{2}< 4\frac{3}{4}\)
\(b,2\frac{4}{5}....3\frac{1}{4}\Rightarrow2\frac{4}{5}=\frac{14}{5};3\frac{1}{4}=\frac{13}{12}\)
\(=2\frac{4}{5}>3\frac{1}{4}\)
\(c,7\frac{2}{9}....5\frac{2}{9}\Rightarrow7\frac{2}{9}=\frac{65}{9};5\frac{2}{9}=\frac{42}{9}\)
\(=7\frac{2}{9}>5\frac{2}{9}\)
\(d,13\frac{5}{6}....13\frac{6}{7}\Rightarrow13\frac{5}{6}=\frac{83}{6};13\frac{6}{7}=\frac{97}{7}\)
\(=13\frac{5}{6}< 13\frac{6}{7}\)
P/s: Quy đồng là bước trung gian nên mk ko ghi bước quy đồng nha
\(\dfrac{3}{4}\times\dfrac{8}{5}:1\dfrac{1}{6}\)
=\(\dfrac{6}{5}:\) \(\dfrac{7}{6}\)
=\(\dfrac{6}{5}\times\dfrac{6}{7}=\dfrac{36}{35}\)
2\(\dfrac{1}{3}\) x 1\(\dfrac{1}{4}\) -\(\dfrac{7}{5}\)
\(\dfrac{7}{3}\times\dfrac{5}{4}-\) \(\dfrac{7}{5}\)
\(\dfrac{35}{12}-\dfrac{7}{5}\)
\(\dfrac{175}{60}-\dfrac{84}{60}=\dfrac{91}{60}\)
4\(\dfrac{2}{3}+1\dfrac{1}{4} +2\dfrac{1}{3}+2\dfrac{3}{7}\)
(4 +2) + \(\left(\dfrac{2}{3}+\dfrac{1}{3}\right)\) +1\(\dfrac{1}{4}\) + \(2\dfrac{3}{7}\)
6 + 1 + \(\dfrac{5}{4}\) + \(\dfrac{17}{7}\)
7 + \(\dfrac{103}{28}\)
\(\dfrac{299}{28}\)