A=(\(4\frac{5}{37}-3\frac{4}{5}+8\frac{15}{29}\)) - \(\left(3\frac{5}{37}-6\frac{15}{29}\right)\)
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\(\left(4\frac{5}{37}-3\frac{4}{5}+8\frac{15}{29}\right)-\left(3\frac{5}{37}-6\frac{14}{29}\right)\)
\(\left(4\frac{5}{37}-3\frac{4}{5}+8\frac{15}{29}\right)-\left(3\frac{5}{37}-6\frac{14}{29}\right)\)
\(a.=\left(\frac{153}{37}-\frac{19}{5}+\frac{199}{23}\right)-\left(\frac{116}{37}-\frac{188}{29}\right)\)
\(=\frac{153}{37}-\frac{19}{5}+\frac{199}{23}-\frac{116}{37}+\frac{188}{29}\)
\(=\frac{37}{37}-\frac{19}{5}+\frac{199}{23}+\frac{188}{29}\)tự giải tiếp ^^
\(b.=\frac{8}{3}.\frac{-15}{4}.\frac{4}{5}\)
\(=\frac{8.\left(-15\right).4}{3.4.5}\)
\(=\frac{-480}{60}=-8\)
\(B=1\frac{6}{41}\cdot\left(\frac{12+\frac{12}{19}-\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}-\frac{3}{37}-\frac{3}{53}}\div\frac{4+\frac{4}{15}+\frac{4}{4}+\frac{4}{2013}}{5+\frac{5}{15}+\frac{5}{4}+\frac{5}{2013}}\right)\cdot\frac{124242423}{237373735}\)
\(B=\frac{47}{41}\cdot\left[\frac{12\left(1+\frac{1}{19}-\frac{1}{37}-\frac{1}{53}\right)}{3\left(1+\frac{1}{19}-\frac{1}{37}-\frac{1}{53}\right)}\div\frac{4\left(1+\frac{1}{15}+\frac{1}{4}+\frac{1}{2013}\right)}{5\left(1+\frac{1}{15}+\frac{1}{4}+\frac{1}{2013}\right)}\right]\cdot\frac{123}{235}\)
\(B=\frac{47}{41}\cdot\left[\frac{12}{3}\div\frac{4}{5}\right]\cdot\frac{123}{235}\)
\(B=\frac{3}{5}\cdot3\cdot\frac{5}{4}\)
\(B=\frac{9}{4}\)
\(\frac{1774}{145}\)
\(\frac{1774}{145}\)