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bài 1:
\(\frac{7}{4}\left(\frac{33}{42}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
\(=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(=\frac{7}{4}.33\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=\frac{231}{4}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{231}{4}\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(=\frac{231}{4}\cdot\frac{4}{21}=11\)
\(\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{1931}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)
= \(\left[\frac{193}{17}.\frac{2}{193}-\frac{193}{17}.\frac{3}{386}+\frac{33}{34}\right]:\left[\frac{1931}{25}.\frac{7}{1931}+\frac{1931}{25}.\frac{11}{3862}+\frac{9}{2}\right]\)
= \(\left[\frac{2}{17}-\frac{3}{17}+\frac{33}{34}\right]:\left[\frac{7}{25}+\frac{11}{50}+\frac{9}{2}\right]\)
= \(\left[\frac{4}{34}-\frac{6}{34}+\frac{33}{34}\right]:\left[\frac{14}{50}+\frac{11}{50}+\frac{225}{50}\right]\)
= \(\frac{31}{34}:2\)
= \(\frac{31}{68}\)
Giải:
Đặt \(A=13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6},B=1\frac{3}{7}+\frac{10}{3},C=12\frac{1}{3}-14\frac{2}{7}\)
Ta có:
\(A=13-2-10+\frac{1}{4}-\frac{5}{27}-\frac{5}{6}=1+\frac{27-20-90}{108}=1-\frac{83}{108}=\frac{25}{108}\)
\(A.230\frac{1}{5}=\frac{25}{108}.230\frac{1}{25}=\frac{25}{108}.\frac{5751}{25}=\frac{5751}{108}=\frac{213}{4}\)
\(A.230\frac{1}{25}+46\frac{3}{4}=\frac{213}{4}+\frac{187}{4}=100\)
\(B=1+\frac{3}{7}+3+\frac{1}{3}=4+\frac{3}{7}+\frac{1}{3}=4+\frac{16}{21}=\frac{100}{21}\)
\(C=12\frac{1}{3}-14\frac{2}{7}=12-14+\frac{1}{3}-\frac{2}{7}=-2+\frac{7-6}{21}=-\frac{41}{21}\)
\(B:C=\frac{100}{21}:\left(-\frac{41}{21}\right)=-\frac{100}{41}\)
\(E=\frac{A.230\frac{1}{25}+46\frac{3}{4}}{B:C}=\frac{100}{-\frac{100}{41}}=-41\)
c.\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
\(\frac{\frac{25}{108}.\frac{5751}{25}+\frac{187}{4}}{\frac{100}{21}:-\frac{41}{21}}\)
\(\frac{\frac{213}{4}+\frac{187}{4}}{-\frac{100}{41}}\)
\(\frac{100}{-\frac{100}{41}}=-41\)
a. \(\frac{4}{9}:-\frac{1}{7}+6\frac{5}{9}:-\frac{1}{7}\)
\(\left(\frac{4}{9}+6\frac{5}{9}\right):-\frac{1}{7}\)
\(7:-\frac{1}{7}=-49\)
\(B=\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(=\dfrac{7}{4}\left(\dfrac{11}{4}+\dfrac{33}{20}+\dfrac{11}{10}+\dfrac{11}{14}\right)\)
\(=\dfrac{7}{4}\cdot\dfrac{11\cdot35+33\cdot7+11\cdot14+11\cdot10}{140}\)
\(=\dfrac{880}{20\cdot4}=11\)
\(C=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right)\cdot\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{100}{21}:\dfrac{-41}{21}}\)
\(=\dfrac{\dfrac{25}{108}\cdot\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{-100}{41}}\)
\(=\dfrac{\dfrac{5751+187\cdot27}{108}}{\dfrac{-100}{41}}=100\cdot\dfrac{-41}{100}=-41\)