Tính B = \(\frac{2}{27}+\frac{10}{18.24}+\frac{14}{48.45}+\frac{4}{90.51}\)
Bài 3 : a) Tính
\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right)\cdot230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
b) Tính :
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+\frac{1}{2011}}\)
Tính A = \(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right)230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
\(\frac{13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}).230.\frac{1}{25}+46\frac{3}{4}}{(1\frac{3}{10}+\frac{10}{3}):(12\frac{1}{3}-14\frac{2}{7})}\)
tính giá trị của biểu thức trên
Tính:
\(E=\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}.\)
Tính:
\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right)\cdot230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
Tính :
\(B=\frac{7}{4}\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
\(C=\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
\(B=\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(=\dfrac{7}{4}\left(\dfrac{11}{4}+\dfrac{33}{20}+\dfrac{11}{10}+\dfrac{11}{14}\right)\)
\(=\dfrac{7}{4}\cdot\dfrac{11\cdot35+33\cdot7+11\cdot14+11\cdot10}{140}\)
\(=\dfrac{880}{20\cdot4}=11\)
\(C=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right)\cdot\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{100}{21}:\dfrac{-41}{21}}\)
\(=\dfrac{\dfrac{25}{108}\cdot\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{-100}{41}}\)
\(=\dfrac{\dfrac{5751+187\cdot27}{108}}{\dfrac{-100}{41}}=100\cdot\dfrac{-41}{100}=-41\)
Tính : a) \(\frac{1}{9}\div\frac{8}{27}\times-3\div\frac{81}{128}\)
b) \(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right)\times230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right)\div\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
Đề bài là tính nha !
Tính: \(E=\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
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:
\(A=13-2-10+\frac{1}{4}-\frac{5}{27}-\frac{5}{6}=1+\frac{27-20-90}{108}=1-\frac{83}{108}=\frac{25}{108}\)
\(A.230\frac{1}{5}=\frac{25}{108}.230\frac{1}{25}=\frac{25}{108}.\frac{5751}{25}=\frac{5751}{108}=\frac{213}{4}\)
\(A.230\frac{1}{25}+46\frac{3}{4}=\frac{213}{4}+\frac{187}{4}=100\)
\(B=1+\frac{3}{7}+3+\frac{1}{3}=4+\frac{3}{7}+\frac{1}{3}=4+\frac{16}{21}=\frac{100}{21}\)
\(C=12\frac{1}{3}-14\frac{2}{7}=12-14+\frac{1}{3}-\frac{2}{7}=-2+\frac{7-6}{21}=-\frac{41}{21}\)
\(B:C=\frac{100}{21}:\left(-\frac{41}{21}\right)=-\frac{100}{41}\)
\(E=\frac{A.230\frac{1}{25}+46\frac{3}{4}}{B:C}=\frac{100}{-\frac{100}{41}}=-41\)