Tính bằng cách hợp lí: \(B=1\dfrac{6}{41}\cdot\left(\dfrac{12+\dfrac{12}{19}-\dfrac{12}{37}-\dfrac{12}{53}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2006}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2006}}\right)\cdot\dfrac{124242423}{237373735}\)
Ta có:B=1\(\dfrac{6}{41}\)( \(\dfrac{12+\dfrac{12}{19}-\dfrac{12}{37}-\dfrac{12}{53}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2006}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2006}}\) )
B=\(\dfrac{47}{41}\) [\(\dfrac{12\left(1+\dfrac{1}{19}-\dfrac{1}{37}-\dfrac{1}{53}\right)}{3\left(1+\dfrac{1}{3}-\dfrac{1}{37}-\dfrac{1}{53}\right)}:\dfrac{4\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2006}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2006}\right)}\) B = \(\dfrac{47}{41}\) [ \(\dfrac{12}{3}:\dfrac{4}{5}\)]
B = \(\dfrac{47}{41}\)[ 4 . \(\dfrac{5}{4}\)]
B = \(\dfrac{47}{41}.5\)
B = \(\dfrac{235}{41}\)
Chúc bn hc tốt!!!
Bài 2: Tính hợp lý:
\(A=\dfrac{63636337-37373763}{1+2+3+...+2006}\)
\(B=1\dfrac{6}{41}\left(\dfrac{12+\dfrac{12}{19}-\dfrac{12}{37}-\dfrac{12}{53}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2006}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2006}}\right)\dfrac{124242423}{237373735}\)
\(A=\dfrac{636363\cdot37-373737\cdot63}{1+2+3+...+2006}\)
\(=\dfrac{37^2\cdot3^3\cdot7^2\cdot13-37^2\cdot3^3\cdot7^2\cdot13}{\left(2006+1\right)\cdot1003}\)
=0
Tính \(S=1\dfrac{6}{41}.\left(\dfrac{12+\dfrac{12}{19}-\dfrac{12}{37}-\dfrac{12}{53}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4}{5}\right).\dfrac{124242423}{237373735}\)
\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)
\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)
\(=\dfrac{-6}{5}.\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}\)
\(=\dfrac{-6}{5}.4:\dfrac{4}{5}\)
\(=\dfrac{-6.4.5}{5.4}=-6\)
\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)
\(=\dfrac{-6}{5}.\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}\)
\(=\dfrac{-6}{5}.\dfrac{4}{3}:\dfrac{4}{5}\)
\(=\dfrac{-6.4.5}{5.3.4}=\dfrac{-6}{3}=-2\)
Vậy...
Thực hiện phép tính một cách hợp lý :
a) \(\dfrac{-12}{7}\) . \(\dfrac{4}{35}\) + \(\dfrac{12}{7}\) . \(\dfrac{-31}{35}\) - \(\dfrac{2}{7}\)
b) 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ... + 97 + 98 - 99 - 100.
c) A = 157 . ( - 37 ) - ( 41 . 53 - 37 . 157 ) + 51 . 53
d) B = \(\left(\dfrac{1}{11}+\dfrac{1}{21}+\dfrac{1}{31}+\dfrac{1}{41}+\dfrac{1}{51}\right)\) \(\left(\dfrac{-41}{123}+\dfrac{31}{-186}-\dfrac{-51}{102}\right)\)
a: \(=\dfrac{-12}{7}\left(\dfrac{4}{35}+\dfrac{31}{35}\right)-\dfrac{2}{7}=\dfrac{-12}{7}-\dfrac{2}{7}=-2\)
b: =(-4)+(-4)+...+(-4)
=-4*25=-100
c: \(=157\cdot\left(-37\right)-41\cdot53+37\cdot157+51\cdot53\)
=10*53
=530
thực hiện phép tính (bằng 1 cách hợp lí)
\(\dfrac{40}{17}\)+\(\dfrac{4}{15}\)-\(\dfrac{3}{37}\)+\(\dfrac{9}{13}\)+1
giúp mình với
\(l,\dfrac{\dfrac{3}{41}-\dfrac{12}{47}+\dfrac{27}{53}}{\dfrac{4}{41}-\dfrac{16}{47}+\dfrac{36}{53}}\)
\(m,\left(3-2\dfrac{1}{3}+\dfrac{1}{4}\right):\left(4-5\dfrac{1}{6}+2\dfrac{1}{4}\right)\)
\(n,F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)
\(p,F=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)
p: \(F=\dfrac{1}{3}\left(\dfrac{3}{3\cdot6}+\dfrac{3}{6\cdot9}+\dfrac{3}{9\cdot12}+...+\dfrac{3}{30\cdot33}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)
n: \(F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)
m: \(=\left(3-\dfrac{7}{3}+\dfrac{1}{4}\right):\left(4-\dfrac{31}{6}+\dfrac{9}{4}\right)\)
\(=\dfrac{36-28+3}{12}:\dfrac{48-62+27}{12}\)
\(=\dfrac{11}{13}\)
Tính :
a, \(\dfrac{3\cdot13-13\cdot18}{15\cdot40-80}\);
b, \(\dfrac{18\cdot34+\left(-18\right)\cdot124}{-36\cdot17+9\cdot\left(-52\right)}\);
c, \(\dfrac{\dfrac{3}{41}-\dfrac{12}{47}+\dfrac{27}{53}}{\dfrac{4}{41}-\dfrac{16}{47}+\dfrac{36}{53}}+\dfrac{-0,25\cdot\dfrac{-2}{3}-0,75:\left(\dfrac{-1}{2}+\dfrac{2}{3}\right)}{\left|-1\dfrac{1}{2}\right|\cdot\left(\dfrac{-2}{3}-75\%:\dfrac{3}{-2}\right)}\).
a: \(=\dfrac{13\left(3-18\right)}{40\left(15-2\right)}=\dfrac{13}{15-2}\cdot\dfrac{-15}{40}=\dfrac{-3}{8}\)
b: \(=\dfrac{18\left(34-124\right)}{36\left(-17-13\right)}=\dfrac{1}{2}\cdot\dfrac{-90}{-30}=\dfrac{3}{2}\)
c: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}+\dfrac{\dfrac{-1}{4}\cdot\dfrac{-2}{3}-\dfrac{3}{4}:\dfrac{1}{6}}{\dfrac{3}{2}\cdot\left(\dfrac{-2}{3}-\dfrac{3}{4}\cdot\dfrac{-2}{3}\right)}\)
\(=\dfrac{3}{4}+\dfrac{\dfrac{2}{12}-\dfrac{9}{2}}{\dfrac{3}{2}\cdot\dfrac{-1}{6}}=\dfrac{3}{4}+\dfrac{-13}{3}:\dfrac{-3}{12}=\dfrac{3}{4}+\dfrac{13}{3}\cdot\dfrac{12}{3}\)
\(=\dfrac{3}{4}+\dfrac{156}{9}=\dfrac{217}{12}\)
\(\text{Thực hiện phép tính một cách hợp lí:}\)
\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}\) \(\left(4-\dfrac{5}{12}\right):2+\dfrac{5}{24}\)
\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}.\dfrac{5}{1}+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{1}{3}.2+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{1}{3}.\left(\dfrac{2}{5}+\dfrac{3}{5}-2\right)\)=\(\dfrac{1}{3}.\left(-1\right)=\dfrac{-1}{3}\)
\(\left(4-\dfrac{5}{12}\right):2+\dfrac{5}{24}\)
=\(\left(4-\dfrac{5}{12}\right).\dfrac{1}{2}+\dfrac{5}{24}\)
=\(4.\dfrac{1}{2}-\dfrac{5}{12}.\dfrac{1}{2}+\dfrac{5}{24}\)
=\(2-\dfrac{5}{24}+\dfrac{5}{24}=2\)
Giải:
\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}\)
\(=\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{3}{5}.\dfrac{1}{3}\)
\(=\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{1}{3}.2+\dfrac{3}{5}.\dfrac{1}{3}\)
\(=\dfrac{1}{3}.\left(\dfrac{2}{5}-2+\dfrac{3}{5}\right)\)
\(=\dfrac{1}{3}.\left(-1\right)\)
\(=\dfrac{-1}{3}\)
\(\left(4-\dfrac{5}{12}\right):2+\dfrac{5}{24}\)
\(=\dfrac{43}{12}:2+\dfrac{5}{24}\)
\(=\dfrac{43}{24}+\dfrac{5}{24}\)
\(=2\)
