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\(B=\dfrac{2-\dfrac{2}{19}+\dfrac{2}{43}-\dfrac{2}{2017}}{3-\dfrac{3}{19}+\dfrac{3}{43}-\dfrac{3}{2017}}:\dfrac{4-\dfrac{4}{29}+\dfrac{4}{41}-\dfrac{4}{2018}}{5-\dfrac{5}{29}+\dfrac{5}{41}-\dfrac{5}{2018}}\)
\(B=\dfrac{2\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\right)}{3\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\right)}:\dfrac{4\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\right)}{5\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\right)}\)
\(B=\dfrac{2}{3}:\dfrac{4}{5}\) ( Do \(\left\{{}\begin{matrix}1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\ne0\\1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\ne0\end{matrix}\right.\))
\(B=\dfrac{2}{3}\cdot\dfrac{5}{4}=\dfrac{2\cdot5}{3\cdot4}=\dfrac{5}{6}\)
\(B=\dfrac{2-\dfrac{2}{19}+\dfrac{2}{43}-\dfrac{2}{2017}}{3-\dfrac{3}{19}+\dfrac{3}{43}-\dfrac{3}{2017}}:\dfrac{4-\dfrac{4}{29}+\dfrac{4}{41}-\dfrac{4}{2018}}{5-\dfrac{5}{29}+\dfrac{5}{41}-\dfrac{5}{2018}}\)
\(\Rightarrow\)\(B=\dfrac{2-\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\right)}{3\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\right)}:\dfrac{4\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\right)}{5\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\right)}\)
\(\Rightarrow B=\dfrac{2}{3}:\dfrac{4}{5}=\dfrac{10}{12}=\dfrac{5}{6}\)
\(A=\dfrac{636363\cdot37-373737\cdot63}{1+2+3+...+2006}\)
\(=\dfrac{37^2\cdot3^3\cdot7^2\cdot13-37^2\cdot3^3\cdot7^2\cdot13}{\left(2006+1\right)\cdot1003}\)
=0
\(A=\frac{2.\left(1-\frac{1}{19}+\frac{1}{43}-\frac{1}{1943}\right)}{3.\left(1-\frac{1}{19}+\frac{1}{43}-\frac{1}{1943}\right)}:\frac{4\left(1-\frac{1}{29}+\frac{1}{41}-\frac{1}{2941}\right)}{5.\left(1-\frac{1}{29}+\frac{1}{41}-\frac{1}{2941}\right)}=\frac{2}{3}:\frac{4}{5}=\frac{2}{3}.\frac{5}{4}=\frac{5}{6}\)
Xin lỗi nha, mình chỉ biết làm bài này nhõn bằng cách qui đồng nhưng số lớn lắm!
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!!!
mk có thắc mắc là bạn để 3 ra ngoài sao 1/3 vẫn giữ nguyên vậy phải bằng 1/9 mới đúng chứ'
đây là tính nhanh à nếu tính bình thường thì tính may tính là ra
a) 17/23 . 8/16 . 23/17. (-80) . 3/4
= (17/23 . 23/17) . (8/16 . 3/4) . (-80)
= 1 . 3/8 . (-80)
= 3/8 . (-80)
= -30
b) 5/11 . 18/29 - 5/11 . 8/29 + 5/11 . 19/29
= 5/11 . (18/29 - 8/29 + 19/29)
= 5/11 . 1
= 5/11
c)(13/23 + 1313/2323 - 131313/232323).(1/3+1/4 -7/12)
= (13/23 + 1313/2323 - 131313/232323).0
= 0
d) 12/2x2 . 22/2x3 . 32/3x4 . 42/4x5 . 52/5x6 . 62/6x7 . 72/7x8 . 82/8x9 . 92/9x10
= 1/2 . 2/3 . 3/4 . 4/5 . 5/6 . 6/7 . 7/8 . 8/9 .9/10
= 1/10
Khó nhìn quá. Bạn thông cảm nhé! 
a: =>19/23>19/x>19/29
=>\(x\in\left\{24;25;26;27;28\right\}\)
b: =>88/132<88/x<88/128
=>132>x>128
=>\(x\in\left\{131;130;129\right\}\)
c: =>\(\left\{{}\begin{matrix}\dfrac{4}{x}-\dfrac{x}{8}< 0\\\dfrac{x}{8}-\dfrac{5}{x}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{32-x^2}{8x}< 0\\\dfrac{x^2-40}{8x}< 0\end{matrix}\right.\)
=>32<x^2<40
=>x=6
a: \(A=\dfrac{19}{9}+\dfrac{4}{11}+\dfrac{2}{3}=\dfrac{209}{99}+\dfrac{44}{99}+\dfrac{66}{99}=\dfrac{319}{99}\)
b: \(B=\dfrac{-50}{60}+\dfrac{-35}{60}+\dfrac{12}{60}=\dfrac{-73}{60}\)
c: \(C=\dfrac{-27}{36}+\dfrac{132}{36}+\dfrac{10}{36}=\dfrac{115}{36}\)
d: \(D=\dfrac{-19}{3}+\dfrac{2}{3}-\dfrac{4}{5}=\dfrac{-17}{3}-\dfrac{4}{5}=\dfrac{-85-12}{15}=-\dfrac{97}{15}\)
\(A=\dfrac{5-\dfrac{5}{29}+\dfrac{5}{41}-\dfrac{5}{2941}}{3-\dfrac{3}{19}+\dfrac{3}{43}-\dfrac{3}{1943}}:\dfrac{4-\dfrac{4}{29}+\dfrac{4}{41}-\dfrac{4}{2941}}{2-\dfrac{2}{19}+\dfrac{2}{43}-\dfrac{2}{1943}}\)
\(=\dfrac{5-\dfrac{5}{29}+\dfrac{5}{41}-\dfrac{5}{2941}}{3-\dfrac{3}{19}+\dfrac{3}{43}-\dfrac{3}{1943}}.\dfrac{2-\dfrac{2}{19}+\dfrac{2}{43}-\dfrac{2}{1943}}{4-\dfrac{4}{29}+\dfrac{4}{41}-\dfrac{4}{2941}}\)
\(=\dfrac{5\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2941}\right)}{3\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{1943}\right)}.\dfrac{2\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{1943}\right)}{4\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2941}\right)}\)
\(=\dfrac{5}{3}.\dfrac{2}{4}=\dfrac{10}{12}=\dfrac{5}{6}\)
Vì đề bài không yêu cầu tính nên bn có thể không tính ra như mk cux đc!
