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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! 
\(=81.\dfrac{12.\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}{4.\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}:\dfrac{5.\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}{6.\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}.\dfrac{158}{711}\)
\(=81.\dfrac{12}{4}:\dfrac{5}{6}.\dfrac{2}{9}\)
\(=243:\dfrac{5}{6}.\dfrac{2}{9}\)
\(=\dfrac{1458}{5}.\dfrac{2}{9}\)
\(=\dfrac{324}{5}\)
\(\left(\dfrac{98}{99}+\dfrac{89}{100}+\dfrac{100}{101}\right)\left(\dfrac{1}{12}-\dfrac{1}{3}+\dfrac{1}{4}\right)\\ =\left(\dfrac{98}{99}+\dfrac{89}{100}+\dfrac{100}{101}\right)\left(-\dfrac{1}{4}+\dfrac{1}{4}\right)\\ =\left(\dfrac{98}{99}+\dfrac{89}{100}+\dfrac{100}{101}\right).0\\ =0\)
a, \(4\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)
\(=\left(-\dfrac{1}{2}\right)\left[\left(4\times-\dfrac{1}{2}\right)-\left(2\times-\dfrac{1}{2}\right)+3\right]+1\)
\(=\left(-\dfrac{1}{2}\right)\left(-2+1+3\right)+1\)
\(=\left(-\dfrac{1}{2}\right)2+1\)
\(=-1+1\)
\(=0\)
@Trịnh Thị Thảo Nhi
a, 4×(−12)3−2×(−12)2+3×(−12)+14×(−12)3−2×(−12)2+3×(−12)+1
=(−12)[(4×−12)−(2×−12)+3]+1=(−12)[(4×−12)−(2×−12)+3]+1
=(−12)(−2+1+3)+1=(−12)(−2+1+3)+1
=(−12)2+1=(−12)2+1
=−1+1=−1+1
=0=0
a) \(5\dfrac{4}{23}.27\dfrac{3}{47}+4\dfrac{3}{47}.\left(-5\dfrac{4}{23}\right)\)
\(=5\dfrac{4}{23}.27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right).5\dfrac{4}{23}\)
\(=5\dfrac{4}{23}.\left[27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right)\right]\)
\(=5\dfrac{4}{23}.\left(27\dfrac{3}{47}-4\dfrac{3}{27}\right)\)
\(=5\dfrac{4}{23}.23\)
\(=\dfrac{119}{23}.23\)
\(=\dfrac{119}{23}\)
b) \(4.\left(\dfrac{-1}{2}\right)^3+\dfrac{3}{2}\)
\(=4.\dfrac{-1}{6}+\dfrac{3}{2}\)
\(=\dfrac{-4}{6}+\dfrac{3}{2}\)
\(=\dfrac{-2}{3}+\dfrac{3}{2}\)
\(=\dfrac{-4}{6}+\dfrac{9}{6}\)
\(=\dfrac{5}{6}\)
c) \(\left(\dfrac{1999}{2011}-\dfrac{2011}{1999}\right)-\left(\dfrac{-12}{1999}-\dfrac{12}{2011}\right)\)
\(=\dfrac{1999}{2011}-\dfrac{2011}{1999}-\dfrac{-12}{1999}+\dfrac{12}{2011}\)
\(=\left(\dfrac{1999}{2011}+\dfrac{12}{2011}\right)-\left(\dfrac{2011}{1999}+\dfrac{-12}{1999}\right)\)
\(=\dfrac{2011}{2011}-\dfrac{1999}{1999}\)
\(=1-1\)
\(=0\)
d) \(\left(\dfrac{-5}{11}+\dfrac{7}{22}-\dfrac{-4}{33}-\dfrac{5}{44}\right):\left(\dfrac{381}{22}-39\dfrac{7}{22}\right)\)
(đợi đã, mình chưa tìm được hướng làm...)
a) \(\dfrac{1}{3}+\dfrac{3}{8}-\dfrac{7}{12}\)
\(=\dfrac{17}{24}-\dfrac{7}{12}\)
\(=\dfrac{1}{8}\)
b) \(\dfrac{-3}{14}+\dfrac{5}{8}-\dfrac{1}{2}\)
\(=\dfrac{23}{56}-\dfrac{1}{2}\)
\(=\dfrac{-5}{56}\)
c) \(\dfrac{1}{4}-\dfrac{2}{3}-\dfrac{11}{18}\)
\(=\dfrac{-5}{12}-\dfrac{11}{18}\)
\(=\dfrac{-37}{36}\)
d) \(\dfrac{1}{4}+\dfrac{5}{12}-\dfrac{1}{13}-\dfrac{7}{8}\)
\(=\dfrac{2}{3}-\dfrac{1}{13}-\dfrac{7}{8}\)
\(=\dfrac{23}{39}-\dfrac{7}{8}\)
\(=\dfrac{-89}{312}\)
\(-2\dfrac{1}{4}.\)\(\left(3\dfrac{5}{12}-1\dfrac{2}{9}\right)\)
=\(\dfrac{-9}{4}\).\(\left(\dfrac{41}{12}-\dfrac{11}{9}\right)\)
=\(\dfrac{-9}{4}.\dfrac{41}{12}-\dfrac{-9}{4}.\dfrac{11}{9}\)
=\(\dfrac{-123}{16}-\dfrac{-11}{4}\)
=\(\dfrac{-123}{16}-\dfrac{-44}{16}\)
=\(\dfrac{-79}{16}\)
\(\left(-25\%+0,75+\dfrac{7}{12}\right)\div\left(-2\dfrac{1}{8}\right)\)
=\(\left(\dfrac{-1}{4}+\dfrac{3}{4}+\dfrac{7}{12}\right)\div\left(\dfrac{-17}{8}\right)\)
=\(\left(\dfrac{-3}{12}+\dfrac{9}{12}+\dfrac{7}{12}\right).\dfrac{-8}{17}\)
=\(\dfrac{13}{12}.\dfrac{-8}{17}=\dfrac{-26}{51}\)
Gọi \(\dfrac{12}{23}+\dfrac{12}{2323}-\dfrac{121212}{232323}\) là A
Ta sẽ tính biểu thức A.\(\left(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{7}{12}\right)\)=A.\(\left(\dfrac{7}{12}-\dfrac{7}{12}\right)=0\)
Vậy \(\left(\dfrac{12}{23}+\dfrac{12}{2323}-\dfrac{121212}{232323}\right).\left(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{7}{12}\right)\)=0