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\(a,=\dfrac{5}{7}-\dfrac{33}{8}=-\dfrac{191}{56}\\ b,=\left(\dfrac{12}{17}+\dfrac{5}{17}\right)+\left(\dfrac{19}{7}+\dfrac{3}{7}\right)=1+3=4\\ c,=\left(0,125\cdot8\right)^{12}-\left(\dfrac{45}{15}\right)^3=1-3^3=-26\\ d,=\left(-\dfrac{1}{3}\right)\left(5\dfrac{2}{7}-2\dfrac{2}{7}\right)=-\dfrac{1}{3}\cdot3=-1\\ e,=\dfrac{3^4\cdot3^6}{3^9}=3\)
\(A=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{64}\)
\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{64}\)
=1/64
a) A = 15/12 + 5/13 + (-3/12) + (-18/13)
= (15/12 - 3/12) + (5/13 - 18/13)
= 1 - 1
= 0
b) B = 11/15 . (-19/13) + (-7/13) . 11/15
= 11/15.(-19/13 - 7/13)
= 11/15 . (-2)
= -22/15
c) C = 2022⁰ - (1/7)⁵ . 7⁵
= 1 - 1/7⁵ . 7⁵
= 1 - 1
= 0
\(a,\dfrac{3}{5}+\dfrac{1}{5}.\dfrac{-17}{9}=\dfrac{3}{5}-\dfrac{17}{45}=\dfrac{27}{45}-\dfrac{17}{45}=\dfrac{10}{45}=\dfrac{2}{9}\\ b,\left(-\dfrac{4}{15}-\dfrac{18}{19}\right)-\left(\dfrac{20}{19}+\dfrac{11}{15}\right)=-\dfrac{4}{15}-\dfrac{18}{19}-\dfrac{20}{19}-\dfrac{11}{15}=\left(-\dfrac{4}{15}-\dfrac{11}{15}\right)-\left(\dfrac{18}{19}+\dfrac{20}{19}\right)=-1-2=-3\)
\(a,=\dfrac{3}{5}+\left(-\dfrac{17}{45}\right)=\dfrac{2}{9}\)
\(b,=-\dfrac{4}{15}-\dfrac{18}{19}-\dfrac{20}{19}-\dfrac{11}{15}=-3\)
\(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)
\(=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{3}{5}-\dfrac{3}{5}\right)+\left(\dfrac{5}{7}-\dfrac{5}{7}\right)-\left(\dfrac{7}{9}-\dfrac{7}{9}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)-\left(\dfrac{11}{13}-\dfrac{11}{13}\right)+\dfrac{13}{15}\)
\(=\dfrac{13}{15}\)
a, \(\left(\dfrac{-5}{11}\right).\dfrac{7}{15}.\left(\dfrac{11}{-5}\right).\left(-30\right)=\dfrac{-5}{11}.\dfrac{7}{15}.\dfrac{-11}{5}.\dfrac{-30}{1}\)= ( - 14 )
b, \(\left(\dfrac{11}{12}:\dfrac{33}{16}\right).\dfrac{3}{5}=\dfrac{11}{12}.\dfrac{16}{33}.\dfrac{3}{5}=\dfrac{1.4.3}{3.3.5}=\dfrac{4}{15}\)
c, \(\dfrac{-7}{15}.\dfrac{5}{8}.\dfrac{15}{-7}.\left(-16\right)=\dfrac{-7}{15}.\dfrac{5}{8}.\dfrac{-15}{7}.\dfrac{-16}{1}\)
\(\dfrac{-1.5.-1.-2}{1.1.1.1}=\left(-10\right)\)
d,\(\left(\dfrac{-1}{2}\right).3\dfrac{1}{5}+\left(\dfrac{-1}{2}\right).-2\dfrac{1}{5}=\left(\dfrac{-1}{2}\right).\left[\dfrac{16}{5}+\left(\dfrac{-11}{5}\right)\right]\)
= \(\left(\dfrac{-1}{2}\right).1=\dfrac{-1}{2}\)
a) (-5/11.11/-5).7/15
=1.7/15=7/15
b)(11/12:33/16).3/5
=(11/12.16/33).3/5
=4/9.3/5=4/15
c)(-7/15.15/-7).5/8
=1.5/8=5/8
d)(-1/2).(16/5.-11/5)
=-1/2.1=-1/2
xg r đó
a)
\(\dfrac{-5}{11}\cdot\dfrac{7}{15}\cdot\dfrac{11}{-5}\cdot\left(-30\right)\\ =\left(\dfrac{-5}{11}\cdot\dfrac{11}{-5}\right)\cdot\left(\dfrac{7}{15}\cdot-30\right)\\ =1\cdot-14\\ =-14\)
b)
\(-\dfrac{1}{3}\cdot\left(-\dfrac{15}{19}\right)\cdot\dfrac{38}{45}\\ =-\dfrac{1}{3}\cdot\left(-\dfrac{15}{29}\cdot\dfrac{38}{45}\right)\\ =-\dfrac{1}{3}\cdot\left(-\dfrac{15}{29}\cdot\dfrac{2\cdot19}{3\cdot15}\right)\\ =-\dfrac{1}{3}\cdot-\dfrac{2}{3}\\ =\dfrac{2}{9}\)