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a)=27+65+346-27-65
=(27-27)+(65-65)+346
=0+0+346
=346
b)=42+69-17-42+17-35
=(42-42)+(17-17)+(69-35)
=0+0+34
=34
A=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2014}{2015}.\frac{2015}{2016}\)
A=\(\frac{1.2.3.4...2015}{2.3.4...2016}=\frac{1}{2016}\)
Hok tốt
A = \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2015}\right).\left(1-\frac{1}{2016}\right)\)
= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2014}{2015}.\frac{2015}{2016}\)
= \(\frac{1}{2016}\)
Vậy ...
a,2.8.5.25.125.4
=(8.125)(4.25)(2.5)
=1000.100.10=1000000
b,-25.3.9.(-4)
=[-25.(-4)](3.9)
=100.27=2700
a) ( 2.5) . ( 8.125 ) . ( 25.4 ) = 10 . 1000 . 100 = 1000000
b) [ -25.(-4) ] . 3 . 9 = ( 25 . 4 ) . 27 = 100 . 27 = 2700
\(B=\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^9.33+3^9.15}{3^9.2^4}\)
\(=\frac{3^9\left(33+15\right)}{3^9.2^4}=\frac{3^9.48}{3^9.16}\)
\(=\frac{48}{16}=3\)
\(B=\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)
\(=\frac{3^{10}.\left(11+5\right)}{3^9.8}\)
\(=\frac{3^{10}.16}{3^9.8}\)
\(=\frac{3.2}{1}\)
\(=6\)
a) \(4,35-\left(2,67-1,65\right)+\left(3,54-6,33\right)\)
\(=4,35-2,67+1,65+3,54-6,33\\ =\left(4,35+1,65\right)-\left(2,67+6,33\right)+3,54\\ =6-9+3,54=3,54-3=0,54\)
b) \(\left(-0,5\right).\left(-0,5\right).\left(-0,8\right)\)
\(=\left(-\dfrac{1}{2}\right)\left(-\dfrac{1}{2}\right)\left(-\dfrac{4}{5}\right)\\ =-\left(\dfrac{1.1.4}{2.2.5}\right)=-\dfrac{1}{5}\)
c) \(3,58.24,45+3,58.75,55\)
\(=3,58\left(24,45+75,55\right)\\ =3,58.100=358\)
d) \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+25\%\right):\dfrac{7}{5}\)
\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{4}\right).\dfrac{5}{7}\\ =\dfrac{4.7.3}{3.5.4}-\dfrac{4}{5}.\dfrac{5}{7}\\ =\dfrac{7}{5}-\dfrac{4}{7}=\dfrac{29}{35}\)
e) \(\left(3,6-2\dfrac{2}{5}\right).\dfrac{-5}{3}+3\left(2\dfrac{1}{2}:50\%\right)\)
\(=\left(\dfrac{18}{5}-\dfrac{12}{5}\right).\dfrac{-5}{3}+3.\left(\dfrac{5}{2}:\dfrac{1}{2}\right)\\ =\dfrac{6}{5}.\dfrac{-5}{3}+3.5=-2+15=13\)