Tính một cách hợp lí.
\(0,65.78 + 2\dfrac{1}{5}.2020 + 0,35.78 - 2,2.2020.\)
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a: \(=\dfrac{28-2-3}{4}:\dfrac{40-2-5}{8}=\dfrac{23}{4}\cdot\dfrac{8}{33}=\dfrac{46}{33}\)
b: =78(0,65+0,35)+2020(2,2-2,2)
=78*1=78
\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}.\dfrac{5}{1}+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{1}{3}.2+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{1}{3}.\left(\dfrac{2}{5}+\dfrac{3}{5}-2\right)\)=\(\dfrac{1}{3}.\left(-1\right)=\dfrac{-1}{3}\)
d: =-2/7-3/11+2/7=-3/11
e: =2+3/7+1+4/7-17/7
=4-17/7=11/7
f: =-2/3*4/5+1/5*11/9
=-8/15+11/45
=-24/45+11/45=-13/45
h: =(-3,1+3,7):0,2=0,6:0,2=3
\(=\left(\dfrac{9}{11}+\dfrac{20}{11}\right)+\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{8}{13}\\ =\dfrac{29}{11}+1+\dfrac{8}{13}\)
Biết đến đó thôi sorry:<
= 11/125 - [(17/18 - 4/9) + (5/7 - 17/14)]
= 11/125 - (1/2 - 1/2)
= 11/125
\(\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}=\dfrac{11}{125}-\left(\dfrac{17}{18}-\dfrac{4}{9}\right)+\left(\dfrac{17}{14}-\dfrac{5}{7}\right)=\dfrac{11}{125}-\dfrac{17-4.2}{18}+\dfrac{17-5.2}{14}=\dfrac{11}{125}-\dfrac{9}{18}+\dfrac{7}{14}=\dfrac{11}{125}-\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{11}{125}\)
Ta có: \(B=\dfrac{5}{13}\cdot\dfrac{8}{15}+\dfrac{5}{13}\cdot\dfrac{26}{15}-\dfrac{5}{13}\cdot\dfrac{8}{15}\)
\(=\dfrac{5}{13}\cdot\dfrac{26}{15}\)
\(=\dfrac{130}{195}=\dfrac{2}{3}\)
\(\begin{array}{l}B = \dfrac{{ - 1}}{9} + \dfrac{8}{7} + \dfrac{{10}}{9} + \dfrac{{ - 29}}{7}\\ = \left( {\dfrac{{ - 1}}{9} + \dfrac{{10}}{9}} \right) + \left( {\dfrac{8}{7} + \dfrac{{ - 29}}{7}} \right)\\ = \dfrac{9}{9} + \dfrac{{ - 21}}{7} = 1 - 3 = - 2\end{array}\)
\(\begin{array}{l}0,65.78 + 2\dfrac{1}{5}.2020 + 0,35.78 - 2,2.2020\\ = 0,65.78 + 2,2.2020 + 0,35.78 - 2,2.2020\\ = (0,65.78 + 0,35.78) + (2,2.2020 - 2,2.2020)\\ = 78.(0,65 + 0,35) + 0\\ = 78.1\\ = 78\end{array}\)