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Ta có: \(\left(\dfrac{1}{57}-\dfrac{1}{5757}+\dfrac{1}{23}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(=\left(\dfrac{1}{57}-\dfrac{1}{5757}+\dfrac{1}{23}\right)\cdot\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\)
=0
\(\left(\dfrac{1}{57}-\dfrac{1}{5757}+\dfrac{1}{23}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)= 0
(1/57 + 1/5757 + 1/23 ) . ( 1/2 - 1/3 - 1/6 )
= ( 1/57 + 1/5757 + 1/23 ) . ( 3/6 - 2/6 - 1/6 )
= (1/57 + 1/5757 + 1/23 ) . 0
= 0
a) \(\frac{53}{101}.\frac{-13}{97}+\frac{53}{101}.\frac{-84}{97}\)
\(=\frac{53}{101}\left(\frac{-13}{97}+\frac{-84}{97}\right)\)
\(=\frac{53}{101}.\frac{-97}{97}\)
\(=\frac{53}{101}.\left(-1\right)\)
\(=\frac{-53}{101}\)
b) \(\left(\frac{1}{57}-\frac{1}{5757}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\left(\frac{1}{57}-\frac{1}{5757}\right)\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)
\(=\left(\frac{1}{57}-\frac{1}{5757}\right).0\)
\(=0\)
c) \(\frac{3^2}{25}.\frac{75}{-21}.\frac{50}{35}\)
\(=\frac{3^2.75.50}{25.\left(-21\right).35}\)
\(=\frac{3.3.25.3.5.5.2}{25.3.\left(-7\right).5.7}\)
\(=\frac{3.3.5.2}{\left(-7\right).7}\)
\(=\frac{90}{-49}\)
d) \(\frac{25.48-25.18}{20.5^3}\)
\(=\frac{25\left(48-18\right)}{10.2.125}\)
\(=\frac{25.10.3}{10.2.25.5}\)
\(=\frac{3}{10}\)
\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\frac{1}{18\cdot19\cdot20}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+\frac{2}{18\cdot19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\cdot\frac{189}{380}=\frac{189}{760}\)
\(C=\frac{52}{1\cdot6}+\frac{52}{6\cdot11}+\frac{52}{11\cdot16}+...+\frac{52}{31\cdot36}\)
\(C=\frac{52}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{6}{31\cdot36}\right)\)
\(C=\frac{52}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{31}-\frac{1}{36}\right)\)
\(C=\frac{52}{5}\cdot\left(1-\frac{1}{36}\right)\)
\(C=\frac{91}{9}\)
a) (-30) + 17 + (-70) + 33
=[-30+(-70)]+(17+33)
=-100+50
=-50
b) 56 - 15 - 6 + (-75)
=(56-6)-[15+(-75)]
=50-(-60)
=110
c) (157 - 56) - (44 + 57)
=157-56-44+57
=(-157-57)-(56+44)
=-100-100
=-200
d) (-24 + 174) - (174 - 24)
=-24+174-174+24
=(-24+24)+(174-174)
=0+0
=0
e) (35 - 47) - (-123 + 35) + [20 - (123 - 57)]
=35-47+123-35+[20-123+57]
=35-47+123-35+20-123+57
=(35-35)+(123-123)+(-47+57)
=0+0+(-10)
=-10
A=\(\dfrac{-2}{9}\)+\(\dfrac{-3}{4}\)+\(\dfrac{3}{5}\)+\(\dfrac{1}{15}\)+\(\dfrac{1}{57}\)+\(\dfrac{1}{3}\)+\(\dfrac{-1}{36}\)
A=(\(\dfrac{-2}{9}\)+\(\dfrac{-3}{4}\)+\(\dfrac{-1}{36}\))+(\(\dfrac{3}{5}\)+\(\dfrac{1}{15}\)+\(\dfrac{1}{3}\))
A=-1+1=0 B=\(\dfrac{1}{2}\)+\(\dfrac{-1}{5}\)+\(\dfrac{-5}{7}\)+\(\dfrac{1}{6}\)+\(\dfrac{-3}{35}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{41}\) B=(\(\dfrac{-1}{5}\)+\(\dfrac{-5}{7}\)+\(\dfrac{-3}{35}\))+(\(\dfrac{1}{2}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{3}\))+\(\dfrac{1}{41}\) B=-1+1+\(\dfrac{1}{41}\)=\(\dfrac{1}{41}\)
câu 1 : A=-2/9+-3/4+3/5+1/15+1/57+1/3+-1/36
=(-2/9+-3/4+-1/36)+(3/5+1/15+1/3)
Vậy p/s 1/57 đâu bạn ?
Đáp án cần chọn là: B
G = 1 57 − 1 5757 + 1 35 1 2 − 1 3 − 1 6 = 1 57 − 1 5757 + 1 35 .0 = 0