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a) \(\dfrac{2}{3}+\dfrac{3}{5}=\dfrac{10}{15}+\dfrac{9}{15}=\dfrac{19}{15}\)
a) \(\dfrac{7}{12}-\dfrac{2}{7}+\dfrac{1}{12}=\dfrac{2}{3}-\dfrac{2}{7}=\dfrac{14}{21}-\dfrac{6}{21}=\dfrac{8}{21}\)
\(\dfrac{2}{5}+\dfrac{4}{9}=\dfrac{18}{45}+\dfrac{20}{45}=\dfrac{18+20}{45}=\dfrac{38}{45}\)
a; A = \(\dfrac{4026\times2014+4030}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2014+2015\right)}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2016-2013\times2+2015\right)}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2016-4026+2015\right)}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2016-2011\right)}{2013\times2016-2011}\)
A = 2
a; (5142 - 17 x 8 + 242 : 11) x (27 - 3 x 9)
= (5142 - 17 x 8 + 242 : 11) x (27 - 27)
= (5142 - 17 x 8 + 242 : 11) x 0
= 0
b;
(1 + \(\dfrac{1}{2}\)) \(\times\) (1 + \(\dfrac{1}{3}\)) \(\times\) ( 1 + \(\dfrac{1}{4}\)) \(\times\) ... \(\times\) (1 + \(\dfrac{1}{2010}\)) \(\times\)(1 + \(\dfrac{1}{2011}\))
= \(\dfrac{2+1}{2}\) \(\times\) \(\dfrac{3+1}{3}\) \(\times\) \(\dfrac{4+1}{4}\)\(\times\) ... \(\times\) \(\dfrac{2010+1}{2010}\)\(\times\) \(\dfrac{2011+1}{2011}\)
= \(\dfrac{3}{2}\)\(\times\)\(\dfrac{4}{3}\)\(\times\)\(\dfrac{5}{4}\)\(\times\)...\(\times\)\(\dfrac{2011}{2010}\)\(\times\)\(\dfrac{2012}{2011}\)
= \(\dfrac{2012}{2}\)
= 1006
Lời giải:
1.
$3\frac{1}{4}-2\frac{1}{3}=3+\frac{1}{4}-(2+\frac{1}{3})$
$=3-2+\frac{1}{4}-\frac{1}{3}$
$=1+\frac{1}{4}-\frac{1}{3}=\frac{5}{4}-\frac{1}{3}=\frac{11}{12}$
2.
$6\frac{5}{9}+2\frac{5}{6}=6+\frac{5}{9}+2+\frac{5}{6}=8+\frac{5}{6}+\frac{5}{9}=8+\frac{25}{18}=8+1+\frac{7}{18}=9+\frac{7}{18}=9\frac{7}{18}$
3.
$6\frac{5}{9}-2\frac{5}{6}=6+\frac{5}{9}-(2+\frac{5}{6})$
$=6+\frac{5}{9}-2-\frac{5}{6}$
$=(6-2)+\frac{5}{9}-\frac{5}{6}$
$=4+\frac{5}{9}-\frac{5}{6}=\frac{41}{9}-\frac{5}{6}=\frac{67}{8}$