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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
\(\frac{22}{7}\)> \(\frac{11}{5}\)vì 22 : 7 = 3,14 ; 11: 5 = 2,2
\(\frac{15}{59}\)< \(\frac{24}{97}\)vì 15 : 59 = 0,21 ; 24 : 97 = 0,24
\(\frac{11}{19}\)< \(\frac{13}{18}\)vì 11 : 19 = 0,57 ; 13 : 18 = 0,72
\(\frac{7}{10}\)> \(\frac{4}{9}\)vì 7 : 10 = 0,7 ; 4 : 9 = 0,44
\(\frac{1}{13}+\frac{3}{13}+\frac{5}{13}+\frac{7}{13}+\frac{9}{13}+\frac{11}{13}\)
\(=\frac{1+3+5+7+9+11}{13}\)
\(=\frac{36}{13}\)
\(\left(a\right)\frac{9}{8}+\frac{15}{32}=\frac{36}{32}+\frac{15}{32}=\frac{36+15}{32}=\frac{51}{32}\)
\(\left(b\right)4+\frac{35}{45}=\frac{4}{1}+\frac{35}{45}=\frac{180}{45}+\frac{35}{45}=\frac{180+35}{45}=\frac{43}{9}\)
\(\left(c\right)\frac{11}{4}-\frac{15}{16}=\frac{44}{16}-\frac{15}{16}=\frac{44-15}{16}=\frac{29}{16}\)
\(\left(d\right)3-\frac{13}{9}=\frac{3}{1}-\frac{13}{9}=\frac{27}{9}-\frac{13}{9}=\frac{27-13}{9}=\frac{14}{9}\)
\(\left(e\right)\frac{5}{6}-\frac{5}{8}=\frac{20}{24}-\frac{15}{24}=\frac{20-15}{24}=\frac{5}{24}\)
\(\left(g\right)\frac{196}{64}-2=\frac{196}{64}-\frac{2}{1}=\frac{196}{64}-\frac{128}{64}=\frac{196-128}{64}=\frac{17}{16}\)
Mình đảm bảo đúng ! Chúc bạn học tốt!
a) \(\frac{9}{8}+\frac{15}{32}=\frac{36}{32}+\frac{15}{32}=\frac{51}{32}\)
b) \(4+\frac{35}{45}=4+\frac{7}{9}=\frac{36}{9}+\frac{7}{9}=\frac{43}{9}\)
c)\(\frac{11}{4}+\frac{15}{16}=\frac{44}{16}+\frac{15}{16}=\frac{59}{16}\)
d )\(3-\frac{13}{9}=\frac{27}{9}-\frac{13}{9}=\frac{14}{9}\)
e ) \(\frac{5}{6}+\frac{5}{8}=\frac{40}{48}+\frac{30}{48}\)\(=\frac{70}{48}\)
Trả lời
\(A=\frac{11}{12+13}+\frac{12}{13+14}+\frac{1}{14+15}\)
Hay
\(A=\frac{11}{12+13}+\frac{12}{13+14}+\frac{13}{14+15}\)
mong xem lại hộ cái
=13/12x14/13x15/14x16/15x...x2006/2005x2007/2006x2008/2007
=2008/12
=502/3
A = 1\(\dfrac{1}{12}\) \(\times\) 1\(\dfrac{1}{13}\) \(\times\) 1\(\dfrac{1}{14}\) \(\times\) 1\(\dfrac{1}{15}\) \(\times\) ... \(\times\) 1\(\dfrac{1}{2005}\) \(\times\) 1\(\dfrac{1}{2006}\) \(\times\) 1\(\dfrac{1}{2007}\)
A = ( 1 + \(\dfrac{1}{12}\)) \(\times\) ( 1 + \(\dfrac{1}{13}\)) \(\times\) ( 1 + \(\dfrac{1}{14}\)) \(\times\)...\(\times\) ( 1 + \(\dfrac{1}{2006}\))\(\times\)(1+\(\dfrac{1}{2007}\))
A = \(\dfrac{13}{12}\) \(\times\) \(\dfrac{14}{13}\) \(\times\) \(\dfrac{15}{14}\) \(\times\) ...\(\times\) \(\dfrac{2007}{2006}\) \(\times\) \(\dfrac{2008}{2007}\)
A = \(\dfrac{13\times14\times15\times...\times2007}{13\times14\times15\times...\times2007}\) \(\times\) \(\dfrac{2008}{12}\)
A = 1 \(\times\) \(\dfrac{502}{3}\)
A = \(\dfrac{502}{3}\)
\(\dfrac{13+x}{20}\) = \(\dfrac{3}{4}\)
13 + \(x\) = 20 \(\times\) \(\dfrac{3}{4}\)
13 + \(x\) = 15
\(x\) = 15 - 13
\(x\) = 2
Cách khác :
\(\dfrac{13+x}{20}=\dfrac{3}{4}\)
\(\dfrac{13+x}{20}=\dfrac{15}{20}\)
\(13+x=15\)
\(x=15-13\)
\(x=2\)
Trả lời:
11/13 + 9
= 11/13 + 117/13
= 128/13
\(\dfrac{11}{13}+9\)
\(=\dfrac{11}{13}+\dfrac{9}{1}\)
\(=\dfrac{11}{13}+\dfrac{117}{13}\)
\(=\dfrac{128}{9}\)