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=(5-10)+(15-20)+...+(2005-2010)+2015
=2015-5-5-...-5
=2015-5*201
=2015-1005
=1010
a. 3/5 -1/5 x 10/13
= 3/5 - 2/13
= 29/65
b. 1/2 + 3/8 : 3/4
= 1/2 + 1/2
= 1
c. (3/5 - 3/20) x4/5
= 9/20 x 4/5
= 9/25
\(a,\frac{3}{5}-\frac{1}{15}x\frac{10}{13}\)
=3/5-2/39
=107/195
b,1/2+3/8:3/4
=1/2+1/2
=1
c,(3/5-3/20)x4/5
=9/20x4/5
=9/25
a: \(\dfrac{5}{3}\cdot\dfrac{7}{10}=\dfrac{5}{10}\cdot\dfrac{7}{3}=\dfrac{14}{3}>2\)
\(\dfrac{5}{7}:\dfrac{7}{10}=\dfrac{5}{7}\cdot\dfrac{10}{7}=\dfrac{50}{49}< 2\)
=>\(\dfrac{5}{3}\cdot\dfrac{7}{10}>\dfrac{5}{7}:\dfrac{7}{10}\)
b: \(\dfrac{4}{5}\cdot\dfrac{5}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)
\(\dfrac{8}{9}:\dfrac{4}{3}=\dfrac{8}{9}\cdot\dfrac{3}{4}=\dfrac{8}{4}\cdot\dfrac{3}{9}=\dfrac{2}{3}\)
Do đó: \(\dfrac{4}{5}\cdot\dfrac{5}{6}=\dfrac{8}{9}:\dfrac{4}{3}\)
c: \(\dfrac{5}{11}\cdot\dfrac{33}{15}=\dfrac{5}{15}\cdot\dfrac{33}{11}=\dfrac{3}{3}=1\)
\(\dfrac{6}{17}\cdot\dfrac{34}{25}=\dfrac{34}{17}\cdot\dfrac{6}{25}=2\cdot\dfrac{6}{25}=\dfrac{12}{25}< 1\)
=>\(\dfrac{5}{11}\cdot\dfrac{33}{15}>\dfrac{6}{17}\cdot\dfrac{34}{25}\)
d: \(\dfrac{15}{19}\cdot\dfrac{38}{5}=\dfrac{15}{5}\cdot\dfrac{38}{19}=3\cdot2=6\)
\(\dfrac{15}{16}:\dfrac{3}{8}=\dfrac{15}{16}\cdot\dfrac{8}{3}=\dfrac{15}{3}\cdot\dfrac{8}{16}=\dfrac{5}{2}\)<6
=>\(\dfrac{15}{19}\cdot\dfrac{38}{5}>\dfrac{15}{16}:\dfrac{3}{8}\)
a, \(\frac{5}{16}+\frac{3}{8}=\frac{5}{16}+\frac{6}{16}=\frac{11}{16}\)
b, \(\frac{15}{25}+\frac{6}{21}=\frac{3}{5}+\frac{2}{7}\)
\(=\frac{21}{35}+\frac{10}{35}=\frac{31}{35}\)
A = 5 + 10 + 15 + ... + 2010 ( tổng này có 402 số hạng )
A = \(\left[\left(5+2010\right).402\right]:2=405015\)
Dãy trên có số số hạng là:
\(\left(2010-5\right):5+1=402\left(số\right)\)
Tổng dãy số trên là:
\(\left(2010+5\right)\times402:2=405015\)