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1) 9 x 10 + 10 x 11 + 11 x 12 + ....+ 2015 x 2016
=9x10x11-8x9x10+10x11x12-9x10x11+...+2015x2016x2017-2014x2015x2013
=2015x2016x2017-8x9x10
=8193537360
2015 x 23 + 55 x 2015 + 11 x 2015 x 2
= 2015 x (23 + 55 + 11 x 2)
= 2015 x (23 + 55 + 22)
= 2015 x (78 + 22)
= 2015 x 100
= 201500
\(2015x23+55x2015+11x2015x2\)
\(=2015x\left(23+55+11x2\right)\)
\(=2015x\left(23+55+22\right)\)
\(=2015x100\)
\(=201500\)
a/
\(\dfrac{4}{15}=1-\dfrac{11}{15}\)
\(\dfrac{5}{16}=1-\dfrac{11}{16}\)
\(\dfrac{11}{15}>\dfrac{11}{16}\Rightarrow1-\dfrac{11}{15}< 1-\dfrac{11}{16}\Rightarrow\dfrac{4}{15}< \dfrac{5}{16}\)
b/
\(\dfrac{2}{113}=\dfrac{4}{226}< \dfrac{4}{115}\)
c/ \(\dfrac{2}{7}=\dfrac{4}{14}< \dfrac{4}{9}\)
d/ \(\dfrac{2}{5}=\dfrac{4}{10}< \dfrac{4}{7}\)
a,\(\dfrac{4}{15}< \dfrac{5}{16}\)
b,\(\dfrac{2}{113}< \dfrac{4}{115}\)
c,\(\dfrac{2}{7}< \dfrac{4}{9}\)
d,\(\dfrac{4}{7}>\dfrac{2}{5}\)
\(\dfrac{x-25-124}{2015}\) + \(\dfrac{x-124-2015}{25}\) + \(\dfrac{x-2015-25}{124}\) = 3
\(\dfrac{x-15-124}{2015}\) - 1 + \(\dfrac{x-124-2015}{25}\) - 1 + \(\dfrac{x-2015-25}{124}\) - 1 = 0
\(\dfrac{x-15-124-2015}{2015}\)+\(\dfrac{x-124-2015-25}{25}\)+\(\dfrac{x-2015-25-124}{124}\) = 0
\(\dfrac{x-\left(15+124+2015\right)}{2015}\)+\(\dfrac{x-\left(124+2015+25\right)}{25}\)+\(\dfrac{x-\left(2015+25+124\right)}{124}\) = 0
(\(x\) - 2164).(\(\dfrac{1}{2015}\)+\(\dfrac{1}{25}\)+\(\dfrac{1}{124}\)) = 0
\(x-2164\) = 0
\(x\) = 2164
2015 + 2015 x 2 + 2015 x 3 + 2015 x 4 = 2015 x ( 1 + 2 + 3 + 4 )
= 2015 x 10 = 20150