12/35...12/3535
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a,\(\frac{2015.2016+2015-1}{2014+2015.2016}=\frac{2015.2016+2014}{2014+2015.2016}=1\)\(1\)
b,\(=1-\frac{1}{5}+\frac{1}{5}...-\frac{1}{2011}+\frac{1}{2011}-\frac{1}{2015}=1-\frac{1}{2015}=\frac{2014}{2015}\)
c,\(=\frac{12}{35}+\frac{12}{35}+\frac{12}{35}+\frac{12}{35}=\frac{12}{35}.4=\frac{48}{35}\)
Ta có : 1212/3535=12/35 ; 1212/9999=12/99
Từ đề bài => = 12/15 +12/35 +12/63 +12/99
= 6 x ( 2/15 + 2/35 +2/63+2/99 )
= 6 x ( 2/3x5 + ....+ 2/9x11)
= 6 x ( 1/3 - 1/5 +....+1/9 - 1/11)
= 6 x ( 1/3 - 1/11)
= 6 x 8/33 = 48/33
a)Ta có : \(\dfrac{1212}{1414}=\dfrac{12.101}{14.101}=\dfrac{12}{14}\)
\(\dfrac{121212}{242424}=\dfrac{12.10101}{24.10101}=\dfrac{12}{24}\)
Vậy \(\dfrac{12}{24}=\dfrac{1212}{2424}=\dfrac{121212}{242424}\)
Ta có : \(\dfrac{2424}{3535}=\dfrac{24.101}{35.101}=\dfrac{24}{35}\)
\(\dfrac{242424}{353535}=\dfrac{24.10101}{35.10101}=\dfrac{24}{35}\)
Vậy \(\dfrac{24}{35}=\dfrac{2424}{3535}=\dfrac{242424}{353535}\)
\(\frac{3}{4}x-14\frac{2}{3}:\left(\frac{11}{15}+\frac{1111}{3535}+\frac{111111}{636363}\right)=12\)
\(\frac{3}{4}x-14\frac{2}{3}:\left(\frac{11}{15}+\frac{1111}{3535}+\frac{111111}{636363}\right)=12\)
\(\frac{3}{4}x-\frac{44}{3}:\left(\frac{11}{15}+\frac{11}{35}+\frac{11}{63}\right)=12\)
\(\frac{3}{4}x-\frac{44}{3}:\frac{11}{9}=12\)
\(\frac{3}{4}x-12=12\)
\(\frac{3}{4}x=12+12\)
\(\frac{3}{4}x=24\)
\(x=24:\frac{3}{4}\)
\(x=32\)
vậy \(x=32\)
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