\(\frac{\frac{7}{35}+\frac{7}{36}-\frac{7}{37}-1}{\frac{35}{36}-\frac{35}{37}-4}\)
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C = 1/3 + -3/4 + 3/5 + 1/57 + -1/36 + 1/15 + -2/9
C = ( 1/3 + 1/57 ) + ( -3/4 + -1/36 ) + ( 3/5 + 1/15 ) + -2/9
C = ( 19/57 + 1/57 ) + ( -27/36 + -1/36 ) + ( 9/15 + 1/15 ) + -2/9
C = 20/57 + -28/36 + 10/15 + -2/9
C = 20/57 + -7/9 + 2/3 + -2/9
C = ( 20/57 + 2/3 ) + ( -7/9 + -2/9 )
C = 58/57 + -1
C = 1/57
D = 1/2 + -1/5 + -5/7 + 1/6 + -3/35 + 1/3 + 1/41
D = ( 1/2 + 1/3 + 1/6 ) + ( -1/5 + -5/7 +-3/35 ) + 1/41
D = ( 3/6 + 2/6 + 1/6 ) + ( -7/35 + -25/35 + -3/35 ) + 1/41
D = 1 + -1 + 1/41
D = 1/41
E = -1/2 + 3/5 + -1/9 + 1/127 + -7/18 + 4/35 + 2/7
E = ( -1/2 + -1/9 + -7/18 ) + ( 3/5 + 4/35 ) + 1/127 + 2/7
E = ( -9/18 + -2/18 + -7/18 ) + ( 21/35 + 4/35 ) + 1/127 + 2/7
E = -1 + 5/7 + 1/257 + 2/7
E = -1 + ( 5/7 + 2/7 ) + 1/127
E = -1 + 1 + 1/127
E = 1/127
\(\frac{x-1}{5}+\frac{x-1}{7}+...+\frac{x-1}{37}=0=>\left(x-1\right).\left(\frac{1}{5}+\frac{1}{7}+...+\frac{1}{37}\right)=0\)
mà \(\frac{1}{5}+\frac{1}{7}+...+\frac{1}{37}>0\)
=> x-1 = 0
=> x = 1
Vậy x=1
đúng nha
a) \(\frac{-1}{2}+\frac{-1}{9}-\frac{-3}{5}+\frac{1}{2006}-\frac{-2}{7}-\frac{7}{18}+\frac{4}{35}\)
\(=\left(\frac{-1}{2}-\frac{1}{9}-\frac{7}{18}\right)+\left(\frac{3}{5}+\frac{4}{35}\right)+\frac{1}{2006}\)
\(=\left(\frac{-9}{18}-\frac{2}{18}-\frac{7}{18}\right)+\left(\frac{21}{35}+\frac{4}{35}\right)+\frac{1}{2006}\)
\(=\left(\frac{-9-2-7}{18}\right)+\left(\frac{21+4}{35}\right)+\frac{1}{2006}\)
\(=\left(\frac{-18}{18}\right)+\left(\frac{25}{35}\right)+\frac{1}{2006}\)
\(=\left(-1\right)+\frac{5}{7}+\frac{1}{2006}\)\(=\frac{-4005}{14042}\)
b) \(\frac{1}{3}-\frac{3}{4}+\frac{3}{5}+\frac{1}{2007}-\frac{1}{36}+\frac{1}{15}-\frac{2}{9}\)
\(=\left(\frac{1}{3}+\frac{1}{2007}-\frac{2}{9}\right)-\left(\frac{3}{4}+\frac{1}{36}\right)+\left(\frac{3}{5}+\frac{1}{15}\right)\)
\(=\left(\frac{669}{2007}+\frac{1}{2007}-\frac{446}{2007}\right)-\left(\frac{27}{36}+\frac{1}{36}\right)+\left(\frac{9}{15}+\frac{1}{15}\right)\)
\(=\frac{224}{2007}-\frac{28}{36}+\frac{10}{15}\)
\(=\frac{224}{2007}-\frac{1561}{2007}+\frac{1338}{2007}\)\(=\frac{1}{2007}\)