Question

Tính nhanh:

b)(1 + \(1\frac{1}{4}\)+ \(1\frac{1}{2}\)+ \(1\frac{3}{4}\)+ 2 + \(2\frac{1}{4}\)+ \(2\frac{1}{2}\)+ \(2\frac{3}{4}\)+ ... + \(4\frac{3}{4}\)) : 23

Answer 1

\(\left(1+1\frac{1}{4}+1\frac{1}{2}+1\frac{3}{4}+2+2\frac{1}{4}+2\frac{1}{2}+2\frac{3}{4}+...+4\frac{3}{4}\right):23\)

= \(\left(\frac{2}{2}+\frac{5}{4}+\frac{3}{2}+\frac{7}{4}+\frac{4}{2}+\frac{9}{4}+\frac{5}{2}+\frac{11}{4}+...+\frac{19}{4}\right):23\)

= \(\left(\frac{4}{4}+\frac{5}{4}+\frac{6}{4}+\frac{7}{4}+\frac{8}{4}+\frac{9}{4}+\frac{10}{4}+\frac{11}{4}+...+\frac{19}{4}\right):23\)

= \(\left(\frac{4+5+6+7+8+9+10+11+...+19}{4}\right):23\)

= \(\left(\frac{\left(19-4\right):1+1x\left(19+4\right):2}{4}\right):23\)

=  \(46:23\)

= \(2\)