\(\dfrac{5}{6}.\dfrac{1}{3}+\dfrac{5}{6}+\dfrac{3}{4}\)

\(=\dfrac{5}{6}.\left(\dfrac{1}{3}+\dfrac{3}{4}\right)=\dfrac{5}{6}.\dfrac{7}{12}=\dfrac{35}{72}\)

= (2/5 + 3/5) + (6/9 + 1/3) + (7/4 + 1/4)

= 1 + (6/9 + 3/9) + 2 = 1 + 1 + 2 = 4

\(\dfrac{2}{5}+\dfrac{6}{9}+\dfrac{7}{4}+\dfrac{3}{5}+\dfrac{1}{3}+\dfrac{1}{4}\\ =\left(\dfrac{2}{5}+\dfrac{3}{5}\right)+\left(\dfrac{7}{4}+\dfrac{1}{4}\right)+\left(\dfrac{2}{3}+\dfrac{1}{3}\right)\\ =1+2+1=4\)

\(\dfrac{2}{5}+\dfrac{6}{9}+\dfrac{7}{4}+\dfrac{3}{5}+\dfrac{1}{3}+\dfrac{1}{4}\)

=\(\dfrac{2}{5}+\dfrac{2}{3}+\dfrac{7}{4}+\dfrac{3}{5}+\dfrac{1}{3}+\dfrac{1}{4}\)

=\(\left(\dfrac{2}{5}+\dfrac{3}{5}\right)+ \left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(\dfrac{7}{4}+\dfrac{1}{4}\right)\)

=1+1+2

=4

a. \(\dfrac{-2}{3}+\dfrac{-1}{5}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)

= \(\dfrac{-4}{6}+\dfrac{-2}{10}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)

= \(\dfrac{-3}{2}+\dfrac{1}{2}+\dfrac{3}{4}\)

= (-1) + \(\dfrac{3}{4}\)

= \(\dfrac{-4}{4}+\dfrac{3}{4}\)

= \(\dfrac{-1}{4}\)

b; 0,5  + \(\dfrac{1}{3}\) + 0,4 + \(\dfrac{5}{7}\) + \(\dfrac{1}{6}\) - \(\dfrac{4}{35}\)

= (\(\dfrac{1}{3}\)+ \(\dfrac{1}{6}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{5}{7}\)- \(\dfrac{4}{35}\)+ \(\dfrac{2}{5}\))

= ( \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{3}{5}\) + \(\dfrac{2}{5}\))

= 1 + 1 

= 2

c; \(\dfrac{1}{3}\) + \(\dfrac{-1}{5}\) - \(\dfrac{-1}{4}\) - \(\dfrac{7}{9}\)- \(\dfrac{-5}{12}\) + \(\dfrac{-1}{45}\) + \(\dfrac{1}{41}\)

  = (\(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) + \(\dfrac{5}{12}\)) - (\(\dfrac{1}{5}\) + \(\dfrac{7}{9}\) + \(\dfrac{1}{45}\)) + \(\dfrac{1}{41}\)

= (\(\dfrac{4}{12}\) + \(\dfrac{3}{12}\) + \(\dfrac{5}{12}\)) - (\(\dfrac{9}{45}+\dfrac{35}{45}\) + \(\dfrac{1}{45}\)) + \(\dfrac{1}{41}\)

= (\(\dfrac{7}{12}\) + \(\dfrac{5}{12}\)) - (\(\dfrac{44}{45}\) + \(\dfrac{1}{45}\)) + \(\dfrac{1}{41}\)

= 1 - 1 + \(\dfrac{1}{41}\)

= 0 + \(\dfrac{1}{41}\)

= \(\dfrac{1}{41}\)

\(13,=-1-1-...-1+101\)

Tổng trên có \(\left(100-1+1\right):2=50\) (số -1)

Vậy tổng trên \(=-50+101=51\)

\(\frac{3-\frac{1}{2}+\frac{1}{4}}{\frac{2}{3}-\frac{5}{6}-\frac{3}{4}}=\frac{\left(3-\frac{1}{2}+\frac{1}{4}\right)\left(3.4\right)}{\left(\frac{2}{3}-\frac{5}{6}-\frac{3}{4}\right)\left(3.4\right)}=\frac{36-6+3}{8-10-9}=\frac{33}{-11}=-3\)

\(A=\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}+\frac{1}{7\times8}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

\(=\frac{1}{2}-\frac{1}{8}\)

\(=\frac{3}{8}\)

Còn một câu hỏi nữa, làm sao để viết chữ phân số trong này thế?

812,5 nha

`a, 4/9 + 2/5 + 5/9= (4/9 +5/9) + 2/5= 1 + 2/5= 5/5 + 2/5= 7/5`

`b, 6/13 + 5/8 + 7/13=(6/13 + 7/13) + 5/8=1+5/8= 8/8 + 5/8= 13/8`

`c, 5/11 + 6/25 + 66/121 = ( 5/11 + 66/121 ) + 6/25=1 + 6/25 = 25/25 + 6/25 = 31/25`

`d, 1/2 + 1/3 + 1/4 + 1/6 + 1/12 + 1/24= (1/2 + 1/4) +(1/3 + 1/6) + (1/12 + 1/24)= 3/4 + 1/2 + 1/8=  6/8 + 4/8 + 1/8= 11/8`

`@ yl`