`= (3/5+6/15) + (3/21 + 6/7) + (8/18+5/9)`
`= 1 + 1 + 1 = 3`
`4/7 xx 7/9 - 4/7 xx 2/9 - 4/7 xx 5/9`
`= 4/7 xx (7/9-2/9-5/9)`
`= 0`
`15/17 : 7/9 + 7/17 : 7/9 - 5/17 : 7/9`
`= (15/17+7/17-5/17) : 7/9`
`= 1 : 7/9`
`= 9/7`
\(a,=\left(\dfrac{3}{5}+\dfrac{6}{15}\right)+\left(\dfrac{3}{21}+\dfrac{6}{7}\right)+\left(\dfrac{8}{18}+\dfrac{5}{9}\right)\\ =\left(\dfrac{3}{5}+\dfrac{2}{5}\right)+\left(\dfrac{1}{7}+\dfrac{6}{7}\right)+\left(\dfrac{4}{9}+\dfrac{5}{9}\right)\\ =1+1+1=3\\ b,=\dfrac{4}{7}\times\left(\dfrac{7}{9}-\dfrac{2}{9}-\dfrac{5}{9}\right)\\ =\dfrac{4}{7}\times0\\ =0\\ c,=\left(\dfrac{15}{17}+\dfrac{7}{17}-\dfrac{5}{17}\right):\dfrac{7}{9}\\ =\dfrac{17}{17}:\dfrac{7}{9}\\ =1:\dfrac{7}{9}\\ =\dfrac{9}{7}\)
\(\dfrac{3}{5}+\dfrac{3}{21}+\dfrac{8}{18}+\dfrac{6}{7}+\dfrac{5}{9}+\dfrac{6}{15}=\dfrac{3}{5}+\dfrac{1}{7}+\dfrac{4}{9}+\dfrac{6}{7}+\dfrac{5}{9}+\dfrac{2}{5}=\dfrac{3}{5}+\dfrac{2}{5}+\dfrac{1}{7}+\dfrac{6}{7}+\dfrac{4}{9}+\dfrac{5}{9}=1+1+1=3\)
*) 3/5 + 3/21 + 8/18 + 6/7 + 5/9 + 6/15
= (3/5 + 6/15) + (3/21 + 6/7) + (8/18 + 5/9)
= (3/5 + 2/5) + (1/7 + 6/7) + (4/9 + 5/9)
= 1 + 1 + 1
= 3
*) 4/7 × 7/9 - 4/7 × 5/9 - 4/7 × 2/9
= 4/7 × (7/9 - 5/9 - 2/9)
= 4/7 × 0
= 0
*) 15/17 : 7/9 + 7/17 : 7/9 - 5/17 : 7/9
= (15/17 + 7/17 - 5/17) : 7/9
= 1 : 7/9
= 1 × 9/7
= 9/7
\(\dfrac{3}{5}+\dfrac{3}{21}+\dfrac{8}{18}+\dfrac{6}{7}+\dfrac{5}{9}+\dfrac{6}{15}=\left(\dfrac{3}{5}+\dfrac{6}{15}\right)+\left(\dfrac{3}{21}+\dfrac{6}{7}\right)+\left(\dfrac{8}{18}+\dfrac{5}{9}\right)\)
\(=1+1+1=3\)
\(\dfrac{4}{7}.\dfrac{7}{9}-\dfrac{4}{7}.\dfrac{2}{9}-\dfrac{4}{7}.\dfrac{5}{9}=\dfrac{4}{7}.\left(\dfrac{7}{9}-\dfrac{2}{9}-\dfrac{5}{9}\right)=\dfrac{4}{7}.0=0\)
\(\dfrac{15}{17}:\dfrac{7}{9}+\dfrac{7}{17}:\dfrac{7}{9}-\dfrac{5}{17}:\dfrac{7}{9}=\left(\dfrac{15}{17}+\dfrac{7}{17}-\dfrac{5}{7}\right):\dfrac{7}{9}=1:\dfrac{7}{9}=\dfrac{9}{7}\)
a)
\(=\left(\dfrac{3}{5}+\dfrac{6}{15}\right)+\left(\dfrac{3}{21}+\dfrac{6}{7}\right)+\left(\dfrac{8}{18}+\dfrac{5}{9}\right)=1+1+1=3\)
b)
\(=\dfrac{4}{7}\cdot\left(\dfrac{7}{9}-\dfrac{2}{9}-\dfrac{5}{9}\right)=\dfrac{4}{7}\cdot0=0\)
c)
\(=\left(\dfrac{15}{17}+\dfrac{7}{17}-\dfrac{5}{17}\right):\dfrac{7}{9}=1:\dfrac{7}{9}=\dfrac{9}{7}\)
