Bài 1:

a/ $\frac{\frac{3}{5}+\frac{3}{25}-\frac{3}{19}-\frac{3}{101}}{\frac{7}{5}+\frac{7}{25}-\frac{7}{19}-\frac{7}{101}}:\frac{\frac{6}{5}+\frac{6}{17}+\frac{6}{97}-\frac{6}{59}}{\frac{3}{5}+\frac{3}{17}+\frac{3}{97}-\frac{3}{59}}$

= $\frac{3\left(\frac{1}{5}+\frac{1}{25}-\frac{1}{19}-\frac{1}{101}\right)}{7\left(\frac{1}{5}+\frac{1}{25}-\frac{1}{19}-\frac{1}{101}\right)}:\frac{6\left(\frac{1}{5}+\frac{1}{17}+\frac{1}{97}-\frac{1}{59}\right)}{3\left(\frac{1}{5}+\frac{1}{17}+\frac{1}{97}-\frac{1}{59}\right)}$

= $\frac{3}{7}:\frac{6}{3}$

= $\frac{3}{7}.\frac{3}{6}=\frac{9}{42}=\frac{3}{14}$

b/ $\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}$

= $\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}$

= $\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}$

= $\frac{1}{3}-\frac{1}{11}$

= $\frac{8}{33}$

Bài 2:

a/ $\overline{ab}+\overline{ba}=10a+b+10b+a=11a+11b⋮11$

Vậy:............

b/ $\overline{abc}-\overline{bca}=100a+10b+c-100b-10c-a=99a-90b-9c⋮9$

Vậy:................

Bài 1:

a) $\frac{\frac{3}5+\frac{3}{25}-\frac{3}{19}-\frac{3}{101}}{\frac{7}5+\frac{7}{25}-\frac{7}{19}-\frac{7}{101}}:$ $\frac{\frac{6}5+\frac{6}{25}-\frac{6}{19}-\frac{6}{101}}{\frac{3}5+\frac{3}{25}-\frac{3}{19}-\frac{3}{101}}=$ $\frac{3(\frac{1}5+\frac{1}{25}-\frac{1}{19}-\frac{1}{101})}{7(\frac{1}5+\frac{1}{25}-\frac{1}{19}-\frac{1}{101})}:$ $\frac{6(\frac{1}5+\frac{1}{25}-\frac{1}{19}-\frac{1}{101})}{3(\frac{1}5+\frac{1}{25}-\frac{1}{19}-\frac{1}{101})}=$ $\frac{3}{7}:\frac{6}3=\frac{3}{14}$ 

b) $\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}$ 

$=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}$

$=\frac{1}3-\frac{1}4+\frac{1}4-\frac{1}5+...+\frac{1}{10}-\frac{1}{11}=\frac{1}3-\frac{1}{11}=\frac{8}{33}$

Bài 2:

a) Ta có: ab+ba=10a+b+10b+a=11a+11b=11(a+b) chia hết 11

b) Ta có: abc-bca=100a+10b+c-100b-10c-a=99a-90b-9c=9(11a-10b-c) chia hết 9

Bài 4:

Gọi hai số lần lượt là a,b

Ta có: 5a/7=2b/9⇔10a/14=10b/45=(10b-10a)/(45-14)=(10×31)/31=10 

            (Áp dụng t/c dãy tỉ số bằng nhau)