Question

Заполните таблицу .....

Answer 1

Пошаговое объяснение:

1.

$a+b=10\dfrac{7}{10} +3\dfrac{1}{5} =(10+3)+\left( \dfrac{7}{10}^{\backslash1}+\dfrac{1}{5}^{\backslash2}\right)=13 +\dfrac{7+2}{10} =13+\dfrac{9}{10} =13\dfrac{9}{10}; \\a-b=10\dfrac{7}{10} -3\dfrac{1}{5} =(10-3)+\left( \dfrac{7}{10}^{\backslash1}-\dfrac{1}{5}^{\backslash2}\right)=7 +\dfrac{7-2}{10} =7+\dfrac{5}{10} =7\dfrac{1}{2}.$

2.

$b=14\dfrac{2}{21} -9\dfrac{3}{7} =(14-9)+\left( \dfrac{2}{21}^{\backslash1}-\dfrac{3}{7}^{\backslash3}\right)=5 +\dfrac{2-9}{21} =4+\dfrac{21+2-9}{21}=\\\\ =4\dfrac{14}{21}=4\dfrac{2}{3} ;$

$a-b=9\dfrac{3}{7} -4\dfrac{2}{3} =(9-4)+\left( \dfrac{3}{7}^{\backslash3}-\dfrac{2}{3}^{\backslash7}\right)=5 +\dfrac{9-14}{21} =4+\dfrac{21+9-14}{21}=\\\\=4+\dfrac{16}{21}=4\dfrac{16}{21}.$

3.

$b=15\dfrac{9}{10} -2\dfrac{3}{100} =(15-2)+\left( \dfrac{9}{10}^{\backslash10}-\dfrac{3}{100}^{\backslash1}\right)=13 +\dfrac{90-3}{100} =\\\\=13+\dfrac{87}{100} =13\dfrac{87}{100}.\\a+b=15\dfrac{9}{10} +13\dfrac{87}{100} =(15+13)+\left( \dfrac{9}{10}^{\backslash10}+\dfrac{87}{100}^{\backslash1}\right)=28 +\dfrac{90+87}{10} =\\\\=28+\dfrac{177}{100} =28+1\dfrac{77}{100}=29\dfrac{77}{100} .$

4.

$a=23- 4\dfrac{3}{5} =22\dfrac{5}{5}-4\dfrac{3}{5}=18\dfrac{2}{5} ;\\\\a-b= 18\dfrac{2}{5}-4\dfrac{3}{5}=14\dfrac{2-3}{5} =13\dfrac{5+2-3}{5} =13\dfrac{4}{5} .$

5.

$a+b=5\dfrac{7}{20} +3\dfrac{3}{10} =(5+3)+\left( \dfrac{7}{20}^{\backslash1}+\dfrac{3}{10}^{\backslash2}\right)=8 +\dfrac{7+6}{10} =8+\dfrac{13}{20} =8\dfrac{13}{20}; \\a-b=5\dfrac{7}{20} -3\dfrac{3}{10} =(5-3)+\left( \dfrac{7}{20}^{\backslash1}-\dfrac{3}{10}^{\backslash2}\right)=2 +\dfrac{7-6}{20} =2+\dfrac{1}{20} =2\dfrac{1}{20}.$

6.

$a=6\dfrac{3}{4} +1\dfrac{5}{8} =(6+1)+\left( \dfrac{3}{4}^{\backslash2}+\dfrac{5}{8}^{\backslash1}\right)=7+\dfrac{6+5}{8} =7+\dfrac{11}{8} =8\dfrac{3}{8}; \\a+b=8\dfrac{3}{8} +1\dfrac{5}{8} =(8+1)+\left( \dfrac{3}{8}^{\backslash1}+\dfrac{5}{8}^{\backslash1}\right)=9+\dfrac{3+5}{8} =9+\dfrac{8}{8} =9+1=10.$

7.

$b=7\dfrac{3}{5} -4\dfrac{3}{10} =(7-4)+\left( \dfrac{3}{5}^{\backslash2}-\dfrac{3}{10}^{\backslash1}\right)=3 +\dfrac{6-3}{10} =3+\dfrac{3}{10} =3\dfrac{3}{10}.\\a- b=4\dfrac{3}{10} -3\dfrac{3}{10} =(4-3)+\left( \dfrac{3}{10}^{\backslash1}-\dfrac{3}{10}^{\backslash1}\right)=1.$