{(1+49)*5/20} + {(51+99)*5/20} = 50/4+150/4= 50

- số số hạng của tổng là:

       (15-1):2+1=8(số hạng)

Tổng trên có giá trị là:

        (15+1).8:2=64

-số số hạng của tổng là:

       (16-2):2+1=8(số hạng)

Tổng trên có giá trị là:

       (16+2).8:2=72

-số số hạng của tổng là:

       (22-1):3+1=8(số hạng)

Tổng trên có giá trị là:

        (22+1).8:2=92

1+3+5+7+9+11+13+15=(1+15)+(3+13)+(5+11)+(7+9)=16x4=64
1+5+9+13+17+21+25+29+33+37=(1+37)+(5+33)+(9+29)+(13+25)+(17+21)=38x5=190

Ta có: \(\dfrac{1}{5}+\dfrac{4}{10}+\dfrac{9}{15}+\dfrac{16}{20}+\dfrac{25}{25}+\dfrac{36}{30}+\dfrac{49}{35}+\dfrac{64}{40}+\dfrac{81}{45}\)

\(=\dfrac{1}{5}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{4}{5}+\dfrac{5}{5}+\dfrac{6}{5}+\dfrac{7}{5}+\dfrac{8}{5}+\dfrac{9}{5}\)

\(=\dfrac{45}{5}=9\)

\(\dfrac{1}{5}+\dfrac{4}{10}+\dfrac{9}{15}+...+\dfrac{81}{45}\\ =\dfrac{1}{5}+\dfrac{2}{5}+\dfrac{3}{5}+...+\dfrac{9}{15}\\ =\dfrac{\left(1+2+3+...+9\right)}{15}=\dfrac{\left(1+9\right)x9:2}{15}\\ =\dfrac{10x9:2}{15}=\dfrac{45}{15}=3\)

a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)

= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + \(\dfrac{8}{1}\)

= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8

=  1 + 1 + 8

=  2 + 8

= 10

b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{10}{20}\)

=  \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (\(\dfrac{2}{2}\) + \(\dfrac{3}{3}\) + \(\dfrac{4}{4}\) + \(\dfrac{5}{5}\)+ \(\dfrac{6}{6}+\dfrac{7}{7}+\dfrac{8}{8}\) + \(\dfrac{10}{10}\))

= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (1 + 1 +1 + 1+ 1+ 1+ 1 +1)

= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x 1 x 8

= \(\dfrac{1}{2}\) + \(\)\(\dfrac{1}{2}\) x 8

= \(\dfrac{1}{2}\) + 4

= \(\dfrac{9}{2}\) 

a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)

  = (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + 8

= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8

= 1 + 1 + 8

= 2 + 8

= 10

b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{9}{18}\) + \(\dfrac{10}{20}\)

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

= \(\dfrac{1}{2}\) x 10

= 5

Đề đúng không bạn, kiểm tra lại đề hộ mình vs.

      \(\frac{7}{10}-\frac{3}{10}x\frac{1}{10}+\frac{4}{25}\)

\(=\frac{7}{10}-\frac{3}{100}+\frac{4}{25}\)

\(=\frac{70}{100}-\frac{3}{100}+\frac{4}{25}\)

\(=\frac{67}{100}+\frac{4}{25}\)

\(=\frac{67}{100}+\frac{16}{100}\)

\(=\frac{83}{100}=0,83\)

      \(\frac{6}{11}x\frac{3}{7}+\frac{3}{7}x\frac{5}{11}\)

\(=\frac{18}{77}+\frac{15}{77}\)

\(=\frac{33}{77}=\frac{3}{7}\)

     \(49x\left(311-4807:23\right)+98210\)

\(=49x\left(311-209\right)+98210\)

\(=49x102+98210\)

\(=4998+98210\)

\(=103208\)

a) \(\frac{7}{10}-\frac{3}{10}\times\frac{1}{10}+\frac{4}{25}\)

\(=\frac{2}{5}\times\frac{1}{10}+\frac{4}{25}\)

\(=\frac{1}{25}+\frac{4}{25}\)

\(=\frac{5}{25}=\frac{1}{5}\)

b)\(\frac{6}{11}\times\frac{3}{7}+\frac{3}{7}\times\frac{5}{11}\)

\(=\frac{3}{7}\times\left(\frac{6}{11}+\frac{5}{11}\right)\)

\(=\frac{3}{7}\times1\)

\(=\frac{3}{7}\)

c)\(49\times\left(311-4807:23\right)+98210\)

\(=49\times\left(311-209\right)+98210\)

\(=49\times102+98210\)

\(=4998+98210\)

\(=103208\)