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

= \(\dfrac{9}{2}\) 

c; \(\dfrac{1}{10}\) + \(\dfrac{4}{20}\) + \(\dfrac{9}{30}\)+\(\dfrac{16}{40}+\dfrac{25}{50}+\dfrac{36}{60}+\dfrac{49}{70}+\dfrac{64}{80}+\dfrac{81}{90}\)

= \(\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}\)

= \(\dfrac{1+2+3+4+5+6+7+8+9}{10}\)

=  \(\dfrac{\left(1+9\right)+\left(2+8\right)+\left(3+7\right)+\left(4+6\right)+5}{10}\)

= \(\dfrac{10+10+10+10+5}{10}\)

= \(\dfrac{\left(10+10+10+10\right)+5}{10}\)

= \(\dfrac{10\times4+5}{10}\)

= \(\dfrac{45}{10}\)

= \(\dfrac{9}{2}\)

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

c; \(\dfrac{1}{10}\) + \(\dfrac{4}{20}\) + \(\dfrac{9}{30}\)+\(\dfrac{16}{40}+\dfrac{25}{50}+\dfrac{36}{60}+\dfrac{49}{70}+\dfrac{64}{80}+\dfrac{81}{90}\)

= \(\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}\)

= \(\dfrac{1+2+3+4+5+6+7+8+9}{10}\)

=  \(\dfrac{\left(1+9\right)+\left(2+8\right)+\left(3+7\right)+\left(4+6\right)+5}{10}\)

= \(\dfrac{10+10+10+10+5}{10}\)

= \(\dfrac{\left(10+10+10+10\right)+5}{10}\)

= \(\dfrac{10\times4+5}{10}\)

= \(\dfrac{45}{10}\)

= \(\dfrac{9}{2}\)

a; 

c; \(\dfrac{1}{10}\) + \(\dfrac{4}{20}\) + \(\dfrac{9}{30}\)+\(\dfrac{16}{40}+\dfrac{25}{50}+\dfrac{36}{60}+\dfrac{49}{70}+\dfrac{64}{80}+\dfrac{81}{90}\)

= \(\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}\)

= \(\dfrac{1+2+3+4+5+6+7+8+9}{10}\)

=  \(\dfrac{\left(1+9\right)+\left(2+8\right)+\left(3+7\right)+\left(4+6\right)+5}{10}\)

= \(\dfrac{10+10+10+10+5}{10}\)

= \(\dfrac{\left(10+10+10+10\right)+5}{10}\)

= \(\dfrac{10\times4+5}{10}\)

= \(\dfrac{45}{10}\)

= \(\dfrac{9}{2}\)

1/10 + 4/20 + 9/30 + 16/40 + 25/50 + 36/60 + 49/70 + 64/80 + 81/90 
= 1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10 
=(1+2+3+....+9) /10 
=(1+9)x9:2 /10 
=45/10=4,5 

chắc chắn đúng đó  

18/15*35/36+18/15*15/18-18/15*20/72

5/6+1/1-1/3

5/6+6/6-2/6

3/2

\(M=\frac{1}{10}+\frac{4}{20}+\frac{9}{30}+\frac{16}{40}+...+\frac{81}{90}\)

\(M=\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\frac{4}{10}+...+\frac{9}{10}\)

\(M=\frac{\left(9+1\right)\cdot\left(9-1+1\right):2}{10}\)

\(M=\frac{10\cdot9:2}{10}=4,5\)

tớ làm giống bạn kia

tớ cũng làm như vậy

\(\dfrac{1}{10}+\dfrac{4}{20}+\dfrac{9}{30}+\dfrac{16}{40}+\dfrac{25}{50}+\dfrac{36}{60}+\dfrac{49}{70}+\dfrac{64}{80}+\dfrac{81}{90}\)
\(=\dfrac{1}{10}+\dfrac{1}{5}+\dfrac{3}{10}+\dfrac{2}{5}+\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{7}{10}+\dfrac{4}{5}+\dfrac{9}{10}\)
\(=\left(\dfrac{1}{10}+\dfrac{3}{10}+\dfrac{7}{10}+\dfrac{9}{10}\right)+\left(\dfrac{1}{5}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{4}{5}\right)+\dfrac{1}{2}\)
\(=2+2+\dfrac{1}{2}\)
\(=4+\dfrac{1}{2}\)
\(=\dfrac{8}{2}+\dfrac{1}{2}=\dfrac{9}{2}\)

\(\frac{1}{10}+\frac{4}{20}+\frac{9}{30}+....+\frac{81}{90}\)

\(=\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+...+\frac{9}{10}\)

\(=\frac{\left(1+2+3+.....+9\right)}{10}\)

\(=\frac{45}{10}=\frac{9}{2}\)

\(S=\frac{1}{10}+\frac{2^2}{20}+\frac{3^2}{30}+....+\frac{9^2}{90}=\frac{1}{10}+\frac{2}{10}+...+\frac{9}{10}=\frac{45}{10}=\frac{9}{2}\)

          Ta có : 

           Đặt \(A=\frac{1}{10}+\frac{4}{20}+\frac{9}{30}+\frac{16}{40}+\frac{25}{50}+\frac{36}{60}+\frac{49}{70}+\frac{64}{80}+\frac{91}{90}\)

                \(A=\frac{1}{10}+\frac{2^2}{20}+\frac{3^2}{30}+\frac{4^2}{40}+\frac{5^2}{50}+\frac{6^2}{60}+\frac{7^2}{70}+\frac{8^2}{80}+\frac{9^2}{90}\)

               \(A=\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\frac{4}{10}+\frac{5}{10}+\frac{6}{10}+\frac{7}{10}+\frac{8}{10}+\frac{9}{10}\)

             \(A=\frac{45}{10}=\frac{9}{2}\)

            Ủng hộ mk nha !!! ^_^

Bài 1:

a; (\(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\)) x \(\dfrac{3}{4}\) = \(\dfrac{1}{4}\)

     \(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) : \(\dfrac{3}{4}\)

      \(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) x \(\dfrac{4}{3}\)

    \(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) =  \(\dfrac{1}{3}\)

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

       \(\dfrac{1}{4}\) \(x\)=  \(\dfrac{8}{24}\) + \(\dfrac{11}{24}\)

          \(\dfrac{1}{4}x=\dfrac{11}{24}\)

           \(x=\dfrac{11}{24}:\dfrac{1}{4}\)

           \(x=\dfrac{11}{24}\times4\)

           \(x=\dfrac{11}{6}\) 

b; \(\dfrac{12}{5}:x\) = \(\dfrac{14}{3}\) x \(\dfrac{4}{7}\)

     \(\dfrac{12}{5}\) : \(x\) = \(\dfrac{8}{3}\)

            \(x\) = \(\dfrac{12}{5}\) : \(\dfrac{8}{3}\)

            \(x\) = \(\dfrac{12}{5}\) x \(\dfrac{3}{8}\)

             \(x\) = \(\dfrac{9}{10}\)

c; \(x+\dfrac{2}{3}\) = 8:4 - 1

   \(x+\dfrac{2}{3}\) = 2 - 1

   \(x+\dfrac{2}{3}\) = 1

  \(x=1-\dfrac{2}{3}\)

  \(x\) = \(\dfrac{1}{3}\)