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6 tháng 8 2018

A = 1 × 2 × 3 + 2 × 3 × 4 + .....+ 48 × 49 × 50

ta có 4 x A = 1 x 2 x 3 x 4 + 2 x 3 x 4 x (5 -1) + .....+ 48 × 49 × 50 x (51 - 47)

= 1 x 2 x 3 x 4 +  2 x 3 x 4 x 5 - 1 x 2 x 3 x 4 + ... + 48 x 49 x 50 x 51 - 47 x 48 x 49 x 50

= 48 x 49 x 50 x 51

suy ra A = (48 x 49 x 50 x 51) : 4

              = 12 x 49 x 50 x 51

nhớ k cho mik nha rùi mik lm nốt cho

12 tháng 10 2024

A = 1 × 2 × 3 + 2 × 3 × 4 + .....+ 48 × 49 × 50

ta có 4 x A = 1 x 2 x 3 x 4 + 2 x 3 x 4 x (5 -1) + .....+ 48 × 49 × 50 x (51 - 47)

= 1 x 2 x 3 x 4 +  2 x 3 x 4 x 5 - 1 x 2 x 3 x 4 + ... + 48 x 49 x 50 x 51 - 47 x 48 x 49 x 50

= 48 x 49 x 50 x 51

suy ra A = (48 x 49 x 50 x 51) : 4

              = 12 x 49 x 50 x 51

 

24 tháng 1 2016

tick trước đi rồi giải thiệt đó

24 tháng 1 2016

49/50 chuẩn ko sai

Ai  ấn Đúng 0 sẽ may mắn cả năm

21 tháng 9 2024

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 \(\dfrac{4}{3}\)

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

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

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

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

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

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

           \(x=\dfrac{11}{6}\) 

   

21 tháng 9 2024

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 \(\dfrac{3}{8}\)

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

 

18 tháng 8 2020

\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+50}\)

\(=\frac{1}{2\times\left(2+1\right):2}+\frac{1}{3\times\left(3+1\right):2}+\frac{1}{4\times\left(4+1\right):2}+...+\frac{1}{50\times\left(50+1\right):2}\)

\(=\frac{1}{2}\times\frac{1}{2\times3}+\frac{1}{2}\times\frac{1}{3\times4}+\frac{1}{2}\times\frac{1}{4\times5}+...+\frac{1}{2}\times\frac{1}{49\times50}\)

\(=\frac{1}{2}\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{49\times50}\right)\)

\(=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{50}\right)=\frac{1}{2}\times\frac{12}{25}=\frac{6}{25}\)

20 tháng 8 2020

\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+..+50}\)

\(=\frac{1}{2.\left(2+1\right):2}+\frac{1}{3.\left(3+1\right):2}+\frac{1}{4.\left(4+1\right):2}+..+\frac{1}{50.\left(50+1\right):2}\)

\(=\frac{1}{2}.\frac{1}{2.3}+\frac{1}{2}.\frac{1}{3.4}+\frac{1}{2}.\frac{1}{4.5}+..+\frac{1}{2}.\frac{1}{49.50}\)

\(=\frac{1}{2}.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{49.50}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)=\frac{1}{2}.\frac{12}{25}=\frac{6}{25}\)

15 tháng 6 2023

3 x 15 + 21 x 15 + 85 x 5

= 45 + 315 + 425

= 785

15 - 30 + 40 

= 25

21 + 19 - 50 + 10

= 0

\(\dfrac{1}{5}-\dfrac{1}{4}+2\)

\(=-\dfrac{1}{20}+2\)

\(=\dfrac{39}{20}\)

\(\left(\dfrac{1}{4}+\dfrac{1}{6}\right)\times\left(\dfrac{1}{2}-\dfrac{1}{4}\right)\)

\(=\dfrac{5}{12}\times\dfrac{1}{4}\)

\(=\dfrac{5}{12}\times\dfrac{3}{12}\)

\(=\dfrac{5}{48}\)

\(\dfrac{1}{10}+\dfrac{1}{5}-\dfrac{3}{4}\)

\(=-\dfrac{9}{20}\)

15 tháng 6 2023

\(3\times15+21\times15+85\times5\\ =15\times\left(3+21\right)+425\\ =15\times24+425\\ =360+425\\ =785\)

\(15-30+40\\ =\left(15+40\right)-30\\ =55-30\\ =25\)

\(21+19-50+10\\ =\left(21+19\right)-\left(50-10\right)\\ =40-40\\ =0\)

\(\dfrac{1}{5}-\dfrac{1}{4}+2\)

\(=\dfrac{4}{20}-\dfrac{5}{20}+\dfrac{40}{20}\)

\(=\dfrac{\left(4+40\right)}{20}-\dfrac{5}{20}\)

\(=\dfrac{44}{20}-\dfrac{5}{20}\)

\(=\dfrac{39}{20}\)

\(\left(\dfrac{1}{4}+\dfrac{1}{6}\right)\times\left(\dfrac{1}{2}-\dfrac{1}{4}\right)\)

\(=\dfrac{5}{12}\times\dfrac{1}{4}\)

\(=\dfrac{5}{48}\)

\(\dfrac{1}{10}+\dfrac{1}{5}-\dfrac{3}{4}\)

\(=\dfrac{2}{20}+\dfrac{4}{20}-\dfrac{15}{20}\)

\(=\dfrac{6}{20}-\dfrac{15}{20}\)

\(=-\dfrac{9}{20}\)

 

25 tháng 1 2016

A=1/1+2+1/1+2+3+1/1+2+3+4+.....+1/1+2+3+4+...+50

Ta có 1/1+2+3+...n=1/[n*(n+1)/2]=2*[1/n(n+1)]=2*[1/n-1/n+1]

Thay n=1;2;3;4;5;6;...;50 Ta có A=2*[1/2-1/51]=49/51

vậy.......................................................

26 tháng 6 2017

Đây mà toán lớp 5 à.

Áp dụng công thức

\(\frac{1}{1+2+...+n}=\frac{1}{\frac{n\left(n+1\right)}{2}}=\frac{2}{n\left(n+1\right)}\)  ta được

\(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+....+50}\)

\(=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{50.51}\)

\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{51}\right)=\frac{49}{51}\)

26 tháng 6 2017

Ta có : \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+.......+\frac{1}{1+2+3+......+50}\)

\(=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+......+\frac{1}{\frac{50.51}{2}}\)

\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+......+\frac{2}{50.51}\)

\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{50.51}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{50}-\frac{1}{51}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{51}\right)\)

\(=2.\frac{1}{2}-2.\frac{1}{51}\)

\(=1-\frac{2}{51}=\frac{49}{51}\)