= 2x4/3x3 x 3x5/4x4 x 4x6/5x5 x.....x 49x51/50x50

= 2x4x3x5x4x6x...49x51/ 3x3x4x4x5x5...50x50

= 2x51/3x50

= 17/25 

A=2.4/3^2 . 3.5/4^2 . 4.6/5^2 ............ . 49.51/50^2

A=2/3-51/50

A=17/25.

Chúc bạn hok tốt.

Bài này cũng dễ ý mà, vô cùng đơn giản.........

Giải:

Ta có: \(A=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.....\dfrac{2499}{2500}.\)

\(=\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.....\dfrac{49.51}{50^2}.\)

\(=\dfrac{\left(2.3.4.....49\right)\left(4.5.6.....51\right)}{\left(3.4.5.....50\right)\left(3.4.5.....50\right)}.\)

\(=\dfrac{2.51}{3.50}.\)

\(=\dfrac{17}{25}.\)

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A= 3^2-1/3.3 . 4^2-1/4.4 . 5^2-1/5.5 . ... 50^2-1/50.50 A= (3+1).(3-1).(4+1).(4-1).(5+1).(5-1). ... (50+1).(50-1) / 3.3.4.4.5.5. ... . 50.50 A=4.2.5.3.6.4. ... 51.49 / 3.3.4.4.5.5....50.50 A=(4.5.6. ... .51).(2.3.4. ... 49)/(3.4.5.... .50).(3.4.5.. ... 50) A= 51.2/3.50 A=17/25

Ta có:

\(A=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}......\dfrac{2499}{2500}\)

= \(\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.\dfrac{4.6}{5.5}......\dfrac{49.51}{50.50}\)

= \(\dfrac{2.4.3.5.4.6......49.51}{3.3.4.4.5.5......50.50}\)

= \(\dfrac{\left(2.3.4....49\right)\left(4.5.6....51\right)}{\left(3.4.5....50\right)\left(3.4.5....50\right)}\)

= \(\dfrac{2}{50}.\dfrac{51}{3}\) = \(\dfrac{17}{25}\)

a) Ta có \(A=\dfrac{n-5}{n-3}=\dfrac{n-3-2}{n-3}=1-\dfrac{2}{n-3}\). Để \(A\inℤ\) thì \(\dfrac{2}{n-3}\inℤ\) hay \(n-3\) là ước của 2. Suy ra \(n-3\in\left\{\pm1;\pm2\right\}\). 

Nếu \(n-3=1\Rightarrow n=4\); \(n-3=-1\Rightarrow n=2\); \(n-3=2\Rightarrow n=5\); \(n-3=-2\Rightarrow n=1\). Vậy để \(A\inℤ\) thì \(n\in\left\{1;2;4;5\right\}\)

 \(A=\dfrac{n+4}{n+1}\) làm tương tự.

b) Dễ thấy các số ở mẫu có thể viết dưới dạng:

\(10=1+2+3+4=\dfrac{4\left(4+1\right)}{2}=\dfrac{4.5}{2}\)

\(15=1+2+3+4+5=\dfrac{5\left(5+1\right)}{2}=\dfrac{5.6}{2}\)

\(21=1+2+...+6=\dfrac{6\left(6+1\right)}{2}=\dfrac{6.7}{2}\)

...

\(120=1+2+...+15=\dfrac{15\left(15+1\right)}{2}=\dfrac{15.16}{2}\)

Do đó \(A=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{15.16}\) 

\(A=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)

\(A=2\left(\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+...+\dfrac{16-15}{15.16}\right)\)

\(A=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(A=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)

\(A=\dfrac{3}{8}\)

\(B=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+\dfrac{24}{25}+...+\dfrac{2499}{2500}\)

\(=1-\dfrac{3}{4}+1-\dfrac{8}{9}+1-\dfrac{15}{16}+1-\dfrac{24}{25}...+1-\dfrac{2499}{2500}\)

\(=49-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+\dfrac{1}{25}+...+\dfrac{1}{2500}\right)\)

Lại có: \(49-\left(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+...+\dfrac{1}{50.50}\right)< 49-\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{50.51}\right)\)

Mà \(49-\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{50.51}\right)\)

\(=49-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{50}-\dfrac{1}{51}\right)\)

\(=49-\left(\dfrac{1}{2}-\dfrac{1}{51}\right)=\dfrac{4942}{102}\) \(\notin Z\)

Vậy B không phải là số nguyên

Mình mới học lớp 5 thôi nha

Mong bạn thông cảm

 👌🏻

A a yamate

\(A=\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}.....\dfrac{899}{30^2}\)

\(A=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.....\dfrac{29.31}{30.30}\)

\(A=\dfrac{1.3.2.4.3.5.....29.31}{2.2.3.3.4.4.....30.30}\)

\(A=\dfrac{1.2.3.....29}{2.3.4....30}.\dfrac{3.4.5.....31}{2.3.4.....30}\)

\(A=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)

\(B=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.....\dfrac{2499}{2500}\)

\(B=\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.\dfrac{4.6}{5.5}.....\dfrac{49.51}{50.50}\)

\(B=\dfrac{2.4.3.5.4.6.....49.51}{3.3.4.4.5.5....50.50}\)

\(B=\dfrac{2.3.4......49}{3.4.5....50}.\dfrac{4.5.6.....51}{3.4.5....50}\)

\(B=\dfrac{2}{50}.\dfrac{51}{3}=\dfrac{17}{25}\)

Giải:

\(A=\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}.....\dfrac{899}{30^2}.\)

\(A=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.....\dfrac{29.31}{30^2}.\)

\(A=\dfrac{1.2.3.....29}{2.3.4.....30}.\dfrac{2.3.4.....31}{2.3.4.....30}.\)

\(A=\dfrac{1}{30}.31=\dfrac{30}{31}.\)

Vậy \(A=\dfrac{30}{31}.\)

\(A=\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}............................\dfrac{899}{30^2}\)

\(\Leftrightarrow A=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}..............................\dfrac{29.31}{30^2}\)

\(\Leftrightarrow A=\dfrac{1.3.2.4.3.5..........29.31}{2.2.3.3.4.4.........30.30}\)

\(\Leftrightarrow A=\dfrac{\left(2.3.........29.30\right).\left(3.4.5......29.31\right)}{\left(2.3....29.30\right).\left(2.3.4.......29.30\right)}\)

\(\Leftrightarrow A=\dfrac{31}{2.30}=\dfrac{31}{60}\)

\(B=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}....................................\dfrac{2499}{2500}\)

\(\Leftrightarrow B=\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.\dfrac{4.6}{5.5}.............................\dfrac{49.51}{50.50}\)

\(\Leftrightarrow B=\dfrac{\left(2.3.4.....49\right).\left(4.5.6......51\right)}{\left(3.4.5....50\right)\left(3.4.5.....50\right)}=\dfrac{2.51}{50.3}=\dfrac{17}{25}\)

Bài 2:

a: -2*(-27)=54

6*9=54

=>Hai phân số này bằng nhau

b: -1/-5=1/5=5/25<>4/25

Bài 3:

a: =>16/x=-4/5

=>x=-20

b: =>(x+7)/15=-2/3

=>x+7=-10

=>x=-17

a) \(\dfrac{-2}{9}\) và \(\dfrac{6}{-27}\)

\(\dfrac{6}{-27}=\dfrac{6:\left(-3\right)}{\left(-27\right):\left(-3\right)}=\dfrac{-2}{9}\)

Vậy \(\dfrac{-2}{9}=\dfrac{6}{-27}\)

b) \(\dfrac{-1}{-5}\) và \(\dfrac{4}{25}\)

\(\dfrac{-1}{-5}=\dfrac{\left(-1\right).\left(-5\right)}{\left(-5\right).\left(-5\right)}=\dfrac{5}{25}\)

Do \(5\ne4\Rightarrow\dfrac{5}{25}\ne\dfrac{4}{25}\)

Vậy \(\dfrac{-1}{-5}\ne\dfrac{4}{25}\)

Bài 3

a) \(\dfrac{-28}{35}=\dfrac{16}{x}\)

\(x=\dfrac{35.16}{-28}\)

\(x=-20\)

b) \(\dfrac{x+7}{15}=\dfrac{-24}{36}\)

\(\left(x+7\right).36=15.\left(-24\right)\)

\(36x+252=-360\)

\(36x=-360-252\)

\(36x=-612\)

\(x=\dfrac{-612}{36}\)

\(x=-17\)