a) A = \(\dfrac{8}{9}\) . \(\dfrac{15}{16}\) . \(\dfrac{24}{25}\). ... .\(\dfrac{2499}{2500}\). Tính
b) Tìm các số nguyên n để phân số \(\dfrac{12}{3n-1}\) có giá trị nguyên.
Các bạn giúp với :<
Bài 1:
a, CMR: A = \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{21}{10^2.11^2}< 1\)
b, Cho B = \(\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+\dfrac{24}{25}+...+\dfrac{2499}{2500}.\) CMR: B không phải là số nguyên.
c, So sánh: C = \(\dfrac{2}{2^1}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{2021}{2^{2020}}\) với 3.
tìm kết quả tích sau:
A= \(\dfrac{8}{9}\)x\(\dfrac{15}{16}\)x\(\dfrac{24}{25}\)x...x\(\dfrac{2499}{2500}\)
= 2x4/3x3 x 3x5/4x4 x 4x6/5x5 x.....x 49x51/50x50
= 2x4x3x5x4x6x...49x51/ 3x3x4x4x5x5...50x50
= 2x51/3x50
= 17/25
Tính \(A=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.............................\dfrac{2499}{2500}\)
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: \(A=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.....\dfrac{2499}{2500}.\)
\(=\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.....\dfrac{49.51}{50^2}.\)
\(=\dfrac{\left(2.3.4.....49\right)\left(4.5.6.....51\right)}{\left(3.4.5.....50\right)\left(3.4.5.....50\right)}.\)
\(=\dfrac{2.51}{3.50}.\)
\(=\dfrac{17}{25}.\)
CHÚC BN HỌC TỐT!!! ^ _ ^
Đừng quên bình luận nếu bài mik sai nhé!!! - _ -
Còn nếu bài mik đúng thì nhớ tick mik để mik lấy SP nha!!! ^ - ^
https://hoc24.vn/hoi-dap/question/214681.html
Tính A = \(\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}...\dfrac{2499}{2500}\)
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) Tìm các số nguyên n để phân số sau có giá trị nguyên:
\(A=\dfrac{n-5}{n-3}\)
\(\dfrac{n+4}{n+1}\)
b) Tính A, biết A= \(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)
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 \(A=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{15.16}\)
\(A=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)
\(A=2\left(\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+...+\dfrac{16-15}{15.16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)
\(A=\dfrac{3}{8}\)
Cho B= \(\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+\dfrac{24}{25}+...+\dfrac{2499}{2500}\). Chứng tỏ B không phải là số nguyên
GIÚP MK
\(B=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+\dfrac{24}{25}+...+\dfrac{2499}{2500}\)
\(=1-\dfrac{3}{4}+1-\dfrac{8}{9}+1-\dfrac{15}{16}+1-\dfrac{24}{25}...+1-\dfrac{2499}{2500}\)
\(=49-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+\dfrac{1}{25}+...+\dfrac{1}{2500}\right)\)
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
Bài 17: Tìm tất cả các số nguyên n sao cho các phân số sau có giá trị là số nguyên.
a) \(\dfrac{12}{3n-1}\) . b) \(\dfrac{2n+3}{7}\) .
c) \(\dfrac{2n+5}{n-3}\) .
Mình mới học lớp 5 thôi nha
Mong bạn thông cảm
tính nhanh
A=\(\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}...\dfrac{899}{30^2}\)
B=\(\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}...\dfrac{2499}{2500}\)
\(A=\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}.....\dfrac{899}{30^2}\)
\(A=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.....\dfrac{29.31}{30.30}\)
\(A=\dfrac{1.3.2.4.3.5.....29.31}{2.2.3.3.4.4.....30.30}\)
\(A=\dfrac{1.2.3.....29}{2.3.4....30}.\dfrac{3.4.5.....31}{2.3.4.....30}\)
\(A=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)
\(B=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.....\dfrac{2499}{2500}\)
\(B=\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.\dfrac{4.6}{5.5}.....\dfrac{49.51}{50.50}\)
\(B=\dfrac{2.4.3.5.4.6.....49.51}{3.3.4.4.5.5....50.50}\)
\(B=\dfrac{2.3.4......49}{3.4.5....50}.\dfrac{4.5.6.....51}{3.4.5....50}\)
\(B=\dfrac{2}{50}.\dfrac{51}{3}=\dfrac{17}{25}\)
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 \(A=\dfrac{30}{31}.\)
\(A=\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}............................\dfrac{899}{30^2}\)
\(\Leftrightarrow A=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}..............................\dfrac{29.31}{30^2}\)
\(\Leftrightarrow A=\dfrac{1.3.2.4.3.5..........29.31}{2.2.3.3.4.4.........30.30}\)
\(\Leftrightarrow A=\dfrac{\left(2.3.........29.30\right).\left(3.4.5......29.31\right)}{\left(2.3....29.30\right).\left(2.3.4.......29.30\right)}\)
\(\Leftrightarrow A=\dfrac{31}{2.30}=\dfrac{31}{60}\)
\(B=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}....................................\dfrac{2499}{2500}\)
\(\Leftrightarrow B=\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.\dfrac{4.6}{5.5}.............................\dfrac{49.51}{50.50}\)
\(\Leftrightarrow B=\dfrac{\left(2.3.4.....49\right).\left(4.5.6......51\right)}{\left(3.4.5....50\right)\left(3.4.5.....50\right)}=\dfrac{2.51}{50.3}=\dfrac{17}{25}\)
\(\dfrac{-2}{9}\)và\(\dfrac{6}{-27}\) b:\(\dfrac{-1}{-5}\)và\(\dfrac{4}{25}\)
Các cặp phân số sau có bằng nhau ko?vì sao?
Bài3: Tìm số nguyên X biết
a)\(\dfrac{-28}{35}\)=\(\dfrac{16}{x}\)
b)\(\dfrac{x+7}{15}\)=\(\dfrac{-24}{36}\)
giúp mình với ae cứu tôi ae cứu tôi :((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((
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) \(\dfrac{x+7}{15}=\dfrac{-24}{36}\)
\(\left(x+7\right).36=15.\left(-24\right)\)
\(36x+252=-360\)
\(36x=-360-252\)
\(36x=-612\)
\(x=\dfrac{-612}{36}\)
\(x=-17\)