Dễ quá bn ^^ 

a, Phân số có tử là 15 , lớn hơn \(\frac{3}{7}\)và nhỏ hơn \(\frac{5}{8}\)có dạng  là : \(\frac{15}{a}\left(a\in Z\right)\)

Vì  \(\frac{15}{a}>\frac{3}{7}\)và \(\frac{15}{a}< \frac{5}{8}\),nên : 

\(\Rightarrow\frac{3}{7}< \frac{15}{a}< \frac{5}{8}\)

\(\Rightarrow\frac{15}{35}< \frac{15}{a}< \frac{15}{24}\)

\(\Rightarrow24< a< 35\Rightarrow a\in\left\{25;26;27;28;29;30;31;32;33;34\right\}\)

Vậy a thuộc {25;26;27;28;29;30;31;32;33;34}

b, Các phân số có mẫu là 12 , lớn hơn \(\frac{-2}{3}\)và nhỏ hơn \(\frac{-1}{4}\)có dạng là : \(\frac{a}{12}\left(a\in Z\right)\)

Vì \(\frac{a}{12}>\frac{-2}{3}\)và \(\frac{a}{12}< \frac{-1}{4}\)

\(\Rightarrow\frac{-2}{3}< \frac{a}{12}< \frac{-1}{4}\)

\(\Rightarrow\frac{-8}{12}< \frac{a}{12}< \frac{-3}{12}\)

\(\Rightarrow-8< a< -3\)

\(\Rightarrow a\in\left\{-7;-6;-5;-4\right\}\)

Vậy a thuộc {-7;-6;-5;-4} 

bạn làm đúng ko

Gọi p/s cần tìm là a

Ta có :

3/7< a <5/8

=> 15/35 <a< 15/24

=> a ∈ {15/34;15/33;...;15/25}

Vậy.......

các bạn giúp mình với mình đang cần đáp án gấp

1) a.Ta có \(A=\frac{3n+9}{n-4}=\frac{3n-12+21}{n-4}=\frac{3\left(n-4\right)}{n-4}+\frac{21}{n-4}=3+\frac{21}{n-4}\)

Vì \(3\inℤ\Rightarrow\frac{21}{n-4}\inℤ\Rightarrow21⋮n-4\Rightarrow n-4\inƯ\left(21\right)\)

=> \(n-4\in\left\{1;-1;3;-3;7;-7;21;-21\right\}\)

=> \(n\in\left\{5;3;8;1;11;-3;25;-17\right\}\)

b) Ta có B = \(\frac{6n+5}{2n-1}=\frac{6n-3+8}{2n-1}=\frac{3\left(2n-1\right)+8}{2n-1}=3+\frac{8}{2n-1}\)

Vì \(3\inℤ\Rightarrow\frac{8}{2n-1}\inℤ\Rightarrow2n-1\inƯ\left(8\right)\Rightarrow2n-1\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)(1)

lại có với mọi n nguyên => 2n \(⋮\)2 => 2n - 1 không chia hết cho 2 (2)

Kết hợp (1) ; (2) => \(2n-1\in\left\{1;-1\right\}\Rightarrow n\in\left\{1;0\right\}\)

2) Ta có : \(\frac{5}{x}+\frac{y}{4}=\frac{1}{8}\)

=> \(\frac{20+xy}{4x}=\frac{1}{8}\)

=> 4x = 8(20 + xy)

=> x = 2(20 + xy)

=> x = 40 + 2xy

=> x - 2xy = 40

=> x(1 - 2y) = 40

Nhận thấy : với mọi y nguyên => 1 - 2y là số không chia hết cho 2 (1)

mà x(1 - 2y) = 40

=> 1 - 2y \(\inƯ\left(40\right)\)(2)

Kết hợp (1) (2) => \(1-2y\in\left\{1;5;-1;-5\right\}\)

Nếu 1 - 2y = 1 => x = 40

=> y = 0 ; x = 40

Nếu 1 - 2y = 5 => x = 8

=> y = -2 ; x = 8 

Nếu 1 - 2y = -1 => x = -40

=> y = 1 ; y = - 40

Nếu 1 - 2y = -5 => x = -8

=> y = 3 ; x =-8

Vậy các cặp (x;y) thỏa mãn là : (40 ; 0) ; (8; - 2) ; (-40 ; 1) ; (-8 ; 3)

4) \(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}=\frac{-\frac{19}{60}.\frac{5}{19}}{\frac{21}{70}.\frac{-4}{3}}=\frac{-\frac{5}{60}}{\frac{2}{5}}=-\frac{5}{60}:\frac{2}{5}=-\frac{5}{24}\)

b) \(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(6,3.12-21.3,6\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{100}}\)

\(=\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).0}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}=0\)

c) \(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}}=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}\right)}\)

\(=\frac{1}{4}+\frac{3}{4}=1\)

1)

a) \(x+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=5\)

\(x+\frac{64}{128}+\frac{32}{128}+\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}=5\)

\(x+\frac{127}{128}=5\)

\(x=5-\frac{127}{128}=\frac{513}{128}\)

b) \(x+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}=3\)

\(x+\frac{729}{2187}+\frac{243}{2187}+\frac{81}{2187}+\frac{27}{2187}+\frac{9}{2187}+\frac{3}{2187}+\frac{1}{2187}=3\)

\(x+\frac{2186}{2187}=3\)

\(x=3-\frac{2186}{2187}=\frac{4375}{2187}\)

2)

a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(=1-\frac{1}{6}=\frac{5}{6}\)

b) \(5\frac{1}{2}+3\frac{5}{6}+\frac{2}{3}\)

\(=\left(5+3\right)+\left(\frac{1}{2}+\frac{2}{3}+\frac{5}{6}\right)\)

\(=8+\left(\frac{3}{6}+\frac{4}{6}+\frac{5}{6}\right)\)

\(=8+2=10\)

c) \(7\frac{7}{8}+1\frac{4}{6}+3\frac{3}{5}\)

\(=\left(7+1+3\right)+\left(\frac{7}{8}+\frac{2}{3}+\frac{3}{5}\right)\)

\(=11+\left(\frac{105}{120}+\frac{80}{120}+\frac{72}{120}\right)\)

\(=11+\frac{257}{120}=\frac{1577}{120}\)

3) Gọi số đó là x. Theo đề ta có :

\(\frac{16-x}{21+x}=\frac{5}{7}\)

\(7\left(16-x\right)=5\left(21+x\right)\)

\(112-7x=105+5x\)

\(112-105=7x-5x\)

\(7=2x\)

\(x=\frac{7}{2}=3,5\) ( vô lí )

Vậy không có số tự nhiên để thõa mãn điều kiện trên.

3/7 = 15/35

5/8 = 15/24

Suy ra các phân số đó lớn hơn 15/35 và nhỏ hơn 15/24

Suy ra các phân số đó có tử là 15 và mẫu lớn hơn 24 và nhỏ hơn 35

Sao nhiều quá vại??

mk lm k nổi đâu

Dài quá nhìn lòi bảng họng lun ak

Bài : 4 

a/ \(\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+....+\frac{1}{24\cdot25}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{24}-\frac{1}{25}\)

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

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

b/ \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+....+\frac{2}{99\cdot101}\)

\(=\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+...+\frac{101-99}{99\cdot101}\)

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\)

\(=\frac{1}{1}-\frac{1}{101}\)

\(=\frac{100}{101}\)

c/ \(\frac{5^2}{1\cdot6}+\frac{5^2}{6\cdot11}+\frac{5^2}{11\cdot16}+\frac{5^2}{16\cdot21}+\frac{5^2}{21\cdot26}+\frac{5^2}{26\cdot31}\)

\(=\frac{25}{1\cdot6}+\frac{25}{6\cdot11}+\frac{25}{11\cdot16}+\frac{25}{16\cdot21}+\frac{25}{21\cdot26}+\frac{25}{26\cdot31}\)

\(=\frac{6-1}{1\cdot6}+\frac{11-6}{6\cdot11}+....+\frac{31-26}{26\cdot31}\)

\(=\frac{25}{5}\cdot\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{26}-\frac{1}{31}\right)\)

\(=\frac{25}{5}\cdot\left(\frac{1}{1}-\frac{1}{31}\right)\)

\(=\frac{25}{5}\cdot\frac{30}{31}\)

\(=\frac{150}{31}\)

d/ \(\frac{3}{1\cdot3}+\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+....+\frac{3}{49\cdot51}\)

\(=\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+....+\frac{51-49}{49\cdot51}\)

\(=\frac{3}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{3}{2}\cdot\left(\frac{1}{1}-\frac{1}{51}\right)\)

\(=\frac{3}{2}\cdot\frac{50}{51}\)

\(=\frac{25}{17}\)

e/ \(\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{775}+\frac{1}{1147}\)

\(=\frac{1}{1\cdot7}+\frac{1}{7\cdot13}+\frac{1}{13\cdot19}+\frac{1}{19\cdot25}+\frac{1}{25\cdot31}+\frac{1}{31\cdot37}\)

\(=\frac{7-1}{1\cdot7}+\frac{13-7}{7\cdot13}+....+\frac{37-31}{31\cdot37}\)

\(=\frac{1}{6}\cdot\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+....+\frac{1}{31}-\frac{1}{37}\right)\)

\(=\frac{1}{6}\cdot\left(1-\frac{1}{37}\right)\)

\(=\frac{1}{6}\cdot\frac{36}{37}\)

\(=\frac{6}{37}\)

Bài 5 : 

b/ \(\left(\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+...+\frac{2}{19\cdot21}\right)-x+4+\frac{221}{231}=\frac{7}{3}\)

\(\left(\frac{13-11}{11\cdot13}+\frac{15-13}{13\cdot15}+...+\frac{21-19}{19\cdot21}\right)-x+4=\frac{7}{3}-\frac{221}{231}\)

\(\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{21}\right)-x+4=\frac{106}{77}\)

\(\left(\frac{1}{11}-\frac{1}{21}\right)-x=\frac{106}{77}-4\)

\(\frac{10}{231}-x=-\frac{202}{77}\)

\(x=\frac{10}{231}-\left(-\frac{202}{77}\right)\)

\(x=\frac{8}{3}\)