Hỏi đáp Phân số

Nếu 2 bạn nữ chuyển và thay vào 2 bạn nam thì số trai hơn số gái lúc này hơn số trai hơn số gái lúc ban đầu : 2 + 2 = 4 ( học sinh )

4 học sinh đó tương ứng với : 2/3 - 4/7 = 2/21 ( số học sinh )

Số học sinh của lớp 5A là : 4 / (2/21) = 42 ( học sinh )

Nhìn ngay là sai đề luôn

2 học sinh nữ chiếm số phần là 

\(\frac{1}{3}-\frac{1}{4}=\frac{1}{12}\) 

Số học sinh lớp 5A là 

\(2:\frac{1}{12}=24\) ( học sinh ) 

\(\frac{1}{n^3}< \frac{1}{\left(n-2\right)n\left(n+1\right)}\Leftrightarrow\frac{\left(n-2\right)n\left(n+1\right)}{n^3}< 1\Leftrightarrow\left(n-2\right)\left(n+1\right)< n^2\)

\(\Leftrightarrow n^2-n-2< n^2\Leftrightarrow-n-2< 0\)Đúng \(\forall n\inℕ\)

--->ĐPCM

\(=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)\cdot...\cdot\left(\frac{1}{100}-1\right)\)  ( có 9 thừa số ) 

\(=-\left(\frac{3}{4}\right)\cdot\left(\frac{8}{9}\right)\cdot\left(\frac{15}{16}\right)\cdot...\left(\frac{99}{100}\right)\) ( có 9 thừa số nên tích sẽ âm ) 

\(=-\left(\frac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot4\cdot6\cdot5\cdot7\cdot6\cdot8\cdot7\cdot9\cdot8\cdot10\cdot9\cdot11}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot6\cdot6\cdot7\cdot7\cdot8\cdot8\cdot9\cdot9\cdot10\cdot10}\right)\) 

\(=-\left(\frac{11}{20}\right)\)

              Bài làm :

Ta có :

\(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{10^2}-1\right)\)

\(=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{100}-1\right)\) 

\(=-\left(\frac{3}{4}\right).\left(\frac{8}{9}\right).\left(\frac{15}{16}\right)...\left(\frac{99}{100}\right)\)

\(=-\left(\frac{1.3.2.4.3.5.4.6.5.7.6.8.7.9.8.10.9.11}{2.2.3.3.4.4.5.5.6.6.7.7.8.8.9.9.10.10}\right)\)

\(=-\left(\frac{11}{20}\right)\)

                                                         Bài giải

\(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\cdot...\cdot\left(\frac{1}{10^2}-1\right)\)

\(=\frac{-3}{4}\cdot\frac{-8}{9}\cdot...\cdot\frac{-99}{100}\)

\(=-\left(\frac{3\cdot8\cdot...\cdot99}{4\cdot9\cdot...\cdot100}\right)=-\frac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot4\cdot6\cdot5\cdot7\cdot6\cdot8\cdot7\cdot9\cdot8\cdot10\cdot9\cdot11}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot6\cdot6\cdot7\cdot7\cdot8\cdot8\cdot9\cdot9\cdot10\cdot10}=-\frac{11}{20}\)

ta có:\(\frac{1}{6}+\frac{y}{3}=\frac{1}{x}\)   <=>\(\frac{1}{6}+\frac{2y}{6}=\frac{1}{x}\)   <=>\(\frac{2y+1}{6}=\frac{1}{x}\)     <=>\(x\left(2y+1\right)=6\)  

                                                                                                                                    <=>\(x\left(2y+1\right)=1.6\)

                                                                                           <=>\(\left(-1\right)\left(-6\right)=2.3=\left(-2\right)\left(-3\right)\)

                                                                  Bài giải

\(\frac{1}{x}=\frac{1}{6}+\frac{y}{3}\)

\(\frac{1}{x}=\frac{1}{6}+\frac{2y}{6}\)

\(\frac{1}{x}=\frac{2y+1}{6}\)

\(\Rightarrow\text{ }x\left(2y+1\right)=6\)

\(\Rightarrow\text{ }x\text{ ; }\left(2y+1\right)\inƯ\left(6\right)=\left\{\pm1\text{ ; }\pm2\text{ ; }\pm3\text{ ; }\pm6\right\}\)

Ta có bảng : 

Vậy ...

\(\frac{1}{x}=\frac{1}{6}+\frac{y}{3}\)

\(\Leftrightarrow\frac{1}{x}=\frac{1}{6}+\frac{2y}{6}\)

\(\Leftrightarrow\frac{1}{x}=\frac{1+2y}{6}\)

\(\Leftrightarrow x\left(1+2y\right)=6\)

6 = 1.6 = (-1).(-6)

   = 2.3 = (-2).(-3)

Ta có bảng sau :

Vì mình chưa biệt x, y thuộc tập gì nên để tạm vầy nhé

\(\frac{-1}{39}+\frac{-1}{52}=\frac{-4}{156}+\frac{-3}{156}=\frac{-7}{156}\)

@Khanh Nguyễn Ngọc

bn lm nhanh ghê rứa? :v

bài này quá dễ

quy đồng ra là tính thôi :>

Sr gui lon

Co j k mik nha

378 x X = 105 

X = 105 / 378 

\(X=\frac{5}{18}\)

\(378\times x=105\)

\(x=105\div378\)

\(x=\frac{5}{18}.\)

378x X=105

x=105:375

x=105/375=0,28

C=\(\frac{-17,5+\frac{5}{3}-2\frac{1}{7}}{7,0-\frac{2}{3}+\frac{6}{7}}\)

  =\(\frac{\frac{-367,5}{21}+\frac{35}{21}-\frac{45}{21}}{\frac{147}{21}-\frac{14}{21}+\frac{18}{21}}\)

  =\(\frac{\frac{-377,5}{21}}{\frac{151}{21}}\)

  =\(-\frac{5}{2}\)

Ta có :\(\frac{-17,5+\frac{5}{3}-2\frac{1}{7}}{7-\frac{2}{3}+\frac{6}{7}}=\frac{-17,5+\frac{5}{3}-\frac{15}{7}}{7-\frac{2}{3}+\frac{6}{7}}=\frac{-2,5\left(.7-\frac{2}{3}+\frac{6}{7}\right)}{7-\frac{2}{3}+\frac{6}{7}}=-2,5\)

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)

\(A=\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+...+\frac{100-99}{99.100}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}\)

     \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

                                          Bài làm :

Ta có :

\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{99\times100}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}\)

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

a) (5/2 + 1/3) : (1 - 2/3)

= 5/3 : 1/3 

= 15/3 = 5

b) (5/2 - 1/3) x 9/2 - 6/7

= 1 x 9/2 - 6/7

= 9/2 - 6/7

= 51/14

Hok tốt

=17/6:(1-2/3)

=17/6:1/3

=17/2

=13/6×9/2-6/7

=39/4-6/7

=249/28

\(\left(\frac{5}{2}+\frac{1}{3}\right):\left(1-\frac{2}{3}\right)\)

\(=\frac{17}{6}:\frac{1}{3}\)

\(=\frac{17}{2}\)

\(\left(\frac{5}{2}-\frac{1}{3}\right)\times\frac{9}{2}-\frac{6}{7}\)

\(=\frac{13}{6}\times\frac{9}{2}-\frac{6}{7}\)

\(=\frac{39}{4}-\frac{6}{7}\)

\(=\frac{239}{28}\)

Đặt C = \(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110}\)

=> 100C = \(\frac{100}{1.101}+\frac{100}{2.102}+...+\frac{100}{10.110}\)

=> 100C = \(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\)

=> 100C = \(\left(1+\frac{1}{2}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\)

=> C = \(\frac{1+\frac{1}{2}+...+\frac{1}{10}-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)}{100}\)

Lại có B = \(\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)

=> 10B = \(\frac{10}{1.11}+\frac{10}{2.12}+...+\frac{10}{100.110}\)

=> 10B = \(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\)

=> 10B = \(\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{110}\right)\)

=> 10B = \(1+\frac{1}{2}+...+\frac{1}{10}-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\)

=> B = \(\frac{1+\frac{1}{2}+...+\frac{1}{10}-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)}{10}\)

Khi đó \(\left(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110}\right)x=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)

<=> C.x = B

<=> \(\frac{1+\frac{1}{2}+...+\frac{1}{10}-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)}{100}x=\frac{1+\frac{1}{2}+...+\frac{1}{10}-\left(\frac{1}{101}+\frac{1}{102}+..+\frac{1}{110}\right)}{10}\)

=> \(x=10\)

Vậy x = 10

ta gọi phần trong ngoặc là A thì ta có

A nhân x = A

     x= A-A

    x=1

Bài làm:

1)

Ta có:

a)  \(1,5\div2,16=\frac{25}{36}\)

b) \(4\div\frac{\frac{2}{7}}{\frac{3}{5}}=\frac{42}{5}\)

c) \(\frac{2}{9}\div0,31=\frac{200}{279}\)

2) Ta có: \(ad=bc\)

\(\Leftrightarrow\frac{ad}{cd}=\frac{bc}{cd}\)

\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)

Trả lời:

\(\frac{2\times4+2\times4\times8+4\times8\times16+8\times16\times32}{3\times4+2\times6\times8+4\times12\times16+8\times24\times32}\)

\(=\frac{2\times4+2\times4\times8+2.2\times2.4\times16+4.2\times4.4\times32}{3\times4+2\times2.3\times2.4+4\times3.4\times4.4+8\times8.3\times8.4}\)

\(=\frac{2\times4\times\left(1+8+2\times2\times16+4\times4\times32\right)}{3\times4\times\left(1+2\times2\times2+4\times4\times4+8\times8\times8\right)}\)

\(=\frac{2\times4\times\left(1+8+64+512\right)}{3\times4\times\left(1+8+64+512\right)}\)

\(=\frac{2}{3}\)

x là nhân

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{72}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)

\(=1-\frac{1}{9}\)

\(=\frac{8}{9}\)

​1/2+1/6+1/12+.......+1/72

1/1.2+1/2.3+1/3.4+............+1/8.9

1-1/2+1/2-1/3+1/3-1/4+..............+1/8-1/9

1-1/9

8/9

HOK TỐT NHA!!!!!!!!!!!!!!!!!!!!!!!

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+........+\frac{1}{72}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.............+\frac{1}{8.9}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...............+\frac{1}{8}-\frac{1}{9}\)

\(=\frac{1}{1}-\frac{1}{9}=\frac{9}{9}-\frac{1}{9}=\frac{8}{9}\)

Chúc bạn học tốt

\(A=\frac{989898.89-898989.98}{2^3+3^4+4^5+.............+2014^{2015}}\)

\(A=\frac{98.10101.89-89.10101.98}{2^3+3^4+4^5+............+2014^{2015}}\)

\(A=\frac{98.10101.\left(89-89\right)}{2^3+3^4+4^5+.............+2014^{2015}}\)

\(A=\frac{98.10101.0}{2^3+3^4+4^5+..........+2014^{2015}}=\frac{0}{2^3+3^4+.................+2014^{2015}}=0\)

Vậy A=0

Chúc bạn học tốt

\(A=\frac{989898.89-898989.98}{2^3+3^4+4^5+.......+2014^{2015}}\)

Xét tử: \(989898.89-898989.98\)ta có:

\(989898.89-898989.98=10101.98.89-10101.89.98\)

\(=10101.\left(98.89-89.98\right)=0\)

mà \(2^3+3^4+4^5+.........+2014^{2015}>0\)

\(\Rightarrow A=0\)

tớ nhầm phần tử 989898.89-898989.98

Bài 2: https://oml.vn/hoi-dap/detail/6465458369.html

Bài 3: https://hoidap247.com/cau-hoi/20162 

Bài 1: https://hoidap247.com/cau-hoi/1009171

Trả lời:

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2018.2019.2020}+\frac{1}{2.2019.2020}\)

\(A=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2018.2019.2020}+\frac{2}{2.2019.2020}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2018.2019}-\frac{1}{2019.2020}+\frac{1}{2019.2020}\right)\)

\(A=\frac{1}{2}.\frac{1}{1.2}\)

\(A=\frac{1}{4}\)