Hỏi đáp Phân số

\(\text{Theo đề ta có :}\)

    \(\frac{\left(-1\right)^6\cdot3^5\cdot4^3}{9^2\cdot2^5}\)

= \(\frac{1\cdot3^5\cdot\left(2^2\right)^3}{\left(3^2\right)^2\cdot2^5}\) =  \(\frac{3^5\cdot2^6}{3^4\cdot2^5}=\frac{3^4\cdot3\cdot2^5\cdot2}{3^4\cdot2^5}=3\cdot2=6\)

mình ko giải theo cách của thương . xin lỗi bạn

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)

đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\) 

\(A\cdot2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)

\(A\cdot2-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)

\(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}\)

\(A=1-\frac{1}{32}\)

\(A=\frac{31}{32}\)

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}=\frac{1\times16}{2\times16}+\frac{1\times8}{4\times8}+\frac{1\times4}{8\times4}+\frac{1\times2}{16\times2}+\frac{1}{32}\)

                                                         \(=\frac{16}{32}+\frac{8}{32}+\frac{4}{32}+\frac{2}{32}+\frac{1}{32}\)

                                                         \(=\frac{16+8+4+2+1}{32}\)\(=\frac{31}{32}\)

a, Cho mk sửa đề

\(\frac{145145}{607607}=\frac{145.1001}{107.1001}=\frac{145}{607}\)

b, \(\frac{19491949}{20062006}=\frac{1949.10001}{2006.10001}=\frac{1949}{2006}\)

mình đang cần vội

\(B=\frac{3737.43-4343.37}{2+4+6+...+100}\)

\(=\frac{37.101.43-43.101.37}{2+4+6+...+100}\)

\(=\frac{0}{2+4+6+...+100}\)

\(=0\)

\(B\)\(=\)\(\frac{3737.43-4343.37}{2+4+6+...+100}\)\(=\)\(\frac{37.101.43-4343.37}{2+4+6+...+100}\)

       \(=\)\(\frac{37.4343-4343.37}{2+4+6+...+100}\)\(=\)\(\frac{0}{2+4+6+...+100}\)\(=\)\(0\)

Xét tử ta có: 

\(101+100+99+98+...........+3+2+1\)

\(=1+2+3+..........+99+100+101\)

\(=\frac{101.102}{2}=5151\)

Xét mẫu ta có:

\(101-100+99-98+.......+3-2+1\)

\(=\left(101-100\right)+\left(99-98\right)+.......+\left(3-2\right)+1\)

\(=1+1+.......+1+1=51\)

\(\Rightarrow A=\frac{5151}{51}=101\)

a, \(\left|x-\frac{2}{3}\right|=\frac{1}{2}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{2}{3}=\frac{1}{2}\\x-\frac{2}{3}=-\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}+\frac{2}{3}\\x=\frac{2}{3}-\frac{1}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{6}\\x=\frac{1}{6}\end{cases}}}\)

b, \(\left|x+\frac{7}{20}\right|=\frac{3}{15}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{7}{20}=\frac{1}{5}\\x+\frac{7}{20}=-\frac{1}{5}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}-\frac{7}{20}\\x=-\frac{1}{5}-\frac{7}{20}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{-3}{20}\\x=\frac{-11}{20}\end{cases}}}\)

c, \(\left|3x+2\right|=\left|7x-4\right|\)

\(\Leftrightarrow\orbr{\begin{cases}3x+2=7-4x\\3x+2=4x-7\end{cases}\Leftrightarrow\orbr{\begin{cases}7x=5\\x=9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{7}\\x=9\end{cases}}}\)

d, \(\left|5-2x\right|=\left|2x-5\right|\)

\(\Leftrightarrow\orbr{\begin{cases}5-2x=2x-5\\5-2x=5-2x\end{cases}\Leftrightarrow\orbr{\begin{cases}-4x=-10\\0x=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x\in Q\end{cases}}}\)

=> Có vô số x thỏa mãn \(\left|5-2x\right|=\left|2x-5\right|\)

e, \(\left|-5-6x\right|=\left|-x-5\right|\)

\(\Leftrightarrow\orbr{\begin{cases}-5-6x=-x-5\\-5-6x=x+5\end{cases}\Leftrightarrow\orbr{\begin{cases}-5x=0\\-7x=10\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=-\frac{10}{7}\end{cases}}}\)

f, \(\left|-x+5\right|=\left|12-3x\right|\) đúng ko ???

\(\Leftrightarrow\orbr{\begin{cases}-x-5=12-3x\\-x+5=3x-12\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\-4x=17\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{17}{4}\end{cases}}}\)

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 giải

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

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

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

              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

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

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

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

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

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

Ta có bảng : 

Vậy ...

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

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

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

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

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

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

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

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