Bài 10: Phép nhân phân số

b) Ta có: \(B=\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}+\dfrac{1}{12\cdot13}+\dfrac{1}{13\cdot14}\)

\(=\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}\)

\(=\dfrac{1}{10}-\dfrac{1}{14}\)

\(=\dfrac{14}{140}-\dfrac{10}{140}\)

\(=\dfrac{4}{140}=\dfrac{1}{35}\)

Ta có: \(A=\dfrac{2^3\cdot5\cdot7\cdot5^2\cdot7^3}{\left(2\cdot5\cdot7^2\right)^2}\)

\(=\dfrac{2^3\cdot5^3\cdot7^4}{2^2\cdot5^2\cdot7^4}=\dfrac{10\cdot2^2\cdot5^2}{2^2\cdot5^2}\)

=10

\(A=\dfrac{(2^3.5.7).(5^2.7^3)}{(2.5.7^2)^2}\)

\(=\dfrac{2^3.5^3.7^4}{2^2.5^2.7^4}\)

\(=2.5\)

\(=10\)

A=2^3.5.7.5^2.7^3/(2.5.7^2)^2

A=2^3.(5.5^2).(7.7^3)/2^2.5^2.7^4

A=2^3.5^3.7^4/2^2.5^2.7^4

A=(2^3/2^2).(5^3/5^2).(7^4/7^4)

A=2.5.1

A=10

Vậy A=10

Mình trả lời rồi bạn nhé!

\(\dfrac{7}{x}<\dfrac{x}{4}<\dfrac{10}{x}\)

\(\Leftrightarrow 7<\dfrac{x^2}{4}<10\)

\(\Leftrightarrow 28< x^2<40\)

Vì x là số nguyên dương \(\Leftrightarrow x^2=36\)

Vậy \(x=6\)

Bài 2: 

a) Ta có: \(A=\dfrac{4}{n-1}+\dfrac{6}{n-1}-\dfrac{3}{n-1}\)

\(=\dfrac{4+6-3}{n-1}\)

\(=\dfrac{7}{n-1}\)

Để A là số tự nhiên thì \(7⋮n-1\)

\(\Leftrightarrow n-1\inƯ\left(7\right)\)

\(\Leftrightarrow n-1\in\left\{1;7\right\}\)

hay \(n\in\left\{2;8\right\}\)

Vậy: \(n\in\left\{2;8\right\}\)

ta có B=2n+9/n+2-3n+5n+1/n+2=4n+10/n+2                                                   Để B là STN thì 4n+10⋮n+2                          4n+8+2⋮n+2                                  4n+8⋮n+2                                                      ⇒2⋮n+2                                     n+2∈Ư(2)                                                        Ư(2)={1;2}                                  Vậy n=0                                                                                  

Ta có: =4+6−3n−1=4+6−3�−1

a) Ta có: \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{2007\cdot2009}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2007\cdot2009}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2007}-\dfrac{1}{2009}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{2009}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2008}{2009}=\dfrac{1004}{2009}\)

a)

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2007.2009}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2007.2009}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2007}-\dfrac{1}{2009}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{2009}\right)=\dfrac{1}{2}.\dfrac{2008}{2009}=\dfrac{1004}{2009}\)

a) \(\dfrac{2}{5}=\dfrac{1}{5}\cdot\dfrac{2}{1}\)

\(\dfrac{15}{12}=\dfrac{5}{4}=\dfrac{5}{2}\cdot\dfrac{1}{2}\)

\(\dfrac{5}{-12}=\dfrac{-5}{12}=\dfrac{-5}{3}\cdot\dfrac{1}{4}\)

\(\dfrac{-3}{-4}=\dfrac{3}{4}=\dfrac{3}{2}\cdot\dfrac{1}{2}\)

Ta có: \(C=c\cdot\dfrac{3}{4}+c\cdot\dfrac{5}{6}-c\cdot\dfrac{19}{12}\)

\(=c\cdot\left(\dfrac{3}{4}+\dfrac{5}{6}-\dfrac{19}{12}\right)\)

\(=c\cdot\left(\dfrac{9}{12}+\dfrac{10}{12}-\dfrac{19}{12}\right)\)

\(=0\)

Vậy: Khi \(c=\dfrac{2002}{2003}\) thì C=0

Ta có: \(B=\dfrac{3}{4}\cdot b+\dfrac{4}{3}\cdot b-\dfrac{1}{2}\cdot b\)

\(=b\left(\dfrac{3}{4}+\dfrac{4}{3}-\dfrac{1}{2}\right)\)

\(=b\left(\dfrac{9}{12}+\dfrac{16}{12}-\dfrac{6}{12}\right)\)

\(=\dfrac{19}{12}\cdot b\)

Thay \(b=\dfrac{6}{19}\) vào B, ta được:

\(B=\dfrac{6}{19}\cdot\dfrac{19}{12}=\dfrac{6}{12}=\dfrac{1}{2}\)

Vậy: Khi \(b=\dfrac{6}{19}\) thì \(B=\dfrac{1}{2}\)

\(A=\dfrac{7}{19}\cdot\dfrac{8}{11}+\dfrac{7}{19}\cdot\dfrac{3}{11}+\dfrac{12}{19}\\ A=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)+\dfrac{12}{19}\\ A=\dfrac{7}{19}\cdot1+\dfrac{12}{19}\\ A=\dfrac{19}{19}=1\)

A = 7/19.8/11+7/19.3/11+12/19

A = 7/19. . (8/11 + 3/11) + 12/19

A = 7/19 . 1 + 12/19

A = 7/19 + 12/19

A = 1

Ta có: \(A=\dfrac{7}{19}\cdot\dfrac{8}{11}+\dfrac{7}{19}\cdot\dfrac{3}{11}+\dfrac{12}{19}\)

\(=\dfrac{7}{19}\cdot\left(\dfrac{8}{11}+\dfrac{3}{11}\right)+\dfrac{12}{19}\)

\(=\dfrac{7}{19}+\dfrac{12}{19}=1\)

\(\dfrac{x+1}{6}+\dfrac{x+8}{3}=\dfrac{x+1}{6}+\dfrac{2.\left(x+8\right)}{6}=\dfrac{3x+9}{6}\\ =\dfrac{3.\left(x+3\right)}{6}=\dfrac{x+3}{2}\)