1. Ta nói phân số đó là tối giản khi bội chung nhỏ nhất của tử số và mẫu số chỉ bằng 1.

a) Gọi \(d=BCNN\left(n+1;2n+3\right)\), ta có:

\(\hept{\begin{cases}n+1⋮d\\2n+3⋮d\end{cases}}\Rightarrow\hept{\begin{cases}2\left(n+1\right)⋮d\\2n+3⋮d\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2n+2⋮d\\2n+3⋮d\end{cases}}\Leftrightarrow\left(2n+2\right)-\left(2n+3\right)⋮d\)

\(\Rightarrow d=1\)

b)  Gọi \(d=BCNN\left(2n+3;4n+8\right)\), ta có:

\(\hept{\begin{cases}2n+3⋮d\\4n+8⋮d\end{cases}}\Rightarrow\hept{\begin{cases}2n+3⋮d\\\left(4n+8\right):2⋮d\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\2n+4⋮d\end{cases}}\Leftrightarrow\left(2n+4\right)-\left(2n+3\right)⋮d\)

\(\Rightarrow d=1\)

c)  Gọi \(d=BCNN\left(3n+1;4n+1\right)\), ta có:

\(\hept{\begin{cases}3n+1⋮d\\4n+1⋮d\end{cases}}\Leftrightarrow\left(4n+1\right)-\left(3n+1\right)⋮d\)

\(\Rightarrow d=n\)

Vậy bài toán c) được chứng minh, bài toán xảy ra khi \(n\)là số nguyên tố

 Gọi \(d=ƯCNN\left(n+1;2n+3\right)\),\(\left(d\ne1,0\right)\) ta có:

\(\hept{\begin{cases}2n+3⋮d\\6n+4⋮d\end{cases}}\Rightarrow\hept{\begin{cases}6n+9⋮d\\6n+4⋮d\end{cases}}\)

\(\Rightarrow\left(6n+9\right)-\left(6n+4\right)⋮d\)

\(\Rightarrow n=5\)

`Answer:`

\(2-\left(\frac{13}{65}+\frac{21}{40}\right)+\left(-\frac{52}{65}+\frac{-1}{-40}\right)\)

\(=2-\frac{13}{65}-\frac{21}{40}-\frac{52}{65}+\frac{1}{40}\)

\(=2+\left(-\frac{13}{65}-\frac{52}{65}\right)+\left(-\frac{21}{40}+\frac{1}{40}\right)\)

\(=2-\frac{65}{65}-\frac{20}{40}\)

\(=2-1-\frac{1}{2}\)

\(=0,5\)

\(X:\frac{2}{-11}=\frac{33}{-4}\)

\(X=\frac{-33}{4}x\frac{-2}{11}\)

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

Vậy \(X=\frac{3}{2}\)

`Answer:`

\(2-\left(\frac{13}{65}+\frac{21}{40}\right)+\left(-\frac{52}{65}+\frac{-1}{-40}\right)\)

\(=2-\frac{13}{65}-\frac{21}{40}-\frac{52}{65}+\frac{1}{40}\)

\(=2+\left(-\frac{13}{65}-\frac{52}{65}\right)+\left(-\frac{23}{40}+\frac{1}{40}\right)\)

\(=2-\frac{65}{65}-\frac{20}{40}\)

\(=2-1-\frac{1}{2}\)

\(=1-\frac{1}{2}\)

\(=0,5\)

Nhớ t cho mik nha

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

\(=\frac{3\cdot2}{1\cdot3\cdot2}+\frac{3\cdot2}{3\cdot5\cdot2}+\frac{3\cdot2}{5\cdot7\cdot2}+...+\frac{3\cdot2}{49\cdot51\cdot2}\)

\(=\frac{3}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{49\cdot51}\right)\)

\(=\frac{3}{2}\cdot\left(\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+...+\frac{51-49}{49\cdot51}\right)\)

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

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

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

\(=\frac{3\cdot50}{2\cdot51}\)

\(=\frac{3\cdot25\cdot2}{2\cdot17\cdot3}\)

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

b, \(\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{97\cdot99}\)

\(=\frac{7-5}{5\cdot7}+\frac{9-7}{7\cdot9}+\frac{11-9}{9\cdot11}+...+\frac{99-97}{97\cdot99}\)

\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{5}-\frac{1}{99}\)

\(=\frac{94}{495}\)

HT

Phân số chỉ phần 44 trang chiếm là :

1 - 1/3 = 2/3

2/3 số trang của quyển sách :

44 : 2/3 = 66 ( trang )

66 trang chiếm :

1 - 1/3 = 2/3 ( số trang )

Cuốn sách có số trang là :

66 : 2/3 = 90 ( trang )

`Answer:`

a. \(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)

\(=\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)

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

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

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

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

b. \(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{97.99}\)

\(=\frac{2}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{97.99}\right)\)

\(=1.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{5}-\frac{1}{99}\)

\(=\frac{94}{495}\)