Bài 1

a)Để A thuộc Z

=>-3 chia hết 2x-1

=>2x-1 thuộc Ư(-3)={1;-1;3;-3}

=>x thuộc {1;0;-1;2}

b)Để B thuộc Z

=>4x+5 chia hết 2x-1

=>2(2x-1)+7 chia hết 2x-1

Ta thấy: 2x-1 chia hết 2x-1 =>2(2x-1) cũng chia hết 2x-1

=>7 chia hết 2x-1

=>2x-1 thuộc Ư(7)={1;-1;7;-7}

=>x thuộc {1;0;-3;4}

Bài 1

a)Để A thuộc Z

=>-3 chia hết 2x-1

=>2x-1 thuộc Ư(-3)={1;-1;3;-3}

=>x thuộc {1;0;-1;2}

b)Để B thuộc Z

=>4x+5 chia hết 2x-1

=>2(2x-1)+7 chia hết 2x-1

Ta thấy: 2x-1 chia hết 2x-1 =>2(2x-1) cũng chia hết 2x-1

=>7 chia hết 2x-1

=>2x-1 thuộc Ư(7)={1;-1;7;-7}

=>x thuộc {1;0;-3;4}

Bài 1a:

A = \(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{2003.2004}\)

A = \(\frac11-\frac12+\frac12-\frac13+\cdots+\frac{1}{2003}-\frac{1}{2004}\)

A = \(\frac11-\frac{1}{2004}\)

A = \(\frac{2003}{2004}\)

Bài 1b:

B = 20 x 15 - 20 x 13 + 20

B = 20 x (15 - 13 + 1)

B = 20 x (2 + 1)

B = 20 x 3

B = 60

1)

A = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+..+\frac{2}{99.101}\)

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

A = \(\frac{1}{1}-\frac{1}{101}\)

A = \(\frac{100}{101}\)

Vậy A = \(\frac{100}{101}\)

B = \(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)

B = \(\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)

B = \(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

B = \(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{101}\right)\)

B = \(\frac{5}{2}.\frac{100}{101}\)

B = \(\frac{250}{101}\)

Vậy B = \(\frac{250}{101}\)

2) 

Gọi ƯCLN ( 2n + 1 ; 3n + 2 ) = d ( d \(\in\)N* )

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

\(\Rightarrow\hept{\begin{cases}6n+3⋮d\\6n+4⋮d\end{cases}\Rightarrow\left(6n+4\right)-\left(6n+3\right)⋮d\Rightarrow1⋮d}\)

\(\Rightarrow d=1\)

Vậy \(\frac{2n+1}{3n+2}\)là p/s tối giản

Gọi ƯCLN ( 2n+3 ; 4n+4 ) = d ( d \(\in\)N* )

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

\(\Rightarrow1⋮d\Rightarrow d=1\)

Vậy ...

Để A có giá trị nguyên

thì 3\(⋮\)(x-1)

mà xeZ nên x-1eZ

x-1e{3;-3}

xe{4;-2}

a) Để \(\frac{3}{x-1}\inℤ\Rightarrow\left(x-1\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow x\in\left\{-2;0;2;4\right\}\)

b) Để \(\frac{4}{2x-1}\inℤ\Rightarrow\left(2x-1\right)\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

=> \(2x\in\left\{-3;-1;0;2;3;5\right\}\)

=> \(x\in\left\{-\frac{3}{2};-\frac{1}{2};0;1;\frac{3}{2};\frac{5}{2}\right\}\)

c) Ta có: \(\frac{3x+7}{x-7}=\frac{\left(3x-21\right)+28}{x-7}=2+\frac{28}{x-7}\)

Xong xét các TH như a,b nhé

thanks nhưng mai mik mới t.i.k đc bạn

 \(a=\frac{2x+4}{x-3}=\frac{2x-6+6+4}{x-3}=\frac{2x-6+10}{x-3}=\frac{2x-6}{x-3}+\frac{10}{x-3}=\)\(2+\frac{10}{x-3}\)                                         Vay de 2x+4 /x-3 la so nguyen thi 2+10/x-3 phai la so nguyen hay 10/x-3 la so nguyen                                                                            Suy ra x-3 thuoc uoc cua 10=(1;-1;2;-2;5;-5;10;-10)                                                                                                                                Roi giai ra tung truong hop

B=\(\frac{2x-5}{x-1}\)

Để \(A\inℤ\) thì \(\left(4x-6\right)⋮\left(2x+1\right)\)

\(\Leftrightarrow\left(4x+2-8\right)⋮\left(2x+1\right)\)

\(\Leftrightarrow\left[2\left(2x+1\right)+8\right]⋮\left(2x+1\right)\)

Vì \(\left[2\left(2x+1\right)\right]⋮\left(2x+1\right)\) nên \(8⋮\left(2x+1\right)\)

\(\Rightarrow2x+1\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

Mà 2x + 1 lẻ nên \(\Rightarrow2x+1\in\left\{\pm1\right\}\)

Lập bảng:

Vậy \(x\in\left\{-1;0\right\}\)

B,C,E tương tự

Bài 1b:

\(\frac31\) + \(\frac33\) + \(\frac36\) + \(\frac{3}{10}\) + ...+\(\frac{3}{x\left(x+1\right):2}\) = \(\frac{2015}{336}\)

3.(\(\frac11+\frac13+\frac16+\frac{1}{10}+\cdots+\frac{1}{x\left(x+1):2\right.})\) = \(\frac{2015}{336}\)

3.2(\(\frac12+\frac16+\frac{1}{12}+\frac{1}{20}+\cdots+\frac{1}{x\left(x+1\right)})=\) \(\frac{2015}{336}\)

6.(\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\cdots+\frac{1}{x.\left(x+1\right)})\) = \(\frac{2015}{336}\)

6.(\(\frac11-\frac12\) + \(\frac12\)-\(\frac14\) +...+ \(\frac{1}{x}\) - \(\frac{1}{x+1}\)) = \(\frac{2015}{336}\)

6.(\(\frac11\) - \(\frac{1}{x+1}\)) = \(\frac{2015}{336}\)

1 - \(\frac{1}{x+1}\) = \(\frac{2015}{336}\) : 6

1 - \(\frac{1}{x+1}\) = \(\frac{2015}{2016}\)

\(\frac{1}{x+1}\) = 1 - \(\frac{2015}{2016}\)

\(\frac{1}{x+1}\) = \(\frac{1}{2016}\)

\(x+1\) = 2016

\(x\) = 2016 - 1

\(x\) = 2015

Bài 2:

A = \(\frac{6n+1}{4n+3}\) (n ∈ Z\(^{-}\))

A ∈ Z khi và chỉ khi:

(6n + 1) ⋮ (4n + 3)

(12n + 2) ⋮ (4n + 3)

[3(4n + 3) - 7] ⋮ (4n + 3)

7 ⋮ (4n + 3)

(4n + 3) ∈ Ư(7) = {-7; -1; 1; 7}

n ∈ {- 5/2; -1; - 1/2; 1}

Nếu n = - 1 thì A = (-6 + 1)/(-4 + 3) = 5 (loại)

Nếu n = 1 thì: A = (6 + 1).(4+3) = 1 (loại)

Không có giá trị nào thỏa mãn đề bài hay n ∈ ∅