a) Để phân số \(\dfrac{26}{x+3}\) nguyên thì \(26⋮x+3\)

\(\Leftrightarrow x+3\in\left\{1;-1;2;-2;13;-13;26;-26\right\}\)

hay \(x\in\left\{-2-4;-1;-5;10;-16;23;-29\right\}\)

b) Để phân số \(\dfrac{x+6}{x+1}\) nguyên thì \(x+6⋮x+1\)

\(\Leftrightarrow5⋮x+1\)

\(\Leftrightarrow x+1\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{0;-2;4;-6\right\}\)

c) Để phân số \(\dfrac{x-2}{x+3}\) nguyên thì \(x-2⋮x+3\)

\(\Leftrightarrow-5⋮x+3\)

\(\Leftrightarrow x+3\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{-2;-4;2;-8\right\}\)

d) Để phân số \(\dfrac{2x+1}{x-3}\) nguyên thì \(2x+1⋮x-3\)

\(\Leftrightarrow7⋮x-3\)

\(\Leftrightarrow x-3\in\left\{1;-1;7;-7\right\}\)

hay \(x\in\left\{4;2;10;-4\right\}\)

Giải:

a) \(\dfrac{26}{x+3}\) 

Để \(\dfrac{26}{x+3}\) là số nguyên thì \(26⋮x+3\) 

\(26⋮x+3\) 

\(\Rightarrow x+3\inƯ\left(26\right)=\left\{\pm1;\pm2;\pm13;\pm26\right\}\) 

Ta có bảng giá trị:

Vậy \(x\in\left\{-29;-16;-5;-4;-2;-1;10;23\right\}\)  

b) \(\dfrac{x+6}{x+1}\) 

Để \(\dfrac{x+6}{x+1}\) là số nguyên thì \(x+6⋮x+1\)  

\(x+6⋮x+1\) 

\(\Rightarrow x+1+5⋮x+1\) 

\(\Rightarrow5⋮x+1\) 

\(\Rightarrow x+1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\) 

Ta có bảng giá trị:

Vậy \(x\in\left\{-6;-2;0;4\right\}\) 

c) \(\dfrac{x-2}{x+3}\) 

Để \(\dfrac{x-2}{x+3}\) là số nguyên thì \(x-2⋮x+3\)   

\(x-2⋮x+3\) 

\(\Rightarrow x+3-5⋮x+3\) 

\(\Rightarrow5⋮x+3\) 

\(\Rightarrow x+3\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\) 

Ta có bảng giá trị:

Vậy \(x\in\left\{-8;-4;-2;2\right\}\) 

d) \(\dfrac{2x+1}{x-3}\) 

Để \(\dfrac{2x+1}{x-3}\) là số nguyên thì \(2x+1⋮x-3\)    

\(2x+1⋮x-3\) 

\(\Rightarrow2x-6+7⋮x-3\) 

\(\Rightarrow7⋮x-3\) 

\(\Rightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\) 

Ta có bảng giá trị:

Vậy \(x\in\left\{-4;2;4;10\right\}\) 

Chúc bạn học tốt!

Để : \(\frac{x-2}{x+3}\in N\) thì x - 2 chia hết cho x + 3

=> x + 3 - 5 chia hết cho x + 3

=> 5 chia hết cho x + 3

=> x + 3 thuộc Ư(5) = {1;5}

=> x = {-2;2}

Vậy x = {-2;2}

Để \(\frac{x-2}{x+3}\)là số tự nhiên thì x - 2 chia hết cho x + 3

=> x - 2 chia hết cho x + 3

     x + 3 - 5 chia hết cho x + 3

=> 5 chia hết x + 3

=> x + 3 thuộc Ư(5) = { 1 ; 5 ; -1 ; -5 }

=> x + 3 thuộc { 1 ;  5 ; -1 ; -5 }

=> x thuộc { -2 ; 2 ; -4 ; -8 }

Vậy x = -2 ; x = 2 ; x = -4 ; x = -8

để A là số nguyên thì 2x+1 chia hết cho x-3

ta có:2x+1 chia hết cho  x-3

       2x-6+7 chia hết cho  x-3

      2x-2.3+7 chia hết cho x-3

    2 (x-3)+7 chia hết cho x-3

2 (x-3) chia hết cho x-3 thì 7 chia hết cho x-3

x-3 thuộc ước của 7. đến đây thì bạn tự làm đc r.

bằng -2;-4;4;-10

A= 2x+1/x-3 = x-3/x-3+7/x-3 = 1+7/x-3

x-3 E Ư(7) = {+-4; 2; 10}

a) \(A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{x+3}\right)\) (ĐK: \(x\ne\pm3\))

\(A=\left[\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x^2-1}{\left(x+3\right)\left(x-3\right)}\right]:\left(2+\dfrac{x+5}{x+3}\right)\)

\(A=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x+3\right)\left(x-3\right)}:\dfrac{2\left(x+3\right)-\left(x+5\right)}{x+3}\)

\(A=\dfrac{-5x-5}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x+3}{x+1}\)

\(A=\dfrac{-5\left(x+1\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)\left(x+1\right)}\)

\(A=\dfrac{-5}{x-3}\)

b) Ta có: \(\left|x\right|=1\)

TH1: \(\left|x\right|=-x\) với \(x< 0\)

Pt trở thành:

\(-x=1\) (ĐK: \(x< 0\)) 

\(\Leftrightarrow x=-1\left(tm\right)\)

Thay \(x=-1\) vào A ta có:

\(A=\dfrac{-5}{x-3}=\dfrac{-5}{-1-3}=\dfrac{5}{4}\)

TH2: \(\left|x\right|=x\) với \(x\ge0\)

Pt trở thành:

\(x=1\left(tm\right)\) (ĐK: \(x\ge0\)) 

Thay \(x=1\) vào A ta có:

\(A=\dfrac{-5}{x-3}=\dfrac{-5}{1-2}=\dfrac{5}{2}\)

c) \(A=\dfrac{1}{2}\) khi:

\(\dfrac{-5}{x-3}=\dfrac{1}{2}\)

\(\Leftrightarrow-10=x-3\)

\(\Leftrightarrow x=-10+3\)

\(\Leftrightarrow x=-7\left(tm\right)\)

d) \(A\) nguyên khi:

\(\dfrac{-5}{x-3}\) nguyên

\(\Rightarrow x-3\inƯ\left(-5\right)\)

\(\Rightarrow x\in\left\{8;-2;2;4\right\}\)

a: \(A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{x+3}\right)\)

\(=\dfrac{x\left(x-3\right)-2\left(x+3\right)-x^2+1}{\left(x-3\right)\left(x+3\right)}:\dfrac{2x+6-x-5}{x+3}\)

\(=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x+1}\)

\(=\dfrac{-5x-5}{\left(x-3\right)}\cdot\dfrac{1}{x+1}=\dfrac{-5}{x-3}\)

b: |x|=1

=>x=-1(loại) hoặc x=1(nhận)

Khi x=1 thì \(A=\dfrac{-5}{1-3}=-\dfrac{5}{-2}=\dfrac{5}{2}\)

c: A=1/2

=>x-3=-10

=>x=-7

d: A nguyên

=>-5 chia hết cho x-3

=>x-3 thuộc {1;-1;5;-5}

=>x thuộc {4;2;8;-2}

a) ĐKXĐ: \(x\ge0,x\ne1\)

\(P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

b) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=-1\)

\(\Leftrightarrow-\sqrt{x}-1=\sqrt{x}-1\Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)

c) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\in Z\)

\(\Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Kết hợp đk:

\(\Leftrightarrow x\in\left\{0\right\}\)

d) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}< 1\)

e) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)

Do \(\sqrt{x}+1\ge1\Leftrightarrow-\dfrac{2}{\sqrt{x}+1}\ge-2\)

\(\Leftrightarrow P=1-\dfrac{2}{\sqrt{x}+1}\ge1-2=-1\)

\(minP=-1\Leftrightarrow x=0\)

\(a,P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\left(x\ge0;x\ne1\right)\\ P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ b,P=-1\Leftrightarrow\sqrt{x}-1=-\sqrt{x}-1\\ \Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\\ c,P\in Z\Leftrightarrow\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\in Z\\ \Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{1;2\right\}\left(\sqrt{x}+1\ge1\right)\\ \Leftrightarrow\sqrt{x}=0\left(x\ne1\right)\\ \Leftrightarrow x=0\)

\(d,P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}< 1\left(\dfrac{2}{\sqrt{x}+1}>0\right)\\ e,P=1-\dfrac{2}{\sqrt{x}+1}\\ \sqrt{x}+1\ge1\Leftrightarrow-\dfrac{2}{\sqrt{x}+1}\ge-\dfrac{2}{1}=-2\\ \Leftrightarrow P=1-\dfrac{2}{\sqrt{x}+1}\ge1-\left(-2\right)=3\)

Dấu \("="\Leftrightarrow x=0\)