ĐKXĐ: \(x\ne\pm1;x\ne0\)

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

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

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

\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{2x}{5x-5}-\dfrac{x^2-1}{x^2+2x+1}\)

\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}.\dfrac{5\left(x-1\right)}{2x}-\dfrac{x^2-1}{x^2+2x+1}\)

\(=\dfrac{10}{x+1}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)^2}\)

\(=\dfrac{10}{x+1}-\dfrac{x-1}{x+1}\)

\(=\dfrac{11-x}{x+1}\)

b) \(A=\dfrac{11-x}{x+1}=2\)

\(\Leftrightarrow11-x=2\left(x+1\right)\)

\(\Leftrightarrow11-x=2x+2\)

\(\Leftrightarrow-x-2x=2-11\)

\(\Leftrightarrow-3x=-9\)

\(\Leftrightarrow x=3\left(nhận\right)\)

c) -Để \(A=\dfrac{11-x}{x+1}\in Z\) thì:

\(\left(11-x\right)⋮\left(x+1\right)\)

\(\Rightarrow\left(12-x-1\right)⋮\left(x+1\right)\)

\(\Rightarrow12⋮\left(x+1\right)\)

\(\Rightarrow\left(x+1\right)\inƯ\left(12\right)\)

\(\Rightarrow\left(x+1\right)\in\left\{1;2;3;4;6;12;-1;-2;-3;-4;-6;-12\right\}\)

\(\Rightarrow x\in\left\{2;3;5;11;-2;-3;-4;-5;-7;-13\right\}\)

a: \(A=\dfrac{x^2-5x+6-x^2+x+2x^2-6}{x\left(x-3\right)}=\dfrac{2x^2-4x}{x\left(x-3\right)}=\dfrac{2x}{x-3}\)

a: \(A=\dfrac{x+2+x^2-2x+x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2}{x^2-4}\)

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

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

a: \(A=\dfrac{x+2+x^2-2x+x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{x^2}{x^2-4}\)

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

a) A xác định \(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne1\end{matrix}\right.\)

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

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

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

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

\(\Rightarrow A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}.\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x+1}\)

b) Để \(A=-\dfrac{1}{2}\)

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

\(\Leftrightarrow2x^2=-\left(x+1\right)\)

\(\Leftrightarrow2x^2+x+1=0\)

\(\Delta=1-8=-7< 0\)

Nên phương trình trên vô nghiệm \(\left(x\in\varnothing\right)\)

c) Để \(A< 1\) 

\(\Leftrightarrow\dfrac{x^2}{x+1}< 1\)

\(\Leftrightarrow x^2< x+1\left(x\ne-1\right)\)

\(\Leftrightarrow x^2-x-1< 0\)

\(\Leftrightarrow x^2-x+\dfrac{1}{4}-\dfrac{1}{4}-1< 0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{4}< 0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2< \dfrac{5}{4}\)

\(\Leftrightarrow-\dfrac{\sqrt[]{5}}{2}< x-\dfrac{1}{2}< \dfrac{\sqrt[]{5}}{2}\)

\(\Leftrightarrow\dfrac{-\sqrt[]{5}+1}{2}< x< \dfrac{\sqrt[]{5}+1}{2}\)

d) Để A nguyên

\(\Leftrightarrow\dfrac{x^2}{x+1}\in Z\)

\(\Leftrightarrow x^2⋮x+1\)

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

\(\Leftrightarrow x^2-x^2+x⋮x+1\)

\(\Leftrightarrow x⋮x+1\)

\(\Leftrightarrow x-x-1⋮x+1\)

\(\Leftrightarrow-1⋮x+1\)

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

\(\Leftrightarrow x\in\left\{-2;0\right\}\left(x\in Z\right)\)

!ERROR 404!

\(A=\dfrac{2x+5}{x+1}\left(x\ne-1\right)\)

Để A nguyên khi và chỉ khi :

\(2x+5⋮x+1\)

\(\Rightarrow2x+5-2\left(x+1\right)⋮x+1\)

\(\Rightarrow2x+5-2x-2⋮x+1\)

\(\Rightarrow3⋮x+1\)

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

\(\Rightarrow x\in\left\{-2;0;-4;2\right\}\left(x\in Z;x\ne-1\right)\)

uuuu

kkdiixj emidos

a) để A là phân số thì 
- 2x+5 là số nguyên => 2x+5 nguyên với mọi x nguyên 
- 2x-1 nguyên va 2x-1#0 => 2x-1 nguyên và 2x-1#0 với mọi x nguyên 
vậy A là phân số với mọi x nguyên. 

b) nhận thấy 2x -1 là số lẻ nên 
(1) <=> A = 1 + 6/(2x-1) để A nguyên thì 1 + 6/(2x-1) nguyên <=> 6/(2x-1) nguyên <=> 
<=> 6 chia hết cho (2x-1) hay (2x-1) là ước lẻ của 6 vậy: 
(2x-1) = { 1 ; 3 ; -1 ; -3 } (*)<=> 2x = { 2 ; 4 ; 0 ; -2 } <=> 
<=> x = { 1 ; 2 ; 0 ; -1} 
vì x nguyên nên x chỉ lấy các giá trị : x = {1 ; 2 ; -1} 

c) A = 1 + 6/(2x-1) để Amax thì 1 + 6/(2x-1) max <=> 6/(2x-1) max 
vì 6 > 0 nên để 6/(2x-1)max thì (2x-1) là ƯSC dương lẻ nhỏ nhất của 6 với x nguyên dương 
<=> 2x-1 = 1 (theo (*)) <=> x = 1 khi đó Amax = 1 + 6/1 = 7 
để Amin thì 1 + 6/(2x-1)min <=> 6/(2x-1)min 
vì 6 > 0 nên để 6/(2x-1)min thì (2x-1) là ƯSC âm lẻ lớn nhất của 6 với x nguyên âm=> (2x-1) = -1 
nhưng (2x-1) = -1 (theo (*)) lại ứng với x = 0 ma x nguyên nên loại trường hợp này nên: 
2x-1 = -3 (theo (*)) <=> x = -1 khi đó Amin = 1 + 6/(-1) = -5.