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\(ĐKXĐ:x\ne1\)

a) \(A=\frac{2\left(x+1\right)}{x^2+x+1}+\frac{2x^2-9x+4}{x^3-1}+\frac{1}{x-1}\)

\(\Leftrightarrow A=\frac{2\left(x+1\right)\left(x-1\right)+2x^2-9x+4+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\Leftrightarrow A=\frac{2\left(x^2-1\right)+3x^2-8x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\Leftrightarrow A=\frac{2x^2-2+3x^2-8x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\Leftrightarrow A=\frac{5x^2-8x+3}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\Leftrightarrow A=\frac{\left(5x-3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\Leftrightarrow A=\frac{5x-3}{x^2+x+1}\)

b) Để \(A=1\)

\(\Leftrightarrow5x-3=x^2+x+1\)

\(\Leftrightarrow x^2-4x+4=0\)

\(\Leftrightarrow\left(x-2\right)^2=0\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Vậy để \(A=1\Leftrightarrow x=2\)

Bài 1 :

a) \(ĐKXĐ:x\ne1\)

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

\(\Leftrightarrow A=\frac{3+x-1}{\left(x-1\right)\left(x+1\right)}\cdot\left(x+1\right)\)

\(\Leftrightarrow A=\frac{x+2}{x-1}\)

b) Thay x = \(\frac{2}{5}\)vào A ta được :

\(A=\frac{\frac{2}{5}+2}{\frac{2}{5}-1}=\frac{\frac{12}{5}}{-\frac{3}{5}}=-4\)

c) Để \(A=\frac{5}{4}\)

\(\Leftrightarrow\frac{x+2}{x-1}=\frac{5}{4}\)

\(\Leftrightarrow4x+8=5x-5\)

\(\Leftrightarrow x=13\)

d) Để \(A>\frac{1}{2}\)

\(\Leftrightarrow\frac{x+2}{x-1}>\frac{1}{2}\)

\(\Leftrightarrow\frac{x+2}{x-1}-\frac{1}{2}>0\)

\(\Leftrightarrow2x+4-x+1>0\)

\(\Leftrightarrow x+5>0\)

\(\Leftrightarrow x>-5\)

Bài 2 :

a) \(ĐKXĐ:\hept{\begin{cases}x\ne-1\\x\ne0\end{cases}}\)

\(A=\frac{x^2}{x^2+x}-\frac{1-x}{x+1}\)

\(A=\frac{x}{x+1}+\frac{x-1}{x+1}\)

\(\Leftrightarrow A=\frac{2x-1}{x+1}\)

b) Để \(A=1\)

\(\Leftrightarrow\frac{2x-1}{x+1}=1\)

\(\Leftrightarrow2x-1=x+1\)

\(\Leftrightarrow x=2\)

b) Để \(A< 2\)

\(\Leftrightarrow\frac{2x-1}{x+1}< 2\)

\(\Leftrightarrow\frac{2x-1}{x+1}-2< 0\)

\(\Leftrightarrow2x-1-2x-1< 0\)

\(\Leftrightarrow-2< 0\)(luôn đúng)

Vậy A < 2 <=> mọi x

a: ĐKXĐ: \(x\ne-1\)

b: \(B=\dfrac{5}{x+1}+\dfrac{10}{x^2-x+1}-\dfrac{15}{x^3-1}\)

\(=\dfrac{5x^2-5x+5+10x+10-15}{\left(x+1\right)\left(x^2-x+1\right)}\)

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

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

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

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

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

b: |x-3|=2

=>x-3=2 hoặc x-3=-2

=>x=5(nhận) hoặc x=1(loại)

Khi x=5 thì \(E=\dfrac{5^2}{5-1}=\dfrac{25}{4}\)

c: Để E=1/2 thì \(\dfrac{x^2}{x-1}=\dfrac{1}{2}\)

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

hay \(x\in\varnothing\)

f) \(A=\dfrac{x^2}{x-1}=\dfrac{x^2-x+x-1+1}{x-1}=\dfrac{x\left(x-1\right)+x-1+1}{x-1}=x+1+\dfrac{1}{x-1}=x-1+\dfrac{1}{x-1}+2\ge2\sqrt{\left(x-1\right).\dfrac{1}{x-1}}+2=4\)\(A=4\Leftrightarrow x=2\)

-Vậy \(A_{min}=4\)

a/ \(A=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)

\(A=x^3+8-\left[x^3+1+3x\left(x+1\right)\right]+3\left(x^2-1\right)\)

\(A=x^3+8-x^3-1-3x\left(x+1\right)+3x^2-3\)

\(A=-3x^2-3x+3x^2+4\)

\(A=4-3x\)

b/ Để \(\left|A\right|=A\)

=> \(A\ge0\)

<=> \(4-3x\ge0\)

<=> \(4\ge3x\)

<=> \(x\ge\frac{3}{4}\)

Vậy khi \(x\ge\frac{3}{4}\)thì \(\left|A\right|=A\).

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

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!