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 \(A_{min}=4\)

BÀi 1:

a: \(\left(a+b\right)^2-\left(a-b\right)^2\)

=(a+b-a+b)(a+b+a-b)

\(=2b\cdot2a=4ab\)

b: \(\left(a+2\right)^2-\left(a+2\right)\left(a-2\right)\)

\(=a^2+4a+4-\left(a^2-4\right)\)

\(=a^2+4a+4-a^2+4=4a+8\)

c: \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)

=>\(4x^2+12x+9-4\left(x^2-1\right)=49\)

=>\(4x^2+12x+9-4x^2+4=49\)

=>12x+13=49

=>12x=36

=>x=3

d: \(Q=\left(x+3\right)^2+\left(x+3\right)\left(x-3\right)-2\left(x+2\right)\left(x-4\right)\)

\(=x^2+6x+9+x^2-9-2\left(x^2-4x+2x-8\right)\)

\(=2x^2+6x-2\left(x^2-2x-8\right)=2x^2+6x-2x^2+4x+16=10x+16\)

Khi x=1/2 thì Q=10*1/2+16=5+16=21

Bài 2:

a: \(A=\left(4x^2+y^2\right)\left(2x+y\right)\left(2x-y\right)\)

\(=\left(4x^2+y^2\right)\left(4x^2-y^2\right)=16x^4-y^4\)

b: \(\left(7x+1\right)^2-\left(x+7\right)^2\)

\(=49x^2+14x+1-\left(x^2+14x+49\right)\)

\(=49x^2+14x+1-x^2-14x-49=48x^2-48=48\left(x^2-1\right)\)

c: \(16x^2-\left(4x-5\right)^2=15\)

=>\(16x^2-\left(16x^2-40x+25\right)=15\)

=>\(16x^2-16x^2+40x-25=15\)

=>40x=40

=>x=1

d: \(A=-x^2+2x+3\)

\(=-x^2+2x-1+4\)

\(=-\left(x-1\right)^2+4\le4\forall x\)

Dấu '=' xảy ra khi x-1=0

=>x=1

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}

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

1111111

1: A=2

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

=>\(2\sqrt{x}-4=\sqrt{x}+1\)

=>\(\sqrt{x}=5\)

=>x=25

2: A<1

=>A-1<0

=>\(\dfrac{\sqrt{x}+1-\sqrt{x}+2}{\sqrt{x}-2}< 0\)

=>\(\dfrac{3}{\sqrt{x}-2}< 0\)

=>\(\sqrt{x}-2< 0\)

=>0<=x<4

3: A<1/3

=>A-1/3<0

=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{1}{3}< 0\)

=>\(\dfrac{3\sqrt{x}+3-\sqrt{x}+2}{3\left(\sqrt{x}-2\right)}< 0\)

=>\(\dfrac{2\sqrt{x}+5}{3\left(\sqrt{x}-2\right)}< 0\)

=>\(\sqrt{x}-2< 0\)

=>0<=x<4

4:
A=căn x

=>\(\sqrt{x}+1=x-2\sqrt{x}\)

=>\(x-3\sqrt{x}-1=0\)

=>\(\left[{}\begin{matrix}\sqrt{x}=\dfrac{3+\sqrt{13}}{2}\left(nhận\right)\\\sqrt{x}=\dfrac{3-\sqrt{13}}{2}\left(loại\right)\end{matrix}\right.\)

=>\(x=\dfrac{11+3\sqrt{13}}{2}\)

1: A={x∈R|x<=2}

=>A=(-∞;2]

B={x∈R|x>5}

=>B=(5;+∞)

A\(\cap\) B=(-∞;2]\(\cap\) (5;+∞)

=∅

A\(\cup\) B=(-∞;2]\(\cup\) (5;+∞)

A\B=(-∞;2]\(5;+∞)

=(-∞;2]

B\A=(5;+∞)\(-∞;2]

=(5;+∞)

\(C_{R}A\) =R\A=R\(-∞;2]

=(2;+∞)

\(C_{R}B=\) R\B=R\(5;+∞)

=(-∞;5]

2: A={x∈R|x<0 hoặc x>=2}

=>A=(-∞;0)\(\cup\) [2;+∞)

B={x∈R|-4<=x<3}

=>B=[-4;3)

A\(\cap\) B=((-∞;0)\(\cup\) [2;+∞))\(\cap\) [-4;3)

=[-4;0)\(\cup\) [2;3)

A\(\cup\) B=((-∞;0)\(\cup\) [2;+∞))\(\cup\) [-4;3)

=(-∞;+∞)

A\B=((-∞;0)\(\cup\) [2;+∞))\[-4;3)

=(-∞;-4)\(\cup\) [3;+∞)

B\A=[-4;3)\((-∞;0)\(\cup\) [2;+∞))

=[0;2)

\(C_{R}A\) =R\A=[0;2)

\(C_{R}B\) =R\B=R\[-4;3)

=(-∞;-4)\(\cup\) [3;+∞)

3: |x-1|<2

=>-2<x-1<2

=>-1<x<3

=>A=(-1;3)

|x+1|<3

=>-3<x+1<3

=>-4<x<2

=>B=(-4;2)

A\(\cap\) B=(-1;3)\(\cap\) (-4;2)

=(-1;2)

A\(\cup\) B=(-1;3)\(\cup\) (-4;2)

=(-4;3)

A\B=(-1;3)\(-4;2)

=[2;3)

B\A=(-4;2)\(-1;3)

=(-4;-1]

\(C_{R}A\) =R\A=R\(-1;3)

=(-∞;-1]\(\cup\) [3;+∞)

\(C_{R}B\) =R\B=R\(-4;2)

=(-∞;-4]\(\cup\) [2;+∞)

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 \(\left|A\right|=A\).