1/x ĐKXĐ là x khác 0;

2/x(x-1) ĐKXĐ là x khác 0;1

4/5x-10 ĐKXĐ là x=2 

2x+4/2x-4 ĐKXĐ là x khác 2 

x+1/x-1 ĐKXĐ là x khác 1 

a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)

b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)

c) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5}{3}\left(x\ne y\right)\)

d) \(\dfrac{x^2-y^2}{x+y}=x-y\left(đk:x\ne-y\right)\)

e) \(\dfrac{x^3-x^2+x-1}{x^2-1}=\dfrac{x^2+1}{x+1}\left(đk:x\ne\pm1\right)\)

f) \(\dfrac{x^2+4x+4}{2x+4}=\dfrac{x+2}{2}\left(đk:x\ne-2\right)\)

a: \(\left(2x+1\right)\left(x^2+x-1\right)\left(2x^2-3x+1\right)=0\)

TH1: 2x+1=0

=>2x=-1

=>\(x=-\frac12\)

TH2: \(x^2+x-1=0\)

=>\(x^2+x+\frac14-\frac54=0\)

=>\(\left(x+\frac12\right)^2=\frac54\)

=>\(\left[\begin{array}{l}x+\frac12=\frac{\sqrt5}{2}\\ x+\frac12=-\frac{\sqrt5}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt5-1}{2}\\ x=\frac{-\sqrt5-1}{2}\end{array}\right.\)

TH3: \(2x^2-3x+1=0\)

=>(x-1)(2x-1)=0

=>x=1/2 hoặc x=1

Do đó: A={-1/2;1/2;1;\(\frac{\sqrt5-1}{2};\frac{-\sqrt5-1}{2}\) }

b: \(6x^2-5x+1=0\)

=>\(6x^2-2x-3x+1=0\)

=>2x(3x-1)-(3x-1)=0

=>(3x-1)(2x-1)=0

=>x=1/3 hoặc x=1/2

=>B={1/3;1/2}

c: \(\left(2x+x^2\right)\cdot\left(x^2+x-2\right)\left(x^2-x-12\right)=0\)

=>\(x\left(x+2\right)\left(x+2\right)\left(x-1\right)\left(x-4\right)\left(x+3\right)=0\)

=>\(x\left(x+2\right)^2\cdot\left(x-1\right)\left(x-4\right)\left(x+3\right)=0\)

=>x∈{0;-2;1;4;-3}

=>C={0;-2;1;4;-3}

g: \(x^2-9=0\)

=>(x-3)(x+3)=0

=>x∈{3;-3}

=>G={3;-3}

h: \(\left(x-1\right)\left(x^2+6x+5\right)=0\)

=>(x-1)(x+1)(x+5)=0

=>x∈{1;-1;-5}

=>H={1;-1;-5}

i: \(x^2-x+2=0\)

=>\(x^2-x+\frac14+\frac74=0\)

=>\(\left(x-\frac12\right)^2+\frac74=0\) (vô lý)

=>x∈∅

=>I=∅

j: \(\left(2x-1\right)\left(x^2-5x+6\right)=0\)

=>(2x-1)(x-2)(x-3)=0

=>x∈{1/2;2;3}

=>J={1/2;2;3}

k: -3<x<13

mà x là số chẵn

nên x∈{-2;0;2;4;6;8;10;12}

=>K={-2;0;2;4;6;8;10;12}

l: \(x^2>4\)

=>x>2 hoặc x<-2

|x|<10

=>-10<x<10

mà x>2 hoặc x<-2

nên x∈{3;4;5;6;7;8;9;-9;-8;-7;-6;-5;-4;-3}

=>L={3;4;5;6;7;8;9;-9;-8;-7;-6;-5;-4;-3}

m: -1<k<5

mà k∈Z

nên k∈{0;1;2;3;4}

=>3k∈{0;3;6;9;12}

=>M={0;3;6;9;12}

n: \(\begin{cases}x^2-1=0\\ x^2-4x+3=0\end{cases}\Rightarrow\begin{cases}x^2=1\\ \left(x-1\right)\left(x-3\right)=0\end{cases}\)

=>x=1

=>N={1}

a: \(\left(2x+1\right)\left(x^2+x-1\right)\left(2x^2-3x+1\right)=0\)

TH1: 2x+1=0

=>2x=-1

=>\(x=-\frac12\)

TH2: \(x^2+x-1=0\)

=>\(x^2+x+\frac14-\frac54=0\)

=>\(\left(x+\frac12\right)^2=\frac54\)

=>\(\left[\begin{array}{l}x+\frac12=\frac{\sqrt5}{2}\\ x+\frac12=-\frac{\sqrt5}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt5-1}{2}\\ x=\frac{-\sqrt5-1}{2}\end{array}\right.\)

TH3: \(2x^2-3x+1=0\)

=>(x-1)(2x-1)=0

=>x=1/2 hoặc x=1

Do đó: A={-1/2;1/2;1;\(\frac{\sqrt5-1}{2};\frac{-\sqrt5-1}{2}\) }

b: \(6x^2-5x+1=0\)

=>\(6x^2-2x-3x+1=0\)

=>2x(3x-1)-(3x-1)=0

=>(3x-1)(2x-1)=0

=>x=1/3 hoặc x=1/2

=>B={1/3;1/2}

c: \(\left(2x+x^2\right)\cdot\left(x^2+x-2\right)\left(x^2-x-12\right)=0\)

=>\(x\left(x+2\right)\left(x+2\right)\left(x-1\right)\left(x-4\right)\left(x+3\right)=0\)

=>\(x\left(x+2\right)^2\cdot\left(x-1\right)\left(x-4\right)\left(x+3\right)=0\)

=>x∈{0;-2;1;4;-3}

=>C={0;-2;1;4;-3}

g: \(x^2-9=0\)

=>(x-3)(x+3)=0

=>x∈{3;-3}

=>G={3;-3}

h: \(\left(x-1\right)\left(x^2+6x+5\right)=0\)

=>(x-1)(x+1)(x+5)=0

=>x∈{1;-1;-5}

=>H={1;-1;-5}

i: \(x^2-x+2=0\)

=>\(x^2-x+\frac14+\frac74=0\)

=>\(\left(x-\frac12\right)^2+\frac74=0\) (vô lý)

=>x∈∅

=>I=∅

j: \(\left(2x-1\right)\left(x^2-5x+6\right)=0\)

=>(2x-1)(x-2)(x-3)=0

=>x∈{1/2;2;3}

=>J={1/2;2;3}

k: -3<x<13

mà x là số chẵn

nên x∈{-2;0;2;4;6;8;10;12}

=>K={-2;0;2;4;6;8;10;12}

l: \(x^2>4\)

=>x>2 hoặc x<-2

|x|<10

=>-10<x<10

mà x>2 hoặc x<-2

nên x∈{3;4;5;6;7;8;9;-9;-8;-7;-6;-5;-4;-3}

=>L={3;4;5;6;7;8;9;-9;-8;-7;-6;-5;-4;-3}

m: -1<k<5

mà k∈Z

nên k∈{0;1;2;3;4}

=>3k∈{0;3;6;9;12}

=>M={0;3;6;9;12}

n: \(\begin{cases}x^2-1=0\\ x^2-4x+3=0\end{cases}\Rightarrow\begin{cases}x^2=1\\ \left(x-1\right)\left(x-3\right)=0\end{cases}\)

=>x=1

=>N={1}

a: ĐKXĐ: x<>0

b: ĐKXĐ: \(x\notin\left\{0;1\right\}\)

c: ĐKXĐ: x<>2

d: ĐKXĐ: x<>2

e: ĐKXĐ: x<>1

Bài 1:

a: ĐKXĐ: \(x+4\ne0\)

=>\(x\ne-4\)

b: ĐKXĐ: \(2x-1\ne0\)

=>\(2x\ne1\)

=>\(x\ne\dfrac{1}{2}\)

c: ĐKXĐ: \(x\left(y-3\right)\ne0\)

=>\(\left\{{}\begin{matrix}x\ne0\\y-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\y\ne3\end{matrix}\right.\)

d: ĐKXĐ: \(x^2-4y^2\ne0\)

=>\(\left(x-2y\right)\left(x+2y\right)\ne0\)

=>\(x\ne\pm2y\)

e: ĐKXĐ: \(\left(5-x\right)\left(y+2\right)\ne0\)

=>\(\left\{{}\begin{matrix}x\ne5\\y\ne-2\end{matrix}\right.\)

 Bài 2:

a: \(\dfrac{-12x^3y^2}{-20x^2y^2}=\dfrac{12x^3y^2}{20x^2y^2}=\dfrac{12x^3y^2:4x^2y^2}{20x^2y^2:4x^2y^2}=\dfrac{3x}{5}\)

b: \(\dfrac{x^2+xy-x-y}{x^2-xy-x+y}\)

\(=\dfrac{\left(x^2+xy\right)-\left(x+y\right)}{\left(x^2-xy\right)-\left(x-y\right)}\)

\(=\dfrac{x\left(x+y\right)-\left(x+y\right)}{x\left(x-y\right)-\left(x-y\right)}=\dfrac{\left(x+y\right)\left(x-1\right)}{\left(x-y\right)\left(x-1\right)}\)

\(=\dfrac{x+y}{x-y}\)

c: \(\dfrac{7x^2-7xy}{y^2-x^2}\)

\(=\dfrac{7x\left(x-y\right)}{\left(y-x\right)\left(y+x\right)}\)

\(=\dfrac{-7x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{-7x}{x+y}\)
d: \(\dfrac{7x^2+14x+7}{3x^2+3x}\)

\(=\dfrac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}\)

\(=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)

e: \(\dfrac{3y-2-3xy+2x}{1-3x-x^3+3x^2}\)

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

\(=\dfrac{\left(3y-2\right)\left(1-x\right)}{\left(1-x\right)^3}=\dfrac{3y-2}{\left(1-x\right)^2}\)

g: \(\dfrac{x^2+7x+12}{x^2+5x+6}\)

\(=\dfrac{\left(x+3\right)\left(x+4\right)}{\left(x+3\right)\left(x+2\right)}\)

\(=\dfrac{x+4}{x+2}\)

a)\(x\in R\)

b)\(x\ne1\)

c) \(x\notin\left\{1;2\right\}\)

d) \(x\notin\left\{3;-3\right\}\)

e) \(x\ne1\)

f) \(x\notin\left\{2;3\right\}\)

a) 

b) 

c) 

d) 

e) 

f) 

Bài 1: 

c: ĐKXĐ: \(x\notin\left\{-1;3\right\}\)