Bài 2: 

a: \(3x^2-3xy=3x\left(x-y\right)\)

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

c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)

d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)

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a: 2xy-3x+2y-61=0

=>2y(x+1)-3x-3-58=0

=>(x+1)(2y-3)=58

mà 2y-3 lẻ

nên (x+1;2y-3)∈{(58;1);(-58;-1);(2;29);(-2;-29)}

=>(x;2y)∈{(57;4);(-59;2);(1;32);(-3;-26)}

=>(x;y)∈{(57;2);(-59;1);(1;16);(-3;13)}

b: \(4y^2-4y+9=x^2\)

=>\(4y^2-4y+1+8-x^2=0\)

=>\(\left(2y-1\right)^2-x^2=-8\)

=>(2y-1-x)(2y-1+x)=-8

=>(2y-1-x;2y-1+x)∈{(1;-8);(-8;1);(-1;8);(8;-1);(2;-4);(-4;2);(-2;4);(4;-2)}

TH1: 2y-1-x=1 và 2y-1+x=-8

=>2y-1-x+2y-1+x=1-8

=>4y-2=-7

=>4y=-5

=>\(y=-\frac54\)

=>LOại

TH2: 2y-1-x=-8 và 2y-1+x=1

=>2y-1-x+2y-1+x=1-8

=>4y-2=-7

=>4y=-5

=>\(y=-\frac54\)

=>LOại

TH3: 2y-1-x=-1 và 2y-1+x=8

=>2y-1-x+2y-1+x=-1+8

=>4y-2=7

=>4y=9

=>y=9/4(loại)

TH4: 2y-1-x=8 và 2y-1+x=-1

=>2y-1-x+2y-1+x=-1+8

=>4y-2=7

=>4y=9

=>y=9/4(loại)

TH5: 2y-1-x=2 và 2y-1+x=-4

=>2y-1-x+2y-1+x=2-4

=>4y-2=-2

=>4y=0

=>y=0(nhận)

2y-1-x=2

=>0-1-x=2

=>-x-1=2

=>-x=3

=>x=-3(nhận)

TH6: 2y-1-x=-4 và 2y-1+x=2

=>2y-1-x+2y-1+x=2-4

=>4y-2=-2

=>4y=0

=>y=0(nhận)

2y-1-x=-4

=>0-1-x=-4

=>x+1=4

=>x=3(nhận)

TH7: 2y-1-x=-2 và 2y-1+x=4

=>2y-1-x+2y-1+x=-2+4

=>4y-2=2

=>4y=4

=>y=1(nhận)

2y-1-x=-2

=>2-1-x=-2

=>1-x=-2

=>x=1+2=3(nhận)

TH8: 2y-1-x=4 và 2y-1+x=-2

=>2y-1-x+2y-1+x=-2+4

=>4y-2=2

=>4y=4

=>y=1(nhận)

2y-1-x=4

=>2-1-x=4

=>1-x=4

=>x=1-4=-3(nhận)

a: \(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)

\(=\frac{2x}{x\left(x+2y\right)}+\frac{y}{y\left(x-2y\right)}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2}{x+2y}+\frac{1}{x-2y}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}=\frac{2\left(x-2y\right)+x+2y+4}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2x-4y+x+2y+4}{\left(x-2y\right)\left(x+2y\right)}=\frac{3x-2y+4}{\left(x-2y\right)\left(x+2y\right)}\)

b: \(\frac{2}{x+2}+\frac{4}{x-2}+\frac{5x+2}{4-x^2}\)

\(=\frac{2\left(x-2\right)+4\left(x+2\right)-5x-2}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{2x-4+4x+8-5x-2}{\left(x-2\right)\left(x+2\right)}=\frac{x+2}{\left(x-2\right)\left(x+2\right)}=\frac{1}{x-2}\)

c: \(\frac{x}{x-2y}+\frac{x}{x+2y}-\frac{4xy}{4y^2-x^2}\)

\(=\frac{x\left(x+2y\right)+x\left(x-2y\right)+4xy}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{x^2+2xy+x^2-2xy+4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x^2+4xy}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2x\left(x+2y\right)}{\left(x+2y\right)\left(x-2y\right)}=\frac{2x}{x-2y}\)

d: \(\frac{3x^2-x}{x-1}+\frac{x+2}{1-x}+\frac{3-2x^2}{x-1}\)

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

\(=\frac{\left(x-1\right)^2}{x-1}=x-1\)

x²+4y²-2x+4y+2=0

<=>x²-2x+1+4y²+4y+1=0

<=>(x-1)²+(2y+1)²=0

<=>x-1=0 và 2y+1=0

<=>x=1 và y=-1/2

a) x²+y²-4x+4y+8=0

⇔ (x-2)²+(y+2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)

b)5x²-4xy+y²=0

⇔ x²+(2x-y)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

c)x²+2y²+z²-2xy-2y-4z+5=0

⇔ (x-y)²+(y-1)²+(z-2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)

b: Ta có: \(5x^2-4xy+y^2=0\)

\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)

\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)