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

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

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

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

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