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

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

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

\(\dfrac{1}{3x+3y}=\dfrac{1}{3\left(x+y\right)}=\dfrac{2\cdot\left(x+y\right)}{6\left(x+y\right)^2}\)

\(\dfrac{1}{2x+2y}=\dfrac{1}{2\left(x+y\right)}=\dfrac{3\left(x+y\right)}{6\left(x+y\right)^2}\)

\(\dfrac{1}{x^2+2xy+y^2}=\dfrac{1}{\left(x+y\right)^2}=\dfrac{6}{6\left(x+y\right)^2}\)

a: 1/x^2y=1/x^2y

3/xy=3x/x^2y

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

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

Bài 2:

a: \(\dfrac{1}{2x^3y}=\dfrac{6yz^3}{12x^3y^2z^3}\)

\(\dfrac{2}{3xy^2z^3}=\dfrac{2\cdot4x^2}{12x^3y^2z^3}=\dfrac{8x^2}{12x^3y^2z^3}\)

MTC : ( x - 1 )( x² + x + 1 )

Ta có : \(\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)

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

\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{6x^2+6x+6}{\left(x-1\right)\left(x^2+x+1\right)}\)

Hnay mới học thì hnay trả lời nhá :P

\(\frac{4x^2-3x+5}{x^3-1};\frac{2x}{x^2+x+1}\)

Ta có : \(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)

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

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

\(\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)

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

\(\frac{2x}{x^2+x+1};\frac{6}{x-1}\)

Ta có : \(x^2+x+1=x^2+x+1\)

\(x-1=x-1\)

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

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

\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{6x^2+6x+6}{x^3-1}\)

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

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

a) \(\dfrac{1}{x-a};\dfrac{2}{x-b}\)

Theo đề bài ta có :

\(\left(x-a\right)\left(x-b\right)=x^2-5x+6\)

\(\Leftrightarrow\left(x-a\right)\left(x-b\right)=\left(x-2\right)\left(x-3\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\)

b) \(\dfrac{1}{x-a}=\dfrac{1}{x-2}=\dfrac{x-3}{\left(x-2\right)\left(x-3\right)}=\dfrac{x-3}{x^2-5x+6}\)

\(\dfrac{2}{x-b}=\dfrac{1}{x-3}=\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x-6}{x^2-5x+6}\)