a: \(\dfrac{3}{x-1}=\dfrac{3\cdot9}{9\cdot\left(x-1\right)}=\dfrac{27}{9\left(x-1\right)}\)

\(\dfrac{4}{3x-3}=\dfrac{12}{9x-9}=\dfrac{12}{9\left(x-1\right)}\)

\(\dfrac{10}{9-9x}=\dfrac{-10}{9x-9}=-\dfrac{10}{9\left(x-1\right)}\)

b: \(\dfrac{3}{2\left(x-3\right)}=\dfrac{3x-9}{2\left(x-3\right)^2}\)

\(\dfrac{3x-2}{x^2-6x+9}=\dfrac{6x-4}{2\left(x-3\right)^2}\)

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

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

a: \(\frac{3x}{2xy^2}=\frac{3x\cdot3}{2xy^2\cdot3}=\frac{9x}{6xy^2}\)

\(\frac{-y}{6y^2x}=\frac{-y}{6xy^2}\)

b: \(\frac{x+4}{x^2+x}=\frac{x+4}{x\left(x+1\right)}\)

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

c: \(\frac{x}{x^2-25}=\frac{x}{\left(x-5\right)\left(x+5\right)}=\frac{x\left(x-5\right)}{\left(x-5\right)^2\cdot\left(x+5\right)}\)

\(\frac{x+2}{x^2-10x+25}=\frac{x+2}{\left(x-5\right)^2}=\frac{\left(x+2\right)\left(x+5\right)}{\left(x+5\right)\left(x-5\right)^2}\)

M T 1 : x 3 – 1 = ( x - 1 ) ( x 2 + x + 1 )

M T 2 : x 2 + x + 1

⇒ M T C : ( x - 1 ) ( x 2 + x + 1 )

⇒ NTP1: 1

⇒ NTP2: x - 1

Quy đồng:

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

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

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

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

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

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

đề viết là 1/2y+2x mà bạn