a) MTC: \(12x^3y^3\)

\(\dfrac{3}{4x^3y^2}=\dfrac{3\cdot3y}{4x^3y^2\cdot3y}=\dfrac{9y}{12x^3y^3}\)

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

b) MTC: \(x\left(x-3\right)^2\)

\(\dfrac{5}{x^2-6x+9}=\dfrac{5}{\left(x-3\right)^2}=\dfrac{5x}{x\left(x-3\right)^2}\)

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

Bài 2:

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

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

\(a,\dfrac{7x-1}{2x^2+6x}=\dfrac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}=\dfrac{7x^2-22x+3}{2x\left(x-3\right)\left(x+3\right)}\\ \dfrac{5-3x}{x^2-9}=\dfrac{2x\left(5-3x\right)}{2x\left(x-3\right)\left(x+3\right)}=\dfrac{10x-6x^2}{2x\left(x-3\right)\left(x+3\right)}\)

a)MTC:\(12x^5y^4\)

\(\dfrac{5}{x^5y^3}=\dfrac{5\cdot12y}{x^5y^3\cdot12y}=\dfrac{60y}{12x^5y^4}\)

\(\dfrac{7}{12x^3y^4}=\dfrac{7\cdot x^2}{12x^3y^4\cdot x^2}=\dfrac{7x^2}{12x^5y^4}\)

b)MTC:\(60x^4y^5\)

\(\dfrac{4}{15x^3y^5}=\dfrac{4\cdot4x}{15x^3y^5\cdot4x}=\dfrac{16x}{60x^4y^5}\)

\(\dfrac{11}{12x^4y^2}=\dfrac{11\cdot5y^3}{12x^4y^2\cdot5y^3}=\dfrac{55y^3}{60x^4y^5}\)

\(\frac{1}{6x^2y^3}=\frac{7x^2}{42x^4y^3},\frac{-5}{21xy^2}=\frac{-10x^3y}{42x^4y^3},\frac{3}{14x^4y}=\frac{3y^2}{14x^4y^3}\)

Ta có: 2 x 2 + 6 x = 2 x x + 3 ;   x 2 - 9 = x + 3 x - 3

Mẫu thức chung: 2x(x + 3)(x – 3)

MT1: 2 x 2 + 6 x = 2 x ( x + 3 ) ;   MT2: x 2 - 9 = ( x - 3 ) ( x + 3 )

MTC: 2x(x - 3)(x + 3)

NTP1: x – 3;    NTP2: 2x

Quy đồng:

\(\dfrac{5}{x^5y^3};\dfrac{7}{12x^3y^4}\)

\(MTC=12x^5y^4\)

Ta có:

\(\dfrac{5}{x^5y^3}=\dfrac{60y}{12x^5y^4}\)

\(\dfrac{7}{12x^3y^4}=\dfrac{7x^2}{12x^5y^4}\)