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

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:

a ) MTC : \(2x\left(x+3\right)\left(x-3\right)\)

\(\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)

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

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

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

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

Ta có: x 3 - 1 = x - 1 x 2 + x + 1

Mẫu thức chung: x 3 - 1

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Ta có:

Suy ra: x 3 - 7 x 2 + 7 x + 15 = x 2 - 4 x - 5 x - 3

Lại có:

Suy ra: x 3 - 7 x 2 + 7 x + 15 = x 2 - 2 x - 3 x - 5