\(=x^2\left(8x-3\right)-\left(8x-3\right)=\left(8x-3\right)\left(x^2-1\right)\)

\(\left(x-4\right)^2-9^2=\left(x-13\right)\left(x+5\right)\)

−5𝑥^2−8𝑥+4

\(=\left[2\left(x-1\right)-3x\right]\left[2\left(x-1\right)+3x\right]\\ =\left(2x-2-3x\right)\left(2x-2+3x\right)\\ =\left(-2-x\right)\left(5x-2\right)\\ =\left(x+2\right)\left(2-5x\right)\)

\(4\left(x-1\right)^2-9x^2=\left(2x-2\right)^2-\left(3x\right)^2=\left(2x-2-3x\right)\left(2x-2+3x\right)=\left(-x-2\right)\left(5x-2\right)=-\left(x+2\right)\left(5x-2\right)=\left(x+2\right)\left(2-5x\right)\)

a: =(x^2-1)^2-2x(x^2-1)+x(x^2-1)-2x^2

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

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

b: \(=\left(x^2+1\right)^2+x\left(x^2+1\right)+2x\left(x^2+1\right)+2x^2\)

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

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

\(=\left(x+1\right)^2\cdot\left(x^2+x+1\right)\)

Lời giải:

$(x^3-2xy+3x^2)(x-2y)=x(x^2-2xy+3x)(x-2y)$

\(4x^3-13x^2+9x-18\)

\(=4x^3-12x^2-x^2+3x+6x-18\)

\(=4x^2\left(x-3\right)-x\left(x-3\right)+6\left(x-3\right)\)

\(=\left(x-3\right)\left(4x^2-x+6\right)\)

x 2 + 3 x + 2 = ( x + 1 ) ( x + 2 )

`

` x^2 – 3x + 2`

`= x^2 – x – 2x + 2` (Tách `–3x = – x – 2x)`

`= (x^2 – x) – (2x – 2)`

1: \(-x^2+2x+8\)

\(=-\left(x^2-2x-8\right)\)

\(=-\left(x-4\right)\left(x+2\right)\)

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

3 x – 3 y + x 2 – y 2 = ( x   –   y ) ( 3   +   x   +   y )