a ) \(2x-1-x^2\)

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

\(=\left(x-1\right)\left(1-x\right)\)

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

b) \(8x^3+y^6\)

\(=\left(2x+y^2\right)\left(4x^2-2xy^2+y^4\right)\)

c) \(x^2-16+4xy+4y^2\)

\(=\left(x^2+4xy+4y^2\right)-16\)

\(=\left(x+2y\right)^2-16\)

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

a) x² + 6x + 9 = x² + 2 . x . 3 + 3² = (x + 3)²

b) 10x – 25 – x² = -(-10x + 25 +x²) = -(25 – 10x + x²)

                         = -(5² – 2 . 5 . x – x²) = -(5 – x)²

c) 8x³ - 1/8 = (2x)³ – (1/2)³ = (2x - 1/2)[(2x)² + 2x . 12 + (1/2)²]

                    ^{= (2x - 1/2)(4x2 + x + 1/4)} 

d)1/25x² – 64y² = (1/5x)2(1/5x)2- (8y)² = (1/5x + 8y)(1/5x - 8y)

x2 + 2x – 3

= x2 + 2x + 1 – 4

= (x + 1)2 – 22

= (x + 1 + 2)(x + 1 – 2)

= (x + 3)(x – 1)

\(x^2+2x+1-16=\left(x+1\right)^2-4^2=\left(x+1-4\right).\left(x+1+4\right)=\left(x-3\right).\left(x+5\right)\)

\(x^2+2x+1-16=\left(x^2+2x+1\right)-4^2=\left(x+1\right)^2-4^2=\left(x+1-4\right)\left(x+1+4\right)=\left(x-3\right)\left(x+5\right)\)

\(x^2+2x+1-16=\left(x+5\right)\left(x-3\right)\)

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

\(=\left(x+1-y\right)\left(x+1+y\right)\)

`x^2+2x+1-y^2+2y-1`

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

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

`=(x+1+y-1)(x+1-y+1)`

`=(x+y)(x-y+2)`

Ta có: \(x^2+2x+1-y^2+2y-1\)

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

\(=\left(x+1-y+1\right)\left(x+1+y-1\right)\)

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

Bài 6:

c: \(9x^2+6x+1=\left(3x+1\right)^2\)

d: \(4x^2-9=\left(2x-3\right)\left(2x+3\right)\)

e: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)

=(x²+2x+1)-y²=(x+1)²-y²=(x-y+1)(x+y+1)

\(x^2+2x+1-y^2\)

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

\(=\left(x+1-y\right)\left(x+1+y\right)\)

\(=\left(x^2-2x+1\right)-\left(y^2-2yz+z^2\right)\)

\(=\left(x-1\right)^2-\left(y-z\right)^2\)

\(=\left(x-1-y+z\right)\left(x-1+y-z\right)\)

\(x^2-2x+1-y^2+2yz-z^2\)

\(=\left(x-1\right)^2-\left(y-z\right)^2\)

\(=\left(x-1-y+z\right)\left(x-1+y-z\right)\)