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)

1, \(a^6+b^3=\left(a^2+b\right)\left(a^4-a^2b+b^2\right)\)

2, \(x^2-10x+25=\left(x-5\right)^2\) 

3, \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

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

1) \(a^6+b^3=\left(a^2\right)^3+b^3=\left(a^2+b\right)\left(a^4-a^2b+b^2\right)\)

2) \(x^2-10x+25=\left(x-5\right)^2\)

3) \(8x^3-\dfrac{1}{8}=\left(2x\right)^3-\left(\dfrac{1}{3}\right)^3=\left(2x-\dfrac{1}{3}\right)\left(4x^2+\dfrac{2x}{3}+\dfrac{1}{4}\right)\)

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

1: \(a^6+b^3=\left(a^2+b\right)\left(a^4-a^2b+b^2\right)\)

2: \(x^2-10x+25=\left(x-5\right)^2\)

3: \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

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

56.B

57.B

58.B

56B

57B

58B

\(a.x^3-2x^2-2x-4\\ =\left(x^3-2x^2\right)-\left(2x-4\right)\\ =x^2\left(x-2\right)-2\left(x-2\right)\\ =\left(x^2-2\right)\left(x-2\right)\)

\(b.xy+1-x-y\\ =\left(xy-x\right)+\left(-y+1\right)\\ =x\left(y-1\right)-\left(y-1\right)\\ =\left(x-1\right)\left(y-1\right)\)

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

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

b: =xy-x-y+1

=x(y-1)-(y-1)

=(x-1)(y-1)

c: =(x-2y)^2-4y

\(=\left(x-2y-2\sqrt{y}\right)\left(x-2y+2\sqrt{y}\right)\)

d: =16-(x^2-2xy+y^2)

=16-(x-y)^2

=(4-x+y)(4+x-y)

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

= \(-\left(x^2+4xy+4y^2\right)\)

= \(-\left(x+2y\right)^2\)

x² - 4xy + 4y² - z² + 2zt - t²

= (x² - 4xy + 4y²) - (z² - 2zt + t²)

= (x - 2y)² - (z - t)²

= (x - 2y + z - t)(x - 2y - z + t)

a: \(=4xy\left(1-5x^2y\right)\)

b: \(=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)

c: \(=x\left(x-a\right)+y\left(x-a\right)=\left(x-a\right)\left(x+y\right)\)

d: \(=\left(x+2y\right)^2-36=\left(x+2y+6\right)\left(x+2y-6\right)\)

`a)x^3-8x^2+16x`

`=x(x^2-8x+16)`

`=x(x-4)^2`

`b)x^2+4y^2+2x-4y-4xy-24`

`=(x-2y)^2+2(x-2y)-24`

`=(x-2y)^2-4(x-2y)+6(x-2y)-24`

`=(x-2y-4)(x-2y+6)`

`c)x^4+x^3-x^2-2x-2`

`=x^4-2x^2+x^3-2x+x^2-2`

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

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

\(a,=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\\ b,=4x^2\left(x^2+2x+1\right)=4x^2\left(x+1\right)^2\\ c,=xy^2\left(x^2-2xy+y^2\right)=xy^2\left(x-y\right)^2\\ d,=\left(x-y\right)\left(x+y\right)-7\left(x-y\right)=\left(x-y\right)\left(x+y-7\right)\\ e,=\left(5x-2y\right)\left(5x+2y\right)\\ f,=x^2+3x+4x+12=\left(x+3\right)\left(x+4\right)\\ i,=x^2+2x-7x-14=\left(x+2\right)\left(x-7\right)\)

Câu 1:

$x^2+4y^2+4xy-16=[x^2+(2y)^2+2.x.2y]-16$

$=(x+2y)^2-4^2=(x+2y-4)(x+2y+4)$

Câu 2:

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

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

Câu 1:

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

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

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

Câu 2:

\(x^3+x^2+y^3+xy\)

\(=\left(x^3+y^3\right)\left(x^2+xy\right)\)

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

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

C1:x^2+4y^2+4xy-16

=[x^2+4xy+(2y)^2]-16

=(x+2y)^2-4^2

=(x+2y-4)(x+2y+4)

C2: x^3+x^2+y^3+xy

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

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

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

bài này ra lâu r nhưng ngứa tay nên giải luôn=)))))