b: 6x^2-24y^2

=6(x^2-4y^2)

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

c: =(4x)^3-(3y)^3

=(4x-3y)(16x^2+12xy+9y^2)

d: x^4-2x^3-x^2

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

Lời giải:

a.

$64x^2-24y^2=8(8x^2-3y^2)=8(\sqrt{8}x-\sqrt{3}y)(\sqrt{8}x+\sqrt{3}y)$

b.

$64x^3-27y^3=(4x)^3-(3y)^3=(4x-3y)(16x^2+12xy+9y^2)$

c.

$x^4-2x^3+x^2=(x^2-x)^2=[x(x-1)]^2=x^2(x-1)^2$

d.

$(x-y)^3+8y^3=(x-y)^3+(2y)^3=(x-y+2y)[(x-y)^2-2y(x-y)+(2y)^2]$

$=(x+y)(x^2-4xy+7y^2)$

a) \(64x^2-24y^2\)

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

b) \(64x^3-27y^3\)

\(=\left(4x\right)^3-\left(3y\right)^3\)

\(=\left(4x-3y\right)\left(16x^2+12xy+9y^2\right)\)

c) \(x^4-2x^3+x^2\)

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

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

d) \(\left(x-y\right)^3+8y^3\)

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

\(=\left(x+y\right)\left(x^2-4xy+7y^2\right)\)

Bài 1 :

\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)

Bài 2 :

 \(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)

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

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

Tick đúng nha 

X^2-6+8

\(=2x\left(x+3\right)\)

\(=6x\left(-y^2+x^2+2x+1\right)\\ =6x\left[\left(x^2+2x+1\right)-y^2\right]\\ =6x\left[\left(x+1\right)^2-y^2\right]\\ =6x\left(x+1-y\right)\left(x+1+y\right)\)

\(6x^2+7x+2\\ =6x^2+3x+4x+2\\ =3x\left(2x+1\right)+2\left(2x+1\right)\\ =\left(3x+2\right)\left(2x+1\right)\)

\(6x^2+7x+2\)

\(=6x^2+3x+4x+2\)

\(=3x\left(2x+1\right)+2\left(2x+1\right)\)

\(=\left(2x+1\right)\left(3x+2\right)\)

[Chorus]

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đặt y=x²+1

=>y²=(x²+1)²

y²=x⁴+2x²+1

đặt P(x)=x^4+6x^3+11x^2+6x+1

=x⁴+2x²+1+6x³+6x+9x²

=x⁴+2x+1+6x(x²+1)+9x²

thay y²=x⁴+2x²+1 và y=x²+1 ta được 

Q(y)=y²+6xy+9x²

=(y+3x)²

thay y=x²+1 ta được:

(x²+3x+1)²

vậy x^4+6x^3+11x^2+6x+1=(x²+3x+1)²