= (x^5-2x^4)-(6x^4-12x^3)+(9x^3-18x^2)-(16x^2-32x)+(48x-96)

 = (x-2).(x^4-6x^3+9x^2-16x+48)

 = (x-2). [ (x^4-3x^3)-(3x^3-9x^2)-(16x-48) ]

 = (x-2).(x-3).(x^3-3x^2-16)

 = (x-2).(x-3).[ (x^3-4x^2)+(x^2-16) ] 

 = (x-2).(x-3).(x-4).(x^2+x+4)

k mk nha

(x² + 2.x.3 + 3² - 1).(x² + 2.x.4 + 16 - 1) - 24

=[(x+3)² - 1]. [(x+4)²-1] -24

=(x+3+1)(x+3-1)(x+4+1)(x+4-1) - 24

=(x+4)(x+2)(x+5)(x-3) - 24

(x²+6x+8)(x²+8x+15)-24

<=>(x²+4x+2x+8)(x²+5x+3x+15)-24

<=> [x(x+4)+2(x+4)][x(x+5)+3(x+5)]-24

<=> (x+4)(x+2)(x+5)(x+3)-24

<=> (x+4)(x+3)(x+2)(x+5)-24

<=>(x²+7x+12)(x²+7x+10)

đặt t=x²+7x+11 ta có:

(t-1)(t+1)-24

<=> t²-1-24

<=>t²-25

<=>(t-5)(t+5)

thay t=x²+7x+11 vào ta có:

(x²+7x+11-5)(x²+7x+11+5)

<=>(x²+7x+6)(x²+7x+16)

x⁴-8x+63=x⁴+4x³+9x²-4x³-16x²-36x+7x²+28x+63

=x².(x²+4x+9)-4x.(x²+4x+9)+7.(x²+4x+9)

=(x²+4x+9)(x²-4x+7)

cái này dễ mà

= (2x)^3-3(2x)^2*1+2*3x*1^2-1^3

= (2x-1)^3

\(a^4-6a^3+27a^2-54a+32\)

\(=\left(a^4-a^3\right)-\left(5a^3-5a^2\right)+\left(22a^2-22a\right)-\left(32a-32\right)\)

\(=\left(a-1\right)\left(a^3-5a^2+22a-32\right)\)

Phân tích đa thức thành nhân tử chứ

\(a,x^2+6x+9\)

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

\(b,10x-25-x^2\)

\(=-\left(x^2-10x+25\right)\)

\(=-\left(x-5\right)^2\)

\(c,8x^3-\frac{1}{8}\)

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

\(=\left(8x-\frac{1}{2}\right)\left(64x^2+4x+\frac{1}{4}\right)\)

\(d,8x^3+12x^2+6xy^2+y^3\)

\(=2\left(4x^3+6x^2+3xy^2+\frac{1}{2}y^3\right)\)

hok tốt!

ai tra loi dung minh cho

Điệp viên 007 sai c

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