a) \(x^2+12x+35\)

\(=x^2+5x+7x+35\)

\(=\left(x^2+5x\right)+\left(7x+35\right)\)

\(=x\left(x+5\right)+7\left(x+5\right)\)

\(=\left(x+5\right)\left(x+7\right)\)

b)\(x^2-x-56\)

\(=x^2+7x-8x-56\)

\(=\left(x^2+7x\right)-\left(8x+56\right)\)

\(=x\left(x+7\right)-8\left(x+7\right)\)

\(=\left(x+7\right)\left(x-8\right)\)

c)\(5x^2-x-4\)

\(=5x^2-5x+4x-4\)

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

\(=5x\left(x-1\right)+4\left(x-1\right)\)

\(=\left(x-1\right)\left(5x+4\right)\)

TL:

a)\(x^2+5x+7x+35\) 

 =\(x\left(x+5\right)+7\left(x+5\right)\) 

=\(\left(x+7\right)\left(x+5\right)\) 

b) \(x^2-x-56\)

  =\(x^2+7x-8x-56\) 

=\(x\left(x+7\right)-8\left(x+7\right)\) 

=\(\left(x-8\right)\left(x+7\right)\) 

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

=\(\left(2x^2+1\right)^2-4x^2\) 

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

.......................(tự lm)

hc tốt

\(4\times^4+81\)

\(=\left(2\times^2\right)^2+36\times^2+9^2-36\times^2\)

\(=\left(2\times^2+9\right)^2-36\times^2\)

\(=\left(2\times^2+9+36\times\right)\left(2\times^2+9-36\times\right)\)

\(4x^4+81=\left(2x\right)^2+2.2x^2.9+9^2-36x^2\)

\(=\left(2x^2+9\right)^2-\left(6x\right)^2=\left(2x^2-6x+9\right)\left(2x^2+6x+9\right)\)

\(64x^4+y^4=\left(8x^2\right)^2+2.8x^2.y^2+\left(y^2\right)^2-16x^2y^2\)

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

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) \(\left(x-y\right)^3+8y^3\)

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

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

  \(A=\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)

Đặt \(x-y=a,y-z=b,z-x=c\Rightarrow a+b+c=0\)

\(\Rightarrow a+b=-c\)

\(\Rightarrow\left(a+b\right)^3=\left(-c\right)^3\)

\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)

\(\Rightarrow a^3+b^3+3ab.\left(-c\right)=-c^3\)

\(\Rightarrow a^3+b^3+c^3=3abc\)

Vậy \(A=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

        \(x^4+4x^2+16\)

\(=\left(x^2\right)^2+2.x^2.4+4^2-4x^2\)

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

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

Lời giải:

a. 

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

$=xy(xy+1)-2(xy+1)=(xy+1)(xy-2)$

b. Bạn xem lại đoạn $-16x^2$ là dấu - hay + vậy?

\(4x^4+4x^3+5x^2+6x+1\)

\(=4x^4+4x^3+5x^2+5x+x+1\)

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

\(=\left(x+1\right).\left(4x+5x+1\right)\)

p/s: tớ nghĩ sai đề nên đổi ạ :))

1: \(6x^2y-9xy^2+3xy\)

\(=3xy\left(2x-3y+1\right)\)

2: \(\left(4-x\right)^2-16\)

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

\(=-x\cdot\left(8-x\right)\)

3: \(x^3+9x^2-4x-36\)

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

\(=\left(x+9\right)\left(x-2\right)\left(x+2\right)\)

1) \(6x^2y-9xy^2+3xy=3xy\left(2x-3y+1\right)\)

2) \(\left(4-x\right)^2-16=\left(4-x\right)^2-4^2=\left(4-x-4\right)\left(4-x+4\right)=-x\left(8-x\right)\)

3) \(x^3+9x^2-4x-36\\ =\left(x^3-2x^2\right)+\left(11x^2-22x\right)+\left(18x-36\right)\\ =x^2\left(x-2\right)+11x\left(x-2\right)+18\left(x-2\right)\\ =\left(x^2+11x+18\right)\left(x-2\right)\\ =\left[\left(x^2+2x\right)+\left(9x+18\right)\right]\left(x-2\right)\\ =\left[x\left(x+2\right)+9\left(x+2\right)\right]\left(x-2\right)\\ =\left(x+2\right)\left(x+9\right)\left(x-2\right)\)

Khó quá , bó tay 

\(\left(x^2-2x+2\right)^4-20x^2\left(x^2-2x+2\right)+64x^4\)

\(=\left[\left(x^2-2x+2\right)^2\right]^2-2.\left(x^2-2x+2\right)^2.10x^2+\left(10x^2\right)^2-36x^4\)

\(=\left[\left(x^2-2x+2\right)^2-10x^2\right]^2-\left(6x^2\right)^2\)\(=\left[\left(x^2-2x+2\right)^2-4x^2\right]\left[\left(x^2-2x+2\right)^2-16x^2\right]\)

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

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