x.x-2.x-1-y.y

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

a) 6x^2-11x+3                              b)2x^2+3x-27                      c)3x^2-8x+4

= 6x^2-2x-9x+3                            =2x^2-6x+9x-27                    =3x^2-6x-2x+4

=2x(3x-1)-3(3x-1)                         =2x(x-3)+9(x-3)                      =3x(x-2)-2(x-2)

=(2x-3)(3x-1)                               =(2x+9)(x-3)                           =(3x-2)(x-2)      

b, A=[(a+1)(a+7)][(a+3)(a+5)]+15

=>A=(a²+8a+7)(a²+8a+15)+15

Đặt a²+8a+11= t

=>a²+8a+7= t-4 và a²+8a+15= t+4

=>A=(t-4)(t+4)+15

=>A=t²-16+15

      =t²-1=(t-1)(t+1)

Thay t = a²+8a+11

=>A=(a²+8a+11-1)(a²+8a+11+1)

=>A=(a²+8a+10)(a²+8a+12)

a)   \(x^2+2xy+7x+7y+y^2+10\)

\(=\left(x+y\right)^2+7\left(x+y\right)+\frac{49}{4}-\frac{9}{4}\)

\(=\left(x+y+\frac{7}{2}\right)^2-\frac{9}{4}\)

\(=\left(x+y+\frac{7}{2}-\frac{3}{2}\right)\left(x+y+\frac{7}{2}+\frac{3}{2}\right)\)

\(=\left(x+y-2\right)\left(x+y+5\right)\)

\(x^4+2017x^2+2016x+2017\)

\(=\left(x^4+x^2+1\right)+2016\left(x^2+x+1\right)\)

\(=\left(x^4+2x^2+1-x^2\right)+2016\left(x^2+x+1\right)\)

\(=\left[\left(x^2+1\right)-x^2\right]+2016\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2016\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2017\right)\)

\(x^4+2017x^2+2016x+2017\)

\(=\left(x^4-x\right)+\left(2007x^2+2007x+2007\right)\)

\(=x.\left(x^3-1\right)+2007.\left(x^2+x+1\right)\)

\(=x.\left(x-1\right)\left(x^2+x+1\right)+2007.\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2007\right)\)

a, x²-7x-14y+2x

=x(x+2)-7(x-2y)

b, x³-4x²y+4xy²-25x

=x³-4x²y+4xy²-y³-25x+y³

=(x-y)³-25x+y³

a ) = x(x+2) - 7(x+2y)

b) =  -4 xy ( x-y) + (x^3-25x)  [ câu này mk , chaqcs là làm đúng đâu ]

hằng đẳng thức số 7 vì 0,125(a+1)^3 - 1 = (0,5a+0,5)^3 - 1^3

x²-4+(x-2)²=(x+2)(x-2)+(x-2)(x-2)=(x-2)(x+2+x-2)=(x-2).2x