\(xz-yz-x^2+2xy-y^2\)

\(=z\left(x-y\right)-\left(x-y\right)^2\)

\(=\left(x-y\right)\left(z-x+y\right)\)

m)xz−yz−x2+2xy−y2

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

= \(z.\left(x-y\right)-\left(x-y\right)^2\)

= (x-y).(z-x+y)

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

2) \(x^2+6x+9+x^2y+3xy\)

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

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

\(x^n+x^{n+3}-x^n=x^{n+3}=x^n\cdot x^3\)

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

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

\(=\left(x+2\right)\left[3\left(x-2\right)-1\right]=\left(x+2\right)\left(3x-7\right)\)

\(x^3+y^3-2(x^2-y^2)\\=(x+y)(x^2-xy+y^2)-2(x-y)(x+y)\\=(x+y)[x^2-xy+y^2-2(x-y)]\\=(x+y)(x^2-xy+y^2-2x+2y)\\=(x+y)(-x^2-xy+2y+y^2)\)

\(\left(x+3\right)\left(x^2-4x+5\right)\)

x^3-x^2-7x+15=0

<=> x^3+3x^2-4x^2-12x+5x+15=0

<=> x^2(x+3)-4x(x+3)+5(x+3)=0

<=> (x+3)(x^2-4x+5)=0

<=> x+3=0 vì x^2-4x+5 khác 0

<=> x=-3

(x+3)(x²-4x+5)

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-8\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)\(=\left(x^2+7x+11-1\right)\left(x^2+7x+11+1\right)-8\)

\(=\left(x^2+7x+11\right)^2-9\)

\(=\left(x^2+7x+11-3\right)\left(x^2+7x+11+3\right)=\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)

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

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

=(x^2+7x+10)(x^2+7x+12)-24

thay x^2+7x+11=y

=> (y-1)(y+1)-24=y^2-1^2-24=y^2-25=(y-5)(y+5)

= (x^2+7x+11-5)(x^2+7x+11+5)=(x^2+7x+6)(x^2+7x+16)=(x^2+x+6x+6)(x^2+7x+16)=[x(x+1)+6(x+1)]((x^2+7x+16)=(x+1)(x+6)(x^2+7x+16)

(x + 2)(x + 3)(x + 5)(x + 7) - 24

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

=(x² + 7x + 10)(x² + 7x +12) - 24

Đặt x² + 7x + 11 = t ; ta có:

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

= t² - 1² - 24

= t² - 25

= (t - 5)(t + 5)

Thay t = x² + 7x + 11 ta được:

(x² + 7x + 11 - 5)(x² + 7x +11 + 5)

= (x² + 7x + 6)(x² + 7x + 16)

= (x + 1)(x + 6)(x² + 7x + 16)

Chúc bn học tốt

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+10+2\right)-24\)( * )

Đặt \(t=x^2+7x+10\), khi đó (*) trở thành:

\(t\left(t+2\right)-24\)

\(=t^2+2t-24\)

\(=\left(t-4\right)\left(t+6\right)\)

Thay \(t=x^2+7x+10\) vào, ta được:

\(\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

Vậy ...