\(x-y-\sqrt{x}-\sqrt{y}\\ =\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-\left(\sqrt{x}+\sqrt{y}\right)\\ =\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}-1\right)\)

a: x^3-7x-6

=x^3-x-6x-6

=x(x-1)(x+1)-6(x+1)

=(x+1)(x^2-x-6)

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

b: =2x^3+x^2-2x^2-x+6x+3

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

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

c: =2x^3-3x^2-2x^2+3x+2x-3

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

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

d: =2x^3+x^2+2x^2+x+2x+1

=(2x+1)(x^2+x+1)

e: =3x^3+x^2-3x^2-x+6x+2

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

f: =27x^3-9x^2-18x^2+6x+12x-4

=(3x-1)(9x^2-6x+4)

a) \(x^3-7x-6\)

\(=x^3-x-6x-6\)

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

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

\(=x\left(x+1\right)\left(x-1\right)-6\left(x+1\right)\)

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

b) \(2x^3-x^2+5x+3\)

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

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

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

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

c) \(2x^3-5x^2+5x+1\)

\(=2x^3-3x^2-2x^2+3x+2x-3\)

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

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

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

d) \(2x^3+3x^2+3x+1\)

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

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

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

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

e) \(3x^3-2x^2+5x+2\)

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

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

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

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

f) \(27x^3-27x^2+18x-4\)

\(=27x^3-9x^2-18x^2+6x+12x-4\)

\(=\left(27x^3-9x^2\right)-\left(18x^2-6x\right)+\left(12x-4\right)\)

\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)

\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)

\(3x\cdot\left(x-y\right)^2-6\cdot\left(y-x\right)\)

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

\(=\left(x-y\right)\left[3x\left(x-y\right)+6\right]\)

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

(x - y)(3x² - 3xy + 6)

Đặt \(x^2+3x+1=t\)

\(\Rightarrow\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6=t.\left(t+1\right)-6\)

\(=t^2+t-6=\left(t^2-2t\right)+\left(3t-6\right)\)

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

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

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

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

Đặt \(x^2+3x+1=a\)ta có :

\(a\left(a+1\right)-6\)

\(=a^2+a-6\)

\(=a^2+6a-a-6\)

\(=\left(a^2+6a\right)-\left(a+6\right)\)

\(=a\left(a+6\right)-\left(a+6\right)\)

\(=\left(a+6\right)\left(a-1\right)\)

Thay \(a=x^2+3x+1\)vào A ta có :

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

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

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

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

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

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

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

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

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

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

Đặt \(\left(x^2+3x+1\right)=a\), ta được:

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

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

Thay \(a=\left(x^2+3x+1\right)\), ta được:

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

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

Cách 1: Tách một hạng tử thành tổng hai hạng tử để xuất hiện nhân tử chung.

a) x2 – 3x + 2

= x2 – x – 2x + 2 (Tách –3x = – x – 2x)

= (x2 – x) – (2x – 2)

= x(x – 1) – 2(x – 1) (Có x – 1 là nhân tử chung)

= (x – 1)(x – 2)

Hoặc: x2 – 3x + 2

= x2 – 3x – 4 + 6 (Tách 2 = – 4 + 6)

= x2 – 4 – 3x + 6

= (x2 – 22) – 3(x – 2)

= (x – 2)(x + 2) – 3.(x – 2) (Xuất hiện nhân tử chung x – 2)

= (x – 2)(x + 2 – 3) = (x – 2)(x – 1)

b) x2 + x – 6

= x2 + 3x – 2x – 6 (Tách x = 3x – 2x)

= x(x + 3) – 2(x + 3) (có x + 3 là nhân tử chung)

= (x + 3)(x – 2)

c) x2 + 5x + 6 (Tách 5x = 2x + 3x)

= x2 + 2x + 3x + 6

= x(x + 2) + 3(x + 2) (Có x + 2 là nhân tử chung)

= (x + 2)(x + 3)

Cách 2: Đưa về hằng đẳng thức (1) hoặc (2)

a) x2 – 3x + 2

(Vì có x2 và  nên ta thêm bớt  để xuất hiện HĐT)

= (x – 2)(x – 1)

b) x2 + x - 6

= (x – 2)(x + 3).

c) x2 + 5x + 6

= (x + 2)(x + 3).

\(x+5\sqrt{x}+6\)

\(=x+2\sqrt{x}+3\sqrt{x}+6\)

\(=\sqrt{x}\left(\sqrt{x}+2\right)+3\left(\sqrt{x}+2\right)\)

\(=\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\)

Đề sai thì phải nhé , phải là \(x\) chứ không phải \(x^2\)