Bài làm:

Ta có: \(-6x+5\sqrt{x}+1\)

\(=\left(-6x+6\sqrt{x}\right)-\left(\sqrt{x}-1\right)\)

\(=-6\sqrt{x}\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)\)

\(=\left(-6\sqrt{x}-1\right)\left(\sqrt{x}-1\right)\)

\(=\left(6\sqrt{x}+1\right)\left(1-\sqrt{x}\right)\)

a) xy+3x-7y-21

=x(y+3)-7(x+3)

=(x-7)(y+3)

b)2xy-15-6x-5y

=2x(y-3)-5(-3+y)

=(2x-5)(y-3)

c)2x^2y+2xy^2-2x-2y

=2x(xy-1)+2y(xy-1)

=(2x+2y)(xy-1)

x(x+3)-5x(x-5)-5(x+3)

=(x-5)(x+3)-5x(x-5)

=(x-5)(x+3-5x)

Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn

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

\(2x+5\sqrt{x}-3=\left(\sqrt{x}+3\right)\left(2\sqrt{x}-1\right)\)

đặt y=x²+1

=>y²=(x²+1)²

y²=x⁴+2x²+1

đặt P(x)=x^4+6x^3+11x^2+6x+1

=x⁴+2x²+1+6x³+6x+9x²

=x⁴+2x+1+6x(x²+1)+9x²

thay y²=x⁴+2x²+1 và y=x²+1 ta được 

Q(y)=y²+6xy+9x²

=(y+3x)²

thay y=x²+1 ta được:

(x²+3x+1)²

vậy x^4+6x^3+11x^2+6x+1=(x²+3x+1)²

= (1 - x³ ) + ( 6x - 6x² ) 

= (1 - x ).(1 + x + x²) + 6x.(1 - x)

= (1 - x).(1+x+x² + 6x)

= (1 - x).(1 + 7x +x² )

2x^5-6x^4-2a^2x^3-6ax^3

 =(2x^5-2a^2x^3)-(6x^4+6ax^3)

 =2x^3(x^2-a^2)-6x^3(x+a)

 =2x^3(x-a)(x+a)-6x^3(x+a)

 =(x+a)(2x^4-2x^3a-6x^3)

 =(x+a) 2x^3 (x-a-3)

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

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

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

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

x(2x^2-x-6)

x(2x^2-4x+3x-6)

x[2x(x-2)+3(x-2)]

x(2x+3)(x-2)