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

a) xy+3x-7y-21=(xy+3x)-(7y+21)= x(y+3)-7(y+3)=(y+3)(x-7)

b)2xy-15-6x+5y=(2xy-6x)+(5y-15)=2x(y-3)+5(y-3)=(y-3)(2x+5)

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

d) x(x+3)-5x(x-5)-5(x+3)=[x(x+3)-5(x+3)]-5x(x-5)=(x+3)(x-5)-5x(x-5)=(x-5)(x+3-5x)=(x-5)(3-4x)

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

\(=x^3-2x^2+4x-x^2+2x-4\)

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

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

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

x^3 - 3x^2 + 6x - 4 

<=> x^3-3x^2+3x-1+3x-3

<=>(x-1)^3+3(x-1)

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

<=>(x-1)+(x^2-2x+4)

\(\Leftrightarrow x^3-3x^2-3x^2+9x-10x+30\)

\(\Leftrightarrow x^2\left(x-3\right)-3x\left(x-3\right)-10\left(x-3\right)\)

\(\Leftrightarrow\left(x-3\right)\left(x^2-3x-10\right)\)

\(\Leftrightarrow\left(x-3\right)\left(x-5\right)\left(x+2\right)\)

\(=\left(x^3-2x^2\right)+\left(8x^2-16x\right)+\left(15x-30\right)\)

\(=x^2\left(x-2\right)+8x\left(x-2\right)+15\left(x-2\right)\)

\(=\left(x-2\right)\left(x^2+8x+15\right)\)

\(=\left(x-2\right)\left(x^2+3x+5x+15\right)\)

\(=\left(x-2\right)\left[x\left(x+3\right)+5\left(x+3\right)\right]\)

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

a) x³-2x²-x+2

=x(x²-1)+2(-x²+1)

=x(x²-1)-2(x²-1)

=(x²-1)(x-2)

b)

x²+6x-y²+9

=x²+6x+9-y²

=(x+3)²-y²

^{=(x+3-y)(x+3+y)}

a) \(=3\left(5y+4x\right)\)

b) \(=\left(x-3\right)^2\)

c) \(=y\left(y^2+2y+3\right)\)

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

x³-6x²-x+30

=x³-5x²-x²+5x-6x+30

=(x-5)(x²-x-6)

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

\(x^3-6x^2-x+30=x^3-3x^2-3x^2+9x-10x+30.\)

\(=x^2\left(x-3\right)-3x\left(x-3\right)-10\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2-3x-10\right)\)

\(=\left(x-3\right)\left(x^2+2x-5x-10\right)\)

\(=\left(x-3\right)\left[x\left(x+2\right)-5\left(x+2\right)\right]\)

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

Vậy \(x^3-6x^2-x+30=\left(x-3\right)\left(x+2\right)\left(x-5\right)\) 

cái này bạn hỏi rồi mà mk đã tl rồi đó bạn ấn vô câu hỏi của tôi rồi kéo xuống mà xem

1.

= (x^3 + 125 ) -(x^2 +5x)

=(x +5) (x^2 -5x +25) -x(x+5)

=(x+5)(x^2 -5x +25 -x)

=(x+5)(x^2 -6x +25)

2.

= (x^3 -27) + (2x^2 -6x)

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

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

=(x-3) (x^2 +5x +9)

^{\(^{x^3-6x^2-x+30=x^3-5x^2-3x^2+15x-2x^2-10x-6x+30}\)}

=x^2(x-5)-3x(x-5)-2x(x-5)-6(x-5)

=(x-5)(x^2-3x-2x-6)

=(x-5)[x(x-3)-2(x-3)]

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

\(x^3-6x^2-x+30\) 

= \(x^3-5x^2-3x^2+15x+2x^2-10x-6x+30\)

= \(x^2\left(x-5\right)-3x\left(x-5\right)+2x\left(x-5\right)-6\left(x-5\right)\)

= \(\left(x-5\right)\left(x^2-3x+2x-6\right)\)

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

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