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

b: \(5x^2+5xy-x-y\)

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

\(=\left(x+y\right)\left(5x-1\right)\)

e) Ta có: \(a^3-a^2-a+1\)

\(=a^2\left(a-1\right)-\left(a-1\right)\)

\(=\left(a-1\right)\left(a^2-1\right)\)

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

f) Ta có: \(x^3-2xy-x^2y+2y^2\)

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

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

a) \(\left(a^2+b^2\right)^2-4a^2b^2=\left(a^2+b^2+2ab\right)\left(a^2+b^2-2ab\right)=\left(a+b\right)^2.\left(a-b\right)^2\)

b) \(3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)

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

d) Đề sai ko ???

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

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

a, \(=\left(a^2+b^2-2ab\right)\left(a^2+b^2+2ab\right)=\left(\left(a-b\right)\left(a+b\right)\right)^2=\left(a^2-b^2\right)^2\)

\(b,=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)

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

\(d,=2\left(x^2+2xy+y^2-4z^2\right)=2\left(\left(x+y\right)^2-4z^2\right)=2\left(x+y-2z\right)\left(x+y+2z\right)\)

\(e,=a^2\left(a-1\right)-\left(a-1\right)=\left(a-1\right)\left(a^2-1\right)\)

\(f,=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x^2-2y\right)\left(x-y\right)\)

Câu 1: A

Câu 21: A

\(16,A\\ 17,C\\ 18,A\\ 19,C\\ 20,A\\ 21,A\)

1a

a) \(3x^2-3xy-5x+5y\)

\(=\left(3x^2-3xy\right)-\left(5x-5y\right)\)

\(=3x\left(x-y\right)-5\left(x-y\right)\)

\(=\left(x-y\right)\left(3x-5\right)\)

b) \(2x^3y-2xy^3-4xy^2-2xy\)

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

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

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

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

c) \(x^2+1+2x-y^2\)

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

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

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

d) \(x^2+4x-2xy-4y+y^2\)

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

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

\(=\left(x-y\right)\left(x-y+4\right)\)

e) \(x^3-2x^2+x\)

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

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

f) \(2x^2+4x+2-2y^2\)

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

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

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

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

a: =3x(x-y)-5(x-y)

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

b: \(=2xy\left(x^2-y^2-2y-1\right)\)

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

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

d:

Sửa đề: x^2+4x-2xy-4y+y^2

=x^2-2xy+y^2+4x-4y

=(x-y)^2+4(x-y)

=(x-y)(x-y+4)

e: =x(x^2-2x+1)

=x(x-1)^2

f: =2(x^2+2x+1-y^2)

=2[(x+1)^2-y^2]

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

a) x⁴ + 2x³ + x²

= x² ( x² + 2x + 1 )

= x² ( x + 1 )² 

b) 5x² - 10xy + 5y² - 20z² 

= 5 [(x² - 2xy + y² ) - 4z² ] 

= 5 [( x - y )² - ( 2z )² ]

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

 c) x³ - x + 3x²y + 3xy²+ y³- y

= ( x³ + 3x²y + 3xy² + y³ ) - ( x + y )

= (x + y )³ - ( x + y)

= ( x + y ) [( x + y )² - 1 ]

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

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

=x^2(x+1)^2

b: =x^3+3x^2y+3xy^2+y^3-x-y

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

=(x+y)[(x+y)^2-1]

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

c: =5(x^2-2xy+y^2-4z^2)

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

Bài 2

a) 5x² + 30y

= 5(x² + 6y)

b) x³ - 2x² - 4xy² + x

= x(x² - 2x - 4y² + 1)

= x[(x² - 2x + 1) - 4y²]

= x[(x - 1)² - (2y)²]

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

Bài 3:

a: \(2x\left(x-3\right)-x+3=0\)

=>\(2x\left(x-3\right)-\left(x-3\right)=0\)

=>(x-3)(2x-1)=0

=>\(\left[{}\begin{matrix}x-3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)

b: \(\left(3x-1\right)\left(2x+1\right)-\left(x+1\right)^2=5x^2\)

=>\(6x^2+3x-2x-1-x^2-2x-1=5x^2\)

=>\(5x^2-x-2=5x^2\)

=>-x-2=0

=>-x=2

=>x=-2

Nhiều thế                  

mình làm bài 1 nhé 

kq sau khi thu gọn là A =x^6 -7+3x

Trời ơi!Làm sao bây giờ,tốt nhất là nên lờ câu hỏi của bạn ấy đi các bạn ạ!