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![](https://rs.olm.vn/images/avt/0.png?1311)
c)x2-2xy+y2+3x-3y-10
=(x-y)2+3(x-y)-10
=(x-y)2+2(x-y).3/2+9/4-49/4
=(x-y+3/2)2-(7/2)2
=(x-y+3/2+7/2)(x-y+3/2-7/2)
=(x-y+5)(x-y-2)
a Đặt \(x^2\)=t[t\(\ge\)0}
6t^2-11t+3=6t^2-3t-9t+3=2t[3t-1] -3[3t-1]=[3t-1][2t-3]=[3x^2-1][2x^2-3]
b Đặt x^2+x=t[t\(\ge\)0]
t^2+3t+2=[t+1][t+2]
Đến đó Dương làm tương tự như câu a nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,x^3+2x^2y+xy^2-4x\)
\(x\left(x^2+2xy+y^2-4\right)\)
\(x\left[\left(x+y\right)^2-2^2\right]\)
\(x\left(x+y+2\right)\left(x+y-2\right)\)
\(2,5x-5y-x^2+2xy-y^2\)
\(5\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
\(5\left(x-y\right)-\left(x-y\right)^2\)
\(\left(x-y\right)\left(5-x+y\right)\)
\(3,x^4-3x^2\)
\(x^2\left(x^2-3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1
a, 2x2+4x+2-2y2 = 2(x2+2x+1-y2)= 2[(x+1)2-y2 ] = 2(x-y+1)(x+y+1)
b, 2x - 2y - x2 + 2xy - y2= 2(x -y) - (x2 - 2xy + y2) = 2(x-y)-(x-y)2=(x-y)(2-x+y)
c, x2-y2-2y-1=x2-(y2+2y+1)=x2-(y+1)2=(x-y-1)(x+y+1)
d, x2-4x-2xy-4y+y2= x2-2xy+y2-4x-4y=(x-y)
2.
a, x2-3x+2=x2-x-2x+2=x(x-1)-2(x-1)=(x-2)(x-1)
b, x2+5x+6=x2+2x+3x+6=x(x+2)+3(x+2)=(x+3)(x+2)
c, x2+6x-6=
Khai triển hằng đẳng thức :
\(x^2+2xy+y^2=\left(x+y\right)^2\)
\(\left(x+y\right)^2=\left(x+y\right)\left(x+y\right)\)