Bài 5:
a) \(3x^2-6xy+3y^2=3\left(x^2-2xy+y^2\right)=3\left(x-y\right)^2\)
b) \(4x^2-6y+8xy-3x=\left(4x^2+8xy\right)-\left(3x+6y\right)=4x\left(x+2y\right)-3\left(x+2y\right)=\left(4x-3\right)\left(x+2y\right)\)
c) \(16x^3+54y^3=2\left(8x^3+27y^3\right)=2\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
d) \(2xy+\dfrac{1}{4}-x^2-y^2=\dfrac{1}{4}-\left(x^2-2xy+y^2\right)=\dfrac{1}{4}-\left(x-y\right)^2=\left(\dfrac{1}{2}-x+y\right)\left(\dfrac{1}{2}+x-y\right)\)
e) \(x^3-y^3-3x^2+3x-1=\left(x-y\right)^3-1=\left(x-y-1\right)\left[\left(x-y\right)^2+x-y+1\right]=\left(x-y-1\right)\left(x^2-2xy+y^2+x-y+1\right)\)
\(1,\\ a,=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\\ b,=x^4-2x^2+2x^3-4x+2x^2-4=\left(x^2-2\right)\left(x^2+2x+2\right)\\ c,=x^2\left(x+2y\right)-\left(x+2y\right)=\left(x-1\right)\left(x+1\right)\left(x+2y\right)\\ d,=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\\ =\left(x-y\right)\left(3x+3y-2x+2y\right)=\left(x-y\right)\left(x+5y\right)\\ e,=x^2\left(x-4\right)-9\left(x-4\right)=\left(x-3\right)\left(x+3\right)\left(x-4\right)\\ f,=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)=\left(x+y\right)\left(x-y-2\right)\)
\(2,\\ a,=\left(x-3\right)\left(x-4\right)\\ b,=\left(x-1\right)\left(2x+1\right)\left(1+3x+6\right)=\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\\ c,=\left(2x+1\right)\left(3-2x+5\right)=2\left(2x+1\right)\left(4-x\right)\\ d,=\left(x-5\right)\left(x-5+x+5+2x+1\right)=\left(x-5\right)\left(4x+1\right)\\ e,=\left(3x-1\right)\left(4x-3+x-1-2x-2\right)=3\left(3x-1\right)\left(x-2\right)\)
Bài 4:
a) \(x^3-1+5x^2-5+3x-3=\left(x-1\right)\left(x^2+x+1\right)+5\left(x-1\right)\left(x+1\right)+3\left(x-1\right)=\left(x-1\right)\left(x^2+x+1+5x+5+3\right)=\left(x-1\right)\left(x^2+6x+9\right)=\left(x-1\right)\left(x+3\right)^2\)
b) \(a^5+a^4+a^3+a^2+a+1=a^4\left(a+1\right)+a^2\left(a+1\right)+\left(a+1\right)=\left(a^4+a^2+1\right)\left(a+1\right)=\left[\left(a^4-a^3+a^2\right)+\left(a^3-a^2+a\right)+\left(a^2-a+1\right)\right]\left(a+1\right)=\left[a^2\left(a^2-a+1\right)+a\left(a^2-a+1\right)+\left(a^2-a+1\right)\right]\left(a+1\right)=\left(a+1\right)\left(a^2+a+1\right)\left(a^2-a+1\right)\)
c) \(x^3-3x^2+3x-1-y^3=\left(x-1\right)^3-y^3=\left(x-1-y\right)\left[\left(x-1\right)^2+\left(x-1\right).y+y^2\right]=\left(x-1-y\right)\left(x^2-2x+1+xy-y+y^2\right)\)
giúp mik với mn ơi 
