a) \(P=3\left(x^2+2xy+y^2\right)-2\left(x+y\right)-100\)

\(P=3\left(x+y\right)^2-2.5-100\)

\(P=3.5^2-110\)

\(P=-35\)

b) \(Q=\left[x^3+y^3+3xy\left(x+y\right)\right]-2\left(x^2+2xy+y^2\right)+3.5+10\)

\(Q=\left(x+y\right)^3-2\left(x+y\right)^2+25\)

\(Q=5^3-2.5^2+25\)

\(Q=100\)

a) \(3x^4y-12x^2y^3=3x^2y\left(x^2-\left(2y\right)^2\right)=3x^2y\left(x+2y\right)\left(x-2y\right)\)

b) Sửa đề: \(x^2-y^2-8x+16=\left(x-4\right)^2-y^2=\left(x-4-y\right)\left(x-4+y\right)\)

c) \(x^3+3x^2+4x+12=x^2\left(x+3\right)+4\left(x+3\right)=\left(x^2+4\right)\left(x+3\right)\)

d) \(3x^2-6xy+3y^2-27=3\left(x^2-2xy+y^2-9\right)=3\left(\left(x-y^2\right)-3^2\right)=3\left(x-y-3\right)\left(x-y+3\right)\) 

\(3x^4y-12x^2y^3=3x^2y\left(x^2-4y^2\right)=3x^2y\left(x-2y\right)\left(x+2y\right)\)

\(x^2-y^2-8y-16=x^2-\left(y^2+8y+16\right)=x^2-\left(y+4\right)^2=\left(x+y+4\right)\left(x-y-4\right)\)

\(x^3+3x^2+4x+12=x^2\left(x+3\right)+4\left(x+3\right)=\left(x^2+4\right)\left(x+3\right)\)

\(3x^2-6xy+3y^2-27=3\left[\left(x-y\right)^2-9\right]=3\left(x-y-3\right)\left(x-y+3\right)\)

k: \(=\left(2-3x\right)\left(4+6x+9x^2\right)\)

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

\(g,27+27x+9x^2+x^3=\left(3+x\right)^3\\ i,2x^2+2y^2-x^2z+z-y^2z-2=\left(2x^2-x^2z\right)+\left(2y^2-y^2z\right)-\left(2-z\right)=x^2\left(2-z\right)+y^2\left(2-z\right)-\left(2-z\right)=\left(x^2+y^2-1\right)\left(2-z\right)\)

\(k,8-27x^2=2^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)

\(l,3x^2-6xy+3y^2=3\left(x^2-2xy+y^2\right)=3\left(x-y\right)^2\)

`@` `\text {Ans}`

`\downarrow`

`a,`

`3x^2 + 6xy + 3y^2 - 3z`

`= 3*x^2 + 3*2xy + 3y^2 - 3z`

`= 3(x^2 + 2xy + y^2 - z)`

`b,`

`x^3 + x^2y - x^2z - xyz`

`= x(x + y)(x-z)`

a) \(3x^2-6xy+3y^2-12x^2=3\left(x^2-2xy+y^2\right)-12x^2=3\left(x-y\right)^2-12x^2=3\left[\left(x-y\right)^2-4x^2\right]=3\left(x-y-2x\right)\left(x-y+2x\right)=3\left(-x-y\right)\left(3x-y\right)\)

b)\(3x^2y^2-6x^2y^3+12x^2y^2=3x^2y^2\left(1-2y+4\right)=3x^2y^2\left(5-2y\right)\)

c) \(3x^2-3y^2+12x-12y=3\left(x^2-y^2\right)+12\left(x-y\right)=3\left(x-y\right)\left(x+y+4\right)\)

a: \(3x^2-6xy+3y^2-12x^2\)

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

\(=3\left[\left(x-y\right)^2-4x^2\right]\)

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

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

b: \(3x^2y^2-6x^2y^3+12x^2y^2\)

\(=3x^2y^2\left(1-2y+4\right)\)

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

c: Ta có: \(3x^2-3y^2+12x-12y\)

\(=3\left(x-y\right)\left(x+y\right)+12\left(x-y\right)\)

\(=3\left(x-y\right)\left(x+y+4\right)\)

\(a,2xy\left(x^2+xy-3y^2\right)=2x^3y+2x^2y^2-6xy^3\)

\(b,\left(x+2\right)\left(3x^2-4x\right)=3x^3-4x^2+6x^2-8x=3x^3+2x^2-8x\)

\(c,\left(3+3x^2-8x-20\right):\left(x+2\right)=3x-14\left(dư:11\right)\)

3x2 + 6xy + 3y2 – 3z2

= 3.(x2 + 2xy + y2 – z2)

(Nhận thấy xuất hiện x2 + 2xy + y2 là hằng đẳng thức nên ta nhóm với nhau)

= 3[(x2 + 2xy + y2) – z2]

= 3[(x + y)2 – z2]

= 3(x + y – z)(x + y + z)

\(3x^2+6xy+3y^2=3\cdot\left(x^2+2xy+y^2\right)=3\cdot\left(x+y\right)^2\)

3x² + 6xy + 3y² 

=3.(x²+2xy+y²)

=3.(x+y)²

3x^2 - 6xy + 3y^2
=3.(x^2 - 2xy + y^2)
=3.(x - y)^2