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a: =3(x^2-y^2-4x+4y)
=3[(x-y)(x+y)-4(x-y)]
=3(x-y)(x+y-4)
b: \(=4x\left(x^2+y^2+2xy-16\right)\)
\(=4x\left[\left(x+y\right)^2-16\right]\)
\(=4x\left(x+y+4\right)\left(x+y-4\right)\)
c: \(=\left(x+4\right)^2-y^2=\left(x+4+y\right)\left(x+4-y\right)\)
d: \(=\left(x^2-1\right)\left(x^2-9\right)=\left(x-1\right)\left(x-3\right)\left(x+1\right)\left(x+3\right)\)
1) \(\left(3x^2-3y^2\right)-\left(12x-12y\right)\)
\(=3xy\left(x-y\right)-12\left(x-y\right)\)
\(=\left(3xy-12\right)\left(x-y\right)\)
2) \(4x^3+4xy^2+8x^2y-16x\)
\(=\left(4x^3-16x\right)+\left(4xy^2+8x^2y\right)\)
\(=4x\left(x^2-4\right)+4xy\left(y+2x\right)\)
Ta có : 3x2 - 3y2 - 12x + 12y
= (3x2 - 3y2) - (12x - 12y)
= 3(x2 - y2) - 12(x - y)
= 3(x - y)(x + y) - 4.3.(x - y)
= 3(x - y)(x + y - 4)
a) 3x2 - 3y2 - 12x + 12x
= 3( x2 - y2- 4x + 4x )
= 3( x - y)( x + y)
b) 4x3 + 4xy2 + 8x2y - 16x
= 4x( x2 + y2 + 2xy - 4)
= 4x[( x + y)2 - 22]
= 4x( x + y - 2)( x + y +2)
c) x4 - 5x2 + 4
= ( x2)2 - 2.2x2 + 22 - x2
= ( x2 - 2)2 - x2
= ( x2 - 2 - x)( x2 - 2 + x)
\(3x^2y-12x^3y^2=3x^2y\left(1-4xy\right)\)
\(2x^2\left(x-y\right)-4xy\left(x-y\right)\)
\(=2x\left(x-y\right)\left(x-2y\right)\)
\(a,A+B-C=16x^4-8x^3y+7x^2y^2-9y^4-15x^4+3x^3y-5x^2y^2-6y^4-5x^3y-3x^2y^2-17y^4-1\)
\(=\left(16x^4-15x^4\right)+\left(-8x^3y+3x^3y-5x^3y\right)+\left(7x^2y^2-5x^2y^2-3x^2y^2\right)+\left(-9y^4-6y^4-17y^4\right)-1\)
\(=x^4-10x^3y-x^2y^2-32y^4-1\)
\(b,A-C+B=A+B-C\) ( giống câu a )
\(a,\)
\(A+B+C\)
\(=16x^4-8x^3y+7x^2y^2-9y^4-15x^4+3x^3y-5x^2y^2-6y^4-\left(5x^3y+3x^2y^2+17y^4+1\right)\)
\(=16x^4-8x^3y+7x^2y^2-9y^4-15x^4+3x^3y-5x^2y^2-6y^4-5x^3y-3x^2y^2-17y^4-1\)
\(=\left(16x^4-15x^4\right)+\left(-9y^4-6y^4-17y^4\right)+\left(-8x^3y+3x^3y-5x^3y\right)+\left(7x^2y^2-5x^2y^2-3x^2y^2\right)-1\)
\(=x^4-32y^4-10x^3y-x^2y^2-1\)
\(b,\)
\(A-C+B=A+B-C=x^4-32y^4-10x^3y-x^2y^2-1\)
\(C=4x^2-4xy+y^2+4x^2-16x+16+1\)
\(=\left(2x-y\right)^2+(2x-4)^2+1\ge1\forall x;y\in R\)
Dấu "=" xảy ra<=> 2x-y=0 và 2x-4=0
<=>2x-y=0 và x=2 <=>y=4 và x=
Vậy....
\(B=3x^2-12x+16\)
\(=x^2-12x+36+2x^2-20\)
\(=\left(x-6\right)^2+2x^2-20\ge-20\forall x\in R\)
Dấu "=" xảy ra <=> \(\left(x-6\right)^2=0\)và \(2x^2=0\)
<=>x1 =6 và x2 =0
Vậy....
a: =>A-B=3x^2y-4xy^2+x^2y-2xy^2=4x^2y-6xy^2
b: =>B-A=-7xy^2+8x^2y-5xy^2+6x^2y=-12xy^2+14x^2y
=>A-B=12xy^2-14x^2y
c: =>B-A=8x^2y^3-4x^3y-3x^2y^3+5x^3y^2=5x^2y^3+x^3y^2
=>A-B=-5x^2y^3-x^3y^2
d: =>A-B=2x^2y^3-7x^3y+6x^2y^3+3x^3y^2=8x^2y^3-7x^3y+3x^3y^2
\(=\dfrac{4x\left(3x^2+4xy-2\right)}{3x^2+4xy-2}=4x\)