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a) x2 - y2 + 4x + 4
= ( x2 + 4x + 4 ) - y2
= ( x + 2 )2 - y2
= ( x + 2 - y )( x + 2 + y )
b) x2 - 2xy + y2 - 1
= ( x2 - 2xy + y2 ) - 1
= ( x - y )2 - 12
= ( x - y - 1 )( x - y + 1 )
c) x2 - 2xy + y2 - 4
= ( x2 - 2xy + y2 ) - 4
= ( x - y )2 - 22
= ( x - y - 2 )( x - y + 2 )
d) x2 - 2xy + y2 - z2
= ( x2 - 2xy + y2 ) - z2
= ( x - y )2 - z2
= ( x - y - z )( x - y + z )
e) 25 - x2 + 4xy - 4y2
= 25 - ( x2 - 4xy + 4y2 )
= 52 - ( x - 2y )2
= ( 5 - x + 2y )( 5 + x - 2y )
f) x2 + y2 - 2xy - 4z2
= ( x2 - 2xy + y2 ) - 4z2
= ( x - y )2 - ( 2z )2
= ( x - y - 2z )( x - y + 2z )
a) \(2x^2y^2-\frac{4}{3}x^2y+2xy\)
\(=xy\left(2xy-\frac{4}{3}x+2\right)\)
b) 2xy2.(x + 5y) - 4xy(5y + x)
= (5y + x)(2xy2 - 4xy)
= 2xy(5y + x)(y - 2)
c) 25 - 4x2 - y2 + 4xy
= 25 - (4x2 - 4xy + y2)
= 52 - (2x + y)2
= (5 - 2x - y)(5 + 2x + y)
d) x2 + 4x - 2xy - 4y +y2
= (x2 - 2xy + y2) + (4x - 4y)
= (x - y)2 + 4(x - y)
= (x - y)(x - y + 4)
e) 12y3 - 3x2y + 12xy - 12y
= 3y(4y2 - x2 + 4x - 4)
= 3y[4y2 - (x - 2)2]
= 3y(2y - x + 2)(2y + x - 2)
f) 64x4 + y4
= (8x2)2 + 16x2y2 + y4 - 16x2y2
= (8x2 + y2)2 - (4xy)2
= (8x2 + y2 - 4xy)(8x2 + y2 + 4xy)
a) \(2x^2y^2-\frac{4}{3}x^2y+2xy\)
b) \(2xy^2\left(x+5y\right)-4xy\left(5y+x\right)\)
\(=\left(x+5y\right)\left(2xy^2-4xy\right)\)
\(=2\left(x+5y\right)\left(xy^2-2xy\right)\)
c) \(25-4x^2-y^2+4xy\)
\(=25-\left(4x^2+y^2-4xy\right)\)
\(=5^2-\left[\left(2x\right)^2-2.2x.y+y^2\right]\)
\(=5^2-\left(2x-y\right)^2\)
\(=\left(5-2x+y\right)\left(5+2x-y\right)\)
d) \(x^2+4x-2xy-4y+y^2\)
\(=\left(x^2-2xy+y^2\right)+\left(4x-4y\right)\)
\(=\left(x-y\right)^2+4\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y\right)+4\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y+4\right)\)
e) \(12y^3-3x^2y+12xy-12y\)
f) \(64x^4+y^4\)
\(=\left(8x^2\right)^2+16x^2y^2+\left(y^2\right)^2-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)
\(=\left(8x^2+y^2+4xy\right)\left(8x^2+y^2-4xy\right)\)
a) x2 - 2xy - 4 + y2
= (x - y)2 - 22
= (x - y - 2)(x - y + 2)
b) x2 + y2 - 1 - 2xy
= (x - y)2 - 12
= (x - y - 1)(x - y + 1)
c) 25 - x2 + 4xy - 4y2
= 52 - (x - 2y)2
= (5 - x + 2y)(5 + x - 2y)
\(\cdot x^2-4x-5=x^2-5x+x-5=\left(x^2-5x\right)+\left(x-5\right)=x\left(x-5\right)+\left(x-5\right)=\left(x-5\right)\left(x+1\right)\)
a/ \(\left(x^2+y^2\right)^2-4x^2y^2=\left(x^2+y^2\right)^2-\left(2xy\right)^2=\left(x^2+y^2-2xy\right)\left(x^2+y^2+2xy\right)=\left(x+y\right)^2\left(x-y\right)^2\)
b/ \(x^2y^2-\left(14y\right)^2=\left(xy-14y\right)\left(xy+14y\right)\)
c/ \(25-x^2+2xy-y^2=25-\left(x^2-2xy+y^2\right)=25-\left(x-y\right)^2=\left(5-x+y\right)\left(5+x-y\right)\)
d/ \(9x^2-x^2+6x=8x^2+6x=2x\left(4x+3\right)\)
x^2 - 25 + y^2 + 2xy
= (x^2 + 2xy + y^2) - 5^2
= ( x + y )^2 - 5^2
= ( x + y - 5 ).( x + y + 5 )
Câu 1:
$(2x^2-3)(x+5)=2x^2(x+5)-3(x+5)=2x^3+10x^2-3x-15$
Câu 2:
a.
$(x+3)^2=x^2+2.x.3+3^2=x^2+6x+9$
b.
$y^2-25=y^2-25$
\(25-x^2-y^2-2xy=25-\left(x^2+2xy+y^2\right)=25-\left(x+y\right)^2=\left(5-x-y\right)\left(5+x+y\right)\)
⇔25-(x2+2xy+y2)
⇔25-(x+y)2
⇔(5-x-y)(5+x+y)