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\(3x^2+6xy+3y^2-3z^2.\)
\(=3.\left(x^2+2xy+y^2-z^2\right)\)
\(=3.\left[\left(x+y\right)^2-z^2\right]\)
\(=3.\left(x+y-z\right).\left(x+y+z\right)\)
32 + 6xy + 3y2 + 3z2
= 3 ( x2 + 2xy + y2 - z2 )
= 3 [ ( x + y )2 - z2 ] = 3
( x + y - z ) ( x + y + z )
a) => 3(x2 + 2xy +y2 - z2 )
=> 3[(x+y)2 - z2]
=> 3(x+y-z)(x+y+z)
b) => (x+y)(x2 - xy +y2) -3(x+y)
=> (x+y)(x2 - xy + y2 -3 )
T I C K cho mình nha camr ơn
_____ CHÚC BẠN HỌC TỐT _________
\(A=3x^2+6xy+3y^2-3z^2\)
\(=3\left(x^2+2xy+y^2-z^2\right)\)
\(=3\left[\left(x^2+2xy+y^2\right)-z^2\right]\)
\(=3\left[\left(x+y\right)^2-z^2\right]\)
\(=3\left(x+y+z\right)\left(x+y-z\right)\)
a.\(xz+yz-5\left(x+y\right)\)
\(=z\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x+y\right)\left(z-5\right)\)
b.\(3x^2-3xy-5x+5y\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
c.\(x^2+6x-y^2-3z^2\)???Sai đề bài ...?
d.\(3x^2+6xy+3y^2-3z^2\)
\(=3\left(x^2+2xy+y^2-z^2\right)\)
\(=3\left[\left(x+y\right)^2-z^2\right]\)'
\(=3\left(x+y-z\right)\left(x+y+z\right)\)
Trả lời:
a, xz + yz - 5 ( x + y )
= ( xz + yz ) - 5 ( x + y )
= z ( x + y ) - 5 ( x + y )
= ( x + y ) ( z - 5 )
b, 3x2 - 3xy - 5x + 5y
= ( 3x2 - 3xy ) - ( 5x - 5y )
= 3x ( x - y ) - 5 ( x - y )
= ( x - y ) ( 3x - 5 )
c, x2 + 6x - y2 - 3z2
= - ( 3x2 - x2 + y2 - 6x )
d, 3x2 + 6xy + 3y2 - 3z2
= 3 ( x2 + 2xy + y2 - x2 )
= 3 [ ( x2 + 2xy + y2 ) - z2 ]
= 3 [ ( x + y )2 - z2 ]
= 3 ( x + y - z ) ( x + y + z )
a) Ta có: \(3x^2+5y-3xy-5x\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
b) Ta có: \(3y^2-3z^2+3x^2+6xy\)
\(=3\left(y^2-z^2+x^2+2xy\right)\)
\(=3\left[\left(x+y\right)^2-z^2\right]\)
\(=3\left(x+y-z\right)\left(x+y+z\right)\)
c) Ta có: \(x^2-25-2xy+y^2\)
\(=\left(x-y\right)^2-5^2\)
\(=\left(x-y-5\right)\left(x-y+5\right)\)
d) Ta có: \(5x^2-10xy+5y^2-20z^2\)
\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
e) Ta có: \(x^2-5x+5y-y^2\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
f) Ta có: \(3x^2-6xy+3y^2-12z^2\)
\(=3\left(x^2-2xy+y^2-4z^2\right)\)
\(=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)
c) \(3x^2-6x+9y^2=3\left(x^2-2x+3y^2\right)\)
h) \(3y^2-3z^2+3x^2+6xy=3\left(x^2-z^2+x^2+2xy\right)\)
\(=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y+z\right)\left(x+y-z\right)\)
g) \(3x^2+5y-3xy-5x=\left(3x^2-3xy\right)+\left(5y-5x\right)=3x\left(x-y\right)+5\left(y-x\right)=\left(x-y\right)\left(3x-5\right)\)
(((3•(x2))+6xy)+(3•(y2)))-3z2
(((3 • (x2)) + 6xy) + 3y2) - 3z2
((3x2 + 6xy) + 3y2) - 3z2
3x2 + 6xy + 3y2 - 3z2 =
3 • (x2 + 2xy + y2 - z2)
3 • (x2 + 2xy + y2 - z2)