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\(5x^2+5xy-x-y\)
\(=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
`5x^2 + 5xy +x +y`
`=(5x^2 + 5xy )+(x+y)`
`=5x(x+y)+(x+y)`
`=(x+y)(5x+1)`
\(5x^2+5xy+x+y\\ =5x\left(x+y\right)+\left(x+y\right)\\ =\left(x+y\right)\left(5x+1\right)\)
5x2 - 5xy + 4y - 4x
= 5x ( x - y ) - 4 ( x - y )
= ( 5x - 4 ) ( x - y )
( x + y )3 + ( x - y )3
= 2x3 + 6xy2
= 2x ( x2 + 3y2 )
5 x^2 - 5xy + 4y - 4x
= 5x ( x - y ) - 4 ( x - y )
= ( x - y ) ( 5x - 4 )
( x + y )^3 + ( x - y )^3
= \(x^3+3x^2y+3xy^2+y^3+x^3-3x^2y+3xy^2-y^3\)
= \(2x^3+6xy^2\)
=\(2x\left(x^2+3y^2\right)\)
a) x( x + 2 )( x + 3 )( x + 5 ) + 5
= [ x( x + 5 ) ][ ( x + 2 )( x + 3 ) ] + 5
= ( x2 + 5x )( x2 + 5x + 6 ) + 5 (1)
Đặt t = x2 + 5x
(1) <=> t( t + 6 ) + 5
= t2 + 6t + 5
= t2 + t + 5t + 5
= t( t + 1 ) + 5( t + 1 )
= ( t + 1 )( t + 5 )
= ( x2 + 5x + 1 )( x2 + 5x + 5 )
b) 6x2 - 5xy + y2 = 6x2 - 3xy - 2xy + y2 = 3x( 2x - y ) - y( 2x - y ) = ( 2x - y )( 3x - y )
a,\(x\left(x+2\right)\left(x+3\right)\left(x+5\right)+5\)
\(=x\left(x+5\right)\left(x+2\right)\left(x+3\right)+5\)
\(=\left(x^2+5x\right)\left(x^2+5x+6\right)+5\)(*)
Đặt \(a=x^2+5x\)ta đc:
(*)=\(a\left(a+6\right)+5\)
\(=a^2+6a+5\)
\(=a^2+a+5a+5\)
\(=a\left(a+1\right)+5\left(a+1\right)\)
\(=\left(a+5\right)\left(a+1\right)\)
\(=\left(x^2+5x+5\right)\left(x^2+5x+1\right)\)
b,\(6x^2-3xy-2xy+y^2\)
\(=3x\left(2x-y\right)-y\left(2x-y\right)\)
\(=\left(3x-y\right)\left(2x-y\right)\)
2x^2-5xy-3y^2
= 2^x + xy - 6xy - 3y^2
= x(2x + y) - 3y(2x + y)
= (2x + y)(x - 3y)
tới đó tách ra thử !! @@