Đáp án:
Giải thích các bước giải:
25$x^{2}$ - $x^{2}$ + 4yx - 4$y^{2}$
= 25$x^{2}$ - ( $x^{2}$ - 4yx + 4$y^{2}$ )
= 25$x^{2}$ - [ $x^{2}$ - 2 . x . 2y + $(2y)^{2}$ ]
= $(5x)^{2}$ - $(x-2y)^{2}$ = ( 5x - x - 2y ) ( 5x + x - 2y )
= (4x - 2y) (6x - 2y )
= 2(2x-y)(3x-y)
.
4$x^{3}$ + 4x$y^{2}$ + 8$x^{2}$ y - 16x
= 4x ( $x^{2}$ + $y^{2}$ + 4xy - 4 )
= 4x [ ($x^{2}$ + 4xy + $y^{2}$ ) - 4 ]
= 4x [ $(x+y)^{2}$ - $2^{2}$ ]
= 4x ( x + y - 2 ) ( x + y + 2 )
.
$x^{2}$ - 5x + 4
= $x^{2}$ - 4x - x + 4
= ($x^{2}$ - x) - ( 4x - 4 )
= x ( x - 1 ) - 4 ( x - 1 )
= ( x - 1 ) ( x - 4 )
.
2$x^{2}$ + 3x - 5
= 2$x^{2}$ + 5x - 2x - 5
= ( 2$x^{2}$ - 2x ) + ( 5x - 5 )
= 2x ( x - 1 ) + 5 ( x - 1 )
= ( x - 1 ) ( 2x + 5 )
.
$x^{5}$ + x + 1
= $x^{5}$ - $x^{2}$ + $x^{2}$ + x + 1
= ( $x^{5}$ - $x^{2}$ ) + ( $x^{2}$ + x + 1 )
= $x^{2}$ ( $x^{3}$ - 1 ) + ( $x^{2}$ + x + 1 )
= $x^{2}$ ( $x^{3}$ - $1^{3}$ ) + ( $x^{2}$ + x + 1 )
= $x^{2}$ ( x - 1 ) ( $x^{2}$ + x + 1 ) + ( $x^{2}$ + x + 1 )
= ($x^{2}$ + x + 1 ) ( $x^{3}$ - $x^{2}$ + 1 )
