Đáp án:
a/ Ta có :
$x^2(x−1)−x+1$
$=x^2(x−1)−(x−1)$
$=(x−1)(x^2−1)$
$=(x−1)(x−1)(x+1)$
$=(x−1)^2(x+1)$
b/ Ta có :
$(a+b)^3−(a−b)^3$
$=[(a+b)−(a−b)][(a+b)^2+(a+b)(a−b)+(a−b)^2]$
$=2b(a^2+2ab+b^2+a^2−b^2+a^2−2ab+b^2)$
$=2b(3a^2+b^2)$
c/ Ta có :
$6x(x−3)+9−3x^2$
$=6x^2−18x+9−3x^2$
$=3x^2−18x+9$
$=3(x^2−6x+3)$
$=3(x^2−6x+9−6)$
$=3[(x−3)^2−6]$
$=3(x−3−√6)(x−3+√6)$
d/ Ta có :
$x(x−y)−5x+5y$
$=x(x−y)−5(x−y)$
$=(x−y)(x−5)$
e/ Ta có
$3(x+4)−x^2−4x$
$=3(x+4)−x(x+4)$
$=(x+4)(3−x)$
f/ Ta có :
$x^2+4x−y^2+4$
$=(x^2+4x+4)−y^2$
$=(x+2)^2−y2$
$=(x+2−y)(x+2+y)$
$=(x−y+2)(x+y+2)$
g/ Ta có
$x^2+5x$
$=x(x+5)$
h/Ta có :
$−x^2+2x+2y+y^2$
$=−(x^2−2x+1)+(y^2+2y+1)$
$=(y+1)^2−(x−1)^2$
$=(y+1−x+1)(y+1+x−1)$
$=(y−x+2)(x+y)$
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