问题
解答题
(1)把 x3+2x2y+y3+2xy2在实数范围内因式分解; (2)已知x=2009-5
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答案
(1)x3+2x2y+y3+2xy2=(x+y)(x2-xy+y2)+2xy(x+y)=(x+y)(x2+xy+y2);
(2)原式=
×(x-1)2 (x+1)(x-1)
+x+1-2×(x+1 2(x-1)
)2=x,3 2
当x=2009-5
时,原式=2009-53
.3
(1)把 x3+2x2y+y3+2xy2在实数范围内因式分解; (2)已知x=2009-5
|
(1)x3+2x2y+y3+2xy2=(x+y)(x2-xy+y2)+2xy(x+y)=(x+y)(x2+xy+y2);
(2)原式=
×(x-1)2 (x+1)(x-1)
+x+1-2×(x+1 2(x-1)
)2=x,3 2
当x=2009-5
时,原式=2009-53
.3