分解因式：3x^2+6x+3=____;-a^2-4b^2+4ab=____;(p^2+q^2)^2-4p^2q^2=____
问题描述：
分解因式：3x^2+6x+3=____;-a^2-4b^2+4ab=____;(p^2+q^2)^2-4p^2q^2=____
答
3x^2+6x+3=3(x²+2x+1)=3(x+1)²
-a^2-4b^2+4ab=-(a²+4b²-4ab)=-(a-2b)²
(p^2+q^2)^2-4p^2q^2=(p²+q²+2pq)(p²+q²-2pq)=(p+q)²(p-q)²
答
3x^2+6x+3
= 3（x^2+2x+1)
= 3(x+1)^2
-a^2-4b^2+4ab
= -(a^2-4ab+4b^2)
= -(a-2b)^2
(p^2+q^2)^2-4p^2q^2
= (p^2+q^2+2pa) (p^2+q^2-2pa)
= (p+q)^2 (p-q)^2
答
3x^2+6x+3
=3(x^2+2x+1)
=3(x+1)^2
-a^2-4b^2+4ab
=-(a^2+4b^2-4ab)
=-(a-2b)^2
(p^2+q^2)^2-4p^2q^2
=[(p^2+q^2)-2pq][(p^2+q^2)+2pq]
=[(p-q)^2][(p+q)]^2
=(p-q)^2(p+q)^2
答
3x^2+6x+3=3(x+1)^2
-a^2-4b^2+4ab= -(a-2b)^2;
(p^2+q^2)^2-4p^2q^2=(p^2-q^2)^2