1樓:ice眖
(1)x4-7x2+1
=x4+2x2+1-9x2
=(x2+1)2-(3x)2
=(x2+3x+1)(x2-3x+1);
(2)x4+x2+2ax+1-a2
=x4+2x2+1-x2+2ax-a2
=(x2+1)2-(x-a)2
=(x2+1+x-a)(x2+1-x+a);
(3)(1+y)2-2x2(1-y2)+x4(1-y)2=(1+y)2-2x2(1-y)(1+y)+x4(1-y)2=(1+y)2-2x2(1-y)(1+y)+[x2(1-y)]2=[(1+y)-x2(1-y)]2
=(1+y-x2+x2y)2
(4)x4+2x3+3x2+2x+1
=x4+x3+x2+x3+x2+x+x2+x+1=x2(x2+x+1)+x(x2+x+1)+x2+x+1=(x2+x+1)2.
把下列各式分解因式:(1)x4-7x2-18; (2)m4n+m3n2-m2n3-mn4
2樓:欽半青
(1)由題意知
x?7x
?18=(x
?9)(x
+2)=(x+3)(x?3)(x
+2)=(x+3)(x-3)(x+
2i)(x-2i)
(2)由題意知
mn+mn?m
n?mn
=mn(m
+mn?mn?n)
=mn[(m
+mn)?(mn
+n)]
=mn[m
(m+n)?n
(m+n)]
=mn(m+n)(m?n)
=mn(m+n)
(m?n)
分解因式:x4-23x2+1
3樓:匿名使用者
(x^2+5x+1)(x^2-5x+1)
思路:一看這個式子,和(x^2±1)^2的樣子很像,肯定要往這上面靠。如果是往(x^2-1)^2上面靠,就多出乙個-21x^2,21這個數字因為不是某個整數的平方,可能性就變小了。
所以往(x^2+1)^2上面靠,配平之後,多出乙個-25x^2,這個數字就非常符合我們的要求。所以,先配方成x^4+2x^2+1-25x^2-----(x^2+1)^2-25x^2-------(x^2+1+5x)(x^2+1-5x),完畢。
4樓:瘋子信譽
原式=(x2)2+2x2+1-25x2
=(x2+1)2-(5x)2
=(x2+5x+1)(x2-5x+1).
因式分解x4+7x3+14x2+7x+1;x4-11x2y2+y2
5樓:三味學堂答疑室
x4+7x3+14x2+7x+1
=(x²+3x+1)(x²+4x+1)
x4-11x2y2+y2
=(x²-y²)²-(3xy)²
=(x²-y²+3xy)(x²-y²-3xy)
把下列各式分解因式:(1)4x2-25y2 (2)x2y-y (3)4x2-(y-z)2 (4)16(a-b)2-(a+b)2 (5)-3xy3
6樓:糠瀁牳
(自1)原
式=(2x+5y)(2x-5y);bai
(du2)原zhi式dao=y(x2-1)=y(x+1)(x-1);
(3)原式=(2x+y-z)(2x-y+z);
(4)原式=(5a-3b)(3a-5b);
(5)原式=-3xy(y2-9)
=-3xy(y+3x)(y-3x);
(6)原式=4a2(x2-4y2 )
=4a2(x+2y)(x-2y);
(7)原式=a2-16+6a-6a
=(a+4)(a-4);
(8)原式=(9x2+y2)(3x+y)(3x-y);
(9)原式=(7p+5q)(p+7q);
(10)原式=-(27a+b)(a+27b).
有下列式子:①-x2-xy-y2;②12a2?ab+12b2;③-4ab-a2+4b4;④4x2+9y2-12xy;⑤3x2+6xy+3y2,其中能用完
7樓:漆雕璟
②12a?ab+12b
=12(a-b)2;
④4x2+9y2-12xy=(2x-3y)2;
⑤3x2+6xy+3y2=3(x+y)2.故選:c.
將下列各式分解因式:(1)ma+mb+mc; (2)x2-25;(3)2x2+4x+2;(4)x4-2x2+1
8樓:灼眼丶夏娜乞
(1)原式=m(a+b+c);
(2)原式=(x+5)(x-5);
(3)原式=2(x2+2x+1)=2(x+1)2;
(4)原式=(x2-1)2=[(x+1)(x-1)]2.
分解因式:(x4-4x2+1)(x4+3x2+1)+10x4=______
9樓:吉格斯
(x4-4x2+1)(x4+3x2+1)+10x4,=[(x4+1)2-x2(x4+1)-12x4]+10x4,=(x4+1)2-x2(x4+1)-2x4,=(x4+1-2x2)(x4+1+x2),=(x2-1)2[(x2+1)2-x2],=(x+1)2(x-1)2(x2+x+1)(x2-x+1).故答案為:(x+1)2(x-1)2(x2+x+1)(x2-x+1).
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