在实数范围内因式分解:(11)3x^2-4x-1(12)x^2-3x-9/4(13)2y^2-6y-3(14)-4a^2

问题描述:

在实数范围内因式分解:(11)3x^2-4x-1(12)x^2-3x-9/4(13)2y^2-6y-3(14)-4a^2+6a+3(15)9y^2-6xy-x^2
(16)-2x^2+7xy-4y^2(17)4x^2-4xy-y^2(18)1/2x^4-5/2x^2+3(19)2x^3+3x^2-2x(20)y^4-9y^2+8(21)3x^4-6x^3+2x^2(22)3x^2-2x-2(23)2a^2-4ab-5b^2(24)-3x^2+4xy+10y^2(25)4x^2+12x+3
1个回答 分类:数学 2014-10-05

问题解答:

我来补答
(11)3x^2-4x-1
3x^2-4x-1
=3(x^2-4/3x-1/3)
令x^2-4/3x-1/3=0
解得:x=(2±√7)/3
即:3x^2-4x-1
=3(x^2-4/3x-1/3)
=3[x-(2+√7)/3][x-(2-√7)/3]
(12)x^2-3x-9/4
令x^2-3x-9/4=0
解得:x=(3±3√2)/2
即:x^2-3x-9/4
=[x-(3+3√2)/2][x-(3-3√2)/2]
(13)2y^2-6y-3
2y^2-6y-3
=2(y^2-3y-3/2)
令:y^2-3y-3/2=0
解得;y=(3±2√3)/2
即:2y^2-6y-3
=2(y^2-3y-3/2)
=2[y-(3+2√3)/2][y-(3-2√3)/2]
(14)-4a^2+6a+3
-4a^2+6a+3
= -4(a^2-3/2-3/4)
令a^2-3/2-3/4=0
解得:a=(3±√21)/4
即;-4a^2+6a+3
=-4[a-(3+√21)/4][a-(3-√21)/4]
(15)9y^2-6xy-x^2
9y^2-6xy-x^2
=9(y^2-2/3xy-1/9x^2)
令y^2-2/3xy-1/9x^2=0
解得:y=(3±3√2)/9 x
即:9y^2-6xy-x^2
=9(y^2-2/3xy-1/9x^2)
=9[y-(3+3√2)/9 x][y-(3-3√2)/9 x]
(16)-2x^2+7xy-4y^2
-2x^2+7xy-4y^2
=-2(x^2-7/2xy+2)
解得:x=(7±√17)/4 y
即:2x^2+7xy-4y^2
=-2(x^2-7/2xy+2)
= - 2[x-(7+√17)/4 y][x-(7-√17)/4 y]
(17)4x^2-4xy-y^2
4x^2-4xy-y^2
=4(x^2-xy-1/4y^2)
解得:x=(1±√2x)/2 y
4x^2-4xy-y^2
=4(x^2-xy-1/4y^2)
=4[x-(1+√2)/2 y][x-(1-√2)/2 y]
=[2x-(1+√2) y][2x-(1-√2) y]
(18)1/2x^4-5/2x^2+3
1/2x^4-5/2x^2+3
=1/2(x^4-5x^2+6)
=1/2(x-2)(x-3)
(19)2x^3+3x^2-2x
2x^3+3x^2-2x
=2x(x^2+3/2x-1)
=2x(x-1/2)(x+2)
(20)y^4-9y^2+8
y^4-9y^2+8
=(y^2-8)(y^2-1)
=(y-2√2)(y+2√2)(y-1)(y+1)
(21)3x^4-6x^3+2x^2
3x^4-6x^3+2x^2
=3x^2(x^2-2x+2/3)
令x^2-2x+2/3=0
解得:x=1±√3/3
即:3x^4-6x^3+2x^2
=3x^2(x^2-2x+2/3)
=3x^2[x-(1+√3/3)][x-(1-√3/3)]
(22)3x^2-2x-2
3x^2-2x-2
=3(x^2-2/3x-2/3)
令x^2-2/3x-2/3=0
解得:x=(1±√7)/3
即:3x^2-2x-2
=3(x^2-2/3x-2/3)
=3[x-(1+√7)/3][x-(1-√7)/3]
(23)2a^2-4ab-5b^2
2a^2-4ab-5b^2
=2(a^2-2ab-5/2b^2)
令a^2-2ab-5/2b^2=0 【把b看成常数】
解得:a=(2±√14)/2 b
即:2a^2-4ab-5b^2
=2(a^2-2ab-5/2b^2)
=2[x-(2-√14)/2 b][x-(2+√14)/2 b]
(24)-3x^2+4xy+10y^2
-3x^2+4xy+10y^2
=-3(x^2-4/3xy-10/3y^2) 【把y当做常数】
令x^2-4/3xy-10/3y^2=0
解得:x=(2±√34)/3 y
即::-3x^2+4xy+10y^2
=-3(x^2-4/3xy-10/3y^2)
=-3[x-(2+√34)/3 y][x-(2-√34)/3 y]
(25)4x^2+12x+3
4x^2+12x+3
=4(x^2+3x+3/4)
令x^2+3x+3/4=0
解得:x=(-3±√6)/2
即:4x^2+12x+3
=4(x^2+3x+3/4)
=4[x-(-3+√6)/2][x-(-3-√6)/2]
=(2x+3-√6)(x+3+√6)
 
 
展开全文阅读
剩余:2000
上一页:质点位移问题