几道因式分解 用一元二次的应用

问题描述:

几道因式分解 用一元二次的应用
2a^2-3a-1
2a^2-3ab-b^2
-3p^2+4p+2
x^2+49+600
x^2-71x+300
6x^2-25x+24
8x^2-189x-72
1个回答 分类:综合 2014-09-26

问题解答:

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2a^2-3a-1
=2(a^2-3a/2)-1
=2(a^2-3a/2+9/16)-1-18/16
=2(a-3/4)^2-34/16
=2[(a-3/4)^2-17/16]
=2(a-3/4-√17/4)(a-3/4+√17/4)
2a^2-3ab-b^2
=2(a^2-3ab/2)-b^2
=2(a^2-3ab/2+9b^2/16)-b^2-b^2/16
=2(a-3b/4)^2-34b^2/16
=2[(a-3b/4)^2-17b^2/16]
=2(a-3b/4-b√17/4)(a-3b/4+b√17/4)
-3p^2+4p+2
=-3(p^2-4p/3)+2
=-3(p^2-4p/3+4/9)+2+12/9
=30/9-3(p-2/3)^2
=3[10/9-(p-2/3)^2]
=3(√10/3-p+2/3)(√10/3+p-2/3)
x^2+49x+600
=x^2+49x+2401/4-2401/4+600
=(x+49/2)^2-1/4
=(x+49/2-1/2)(x+49/2+1/2)
=(x+24)(x+25)
x^2-71x+300
=x^2-71x+5041/4-5041/4+300
=(x-71/2)^2-3841/4
=(x+49/2-√3841/2)(x+49/2+√3841/2)
6x^2-25x+24
=(3x-8)(2x-3)
8x^2-189x-72
=(x-24)(8x+3)
 
 
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