在实数范围内因式分解,要具体过程,谢谢

\u5728\u5b9e\u6570\u8303\u56f4\u5185\u56e0\u5f0f\u5206\u89e3

1\uff092x²+3x-6

\u6709\u4e24\u79cd\u65b9\u6cd5

\uff081\uff09\u6c42\u6839\u6cd5
\u5bf9\u4e8e\u5173\u4e8ex\u7684\u65b9\u7a0bax²+bx+c=0\uff0c\u82e5x1\uff0cx2\u662f\u5176\u4e24\u6839
\u90a3\u4e48ax²+bx+c=a(x-x1)(x-x2)

\u8bbe2x²+3x-6=0\uff0c
\u89e3\u5f97\uff0cx1=\u3010(-3+\u221a57)/4\u3011\uff0cx2=\u3010(-3-\u221a57)/4\u3011
\u6240\u4ee52x²+3x-6= 2[x-\u3010(-3+\u221a57)/4\u3011][x-\u3010(-3-\u221a57)/4\u3011]

\uff082\uff09\u914d\u65b9\u6cd5
2x²+3x-6
=2[x²+(3/2)x-3]
=2[x²+(3/2)x+(3/4)²-(57/16)]
=-[x²-2\u00d7(3/4)\u00d7x+(3/4)²-(\u221a57/4)²]
=-\u3010[x-(3/4)]²-(\u221a57/4)²\u3011
=-[x-(3/4)-(\u221a57/4)][x-(3/4)+(\u221a57/4)]


2\uff09-5x²+6xy+y²
\uff081\uff09\u6c42\u6839\u6cd5
\u8bbe-5x²+6xy+y²=0\uff0c\u628ay\u770b\u6210\u5e38\u6570
x1=\u3010(6+2\u221a14)/10\u3011y\uff0cx2=\u3010(6-2\u221a14)/10\u3011y

\u6240\u4ee5 -5x²+6xy+y²=-5[x-\u3010(6+2\u221a14)/10\u3011y][x-\u3010(6-2\u221a14)/10\u3011y]

\uff082\uff09\u914d\u65b9\u6cd5
-5x²+6xy+y²
=y²+6xy-5x²
=(y²+6xy+9x²)-14x²
=(y+3x)²-(\u221a14x)²
=(y+3x-\u221a14x)(y+3x+\u221a14x)

\u3010\u5e0c\u671b\u5bf9\u4f60\u6709\u5e2e\u52a9\u3011

\u5c31\u662f\u5206\u89e3\u56e0\u5f0f,\u6bcf\u4e2a\u56e0\u5f0f\u90fd\u662f\u5b9e\u6570\u4e14\u4e0d\u53ef\u518d\u5206,\u59824-2=2(2-1)=2(\u6839\u53f72+1)(\u6839\u53f72-1).(\u56e0\u4e3a\u53ef\u5206\u89e3\u4e3a\u6839\u53f7\u4e8c\u7684\u5e73\u65b9,\u4e14\u6839\u53f7\u4e8c\u662f\u5b9e\u6570)









1)x^2+3√2x-2
=(x+3√2/2)^2-13/2
=(x+3√2/2)^2-(√26/2)^2
=[(x+3√2/2)+√26/2][(x+3√2/2)-√26/2]
=[x+√2(3+√13)/2][x+√2(3-√13)/2]
2)2x^2-4xy-3y^2
=2x^2-4xy+2y^2]-5y^2
=[√2(x-y)]^2-(√5y)^2
=[√2(x-y)+√5y][√2(x-y)-√5y]
=[√2x+(√5-√2)y][√2x-(√5+√2)y]

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