作业帮一元二次方程ax^2+bx+c=0中,abc都是实数,且(2-a)^2+√下(a^2+b+c)+|c+8|=0,求代数式
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一元二次方程ax^2+bx+c=0中,abc都是实数,且(2-a)^2+√下(a^2+b+c)+|c+8|=0,求代数式x^2+x+1的值
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平方和绝对值和根号都是大于或等于0的 所以只能都等于0才能相加为0
2-a＝0得 a＝2 c+8＝0 得c＝-8
a²+b+c＝0 得b＝4
2x²+4x-8＝0 得x²+2x-4＝0
（x+1）²＝5 x＝±√5-1
x²+x+1＝4-2x+x+1＝5-x＝6±√5
2-a＝0得 a＝2 c+8＝0 得c＝-8
a²+b+c＝0 得b＝4
2x²+4x-8＝0 得x²+2x-4＝0
（x+1）²＝5 x＝±√5-1
x²+x+1＝4-2x+x+1＝5-x＝6±√5
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