Re: [代數] 資優班考題1題
※ 引述《stu2005131 (自由幻夢)》之銘言:
: 解方程式[5-(5-x)^(2)]^2=x
可觀察當(5-x)^2=x時
該式成立
故移項必可分解出x^2-11x+25
則(x^2-10x+20)^2-x=0
x^4-20x^3+140x^2-401x+400=0
(x^2-11x+25)(x^2-9x+16)=0
再用公式解即可
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