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當m為何值時,方程組y=x+m,x平方+y平方=2有兩個相等實數解 把Y=x+m代入到x^2+y^2=2: x^2+x^2+2mx+m^2=2 2x^2+2mx+m^2-2=0 有二個相等的實數解,則有判別式=0,即有:4m^2-8(m^2-2)=0 -4m^2=-16 m^2=4 m=(+/-)2 即m=+2或-2.
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