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设函数f(x)=x2+ax+b,集合A={x|f(x)=x}={a},求实数a,b的值
人气:173 ℃ 时间:2020-05-08 00:50:53
解答
x=f(x)x=x^2+ax+bx^2+(a-1)x+b = 0a is only solutionthen △ = 0(a-1)^2 - 4b = 0 => b =(a-1)^2/4also a is solution,thena^2 +(a-1)a + b = 0a^2+ (a-1)a +(a-1)^2/4 =08a^2-4a +(a-1)^2 = 09a^2 -6a + 1 = 0(3a-...
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