求y=(x^2-x+2)/(x^2+x+2)的值域
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求y=(x^2-x+2)/(x^2+x+2)的值域求y=(x^2-x+2)/(x^2+x+2)的值域求y=(x^2-x+2)/(x^2+x+2)的值域分母x²+x+2>0对于x∈R恒成立故用
求y=(x^2-x+2)/(x^2+x+2)的值域
求y=(x^2-x+2)/(x^2+x+2)的值域
求y=(x^2-x+2)/(x^2+x+2)的值域
分母x²+x+2>0对于x∈R恒成立
故用判别式法,两边去分母
y(x²+x+2)=x²-x+2
整理得(y-1)x²+(y+1)x+2y-2=0
判别式(y+1)²-4(y-1)(2y-2)≥0
即7y²-18y+7≤0
解得(9-4√2)/7≤y≤(9+4√2)/7
故函数值域为[(9-4√2)/7,(9+4√2)/7]
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