函数y=(2x²-x+2)/(x²+x+1)的值域为?
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函数y=(2x²-x+2)/(x²+x+1)的值域为?
函数y=(2x²-x+2)/(x²+x+1)的值域为?
函数y=(2x²-x+2)/(x²+x+1)的值域为?
解析:y=(2x²-x+2)/(x²+x+1)
y(x²+x+1)=(2x²-x+2)
yx²+yx+y=2x²-x+2
(2-y)x²-(y+1)x+2-y=0
⊿=(y+1)^2-4(2-y)^2>0
3y^2-18y+15
y=(2x²-x+2)/(x²+x+1)
y=2-3x/(x²+x+1)
y=2-3/(x+1/x+1)x+1/x≥2√x×1/x
y≤2-3/(2+1)=2-1=1
函数y=(2x²-x+2)/(x²+x+1)的值域为(负无穷,1]可以用高一的只是解答吗?没看懂(a+b)^2≥2ab还没学么?!没高一应该学过了吧,...
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y=(2x²-x+2)/(x²+x+1)
y=2-3x/(x²+x+1)
y=2-3/(x+1/x+1)x+1/x≥2√x×1/x
y≤2-3/(2+1)=2-1=1
函数y=(2x²-x+2)/(x²+x+1)的值域为(负无穷,1]
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分子分母先因式分解,然后约分。同时注意分子不等于0