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求(x²+2x+3)/(x+2)(x>-2)的值域
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(x²+2x+3)/(x+2)
=[x(x+2)+3]/(x+2)
=x+[3/(x+2)]
=(x+2)+[3/(x+2)]-2
令x+2=t,则t>0,且原式=t+(3/t)-2
≥2√[t·(3/t)]-2【当且仅当t=√3,即x=√3-2时取等号】
=2√3-2
所以原式的值域为[2√3-2,+∞)
=[x(x+2)+3]/(x+2)
=x+[3/(x+2)]
=(x+2)+[3/(x+2)]-2
令x+2=t,则t>0,且原式=t+(3/t)-2
≥2√[t·(3/t)]-2【当且仅当t=√3,即x=√3-2时取等号】
=2√3-2
所以原式的值域为[2√3-2,+∞)
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