已知x>-1,则f(x)=(x^2-3x+1)/(x+1)的最小值回答 共1条 f(x)=(x²-3x-4+5)/(x+1)=[(x+1)(x-4)+5]/(x+1)=x-4+5/(x+1)=(x+1)+5/(x+1)-5x>-1x+1>1所以f(x)>=2√[(x+1)*5/(x+1)]-5=2√5-5所以最小值是2√5-5 由f(x)=(x+1)+5/(x+1)-5变为f(x)>=2
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![已知x>-1,则f(x)=(x^2-3x+1)/(x+1)的最小值回答 共1条 f(x)=(x²-3x-4+5)/(x+1)=[(x+1)(x-4)+5]/(x+1)=x-4+5/(x+1)=(x+1)+5/(x+1)-5x>-1x+1>1所以f(x)>=2√[(x+1)*5/(x+1)]-5=2√5-5所以最小值是2√5-5 由f(x)=(x+1)+5/(x+1)-5变为f(x)>=2](/uploads/image/z/995146-34-6.jpg?t=%E5%B7%B2%E7%9F%A5x%3E-1%2C%E5%88%99f%28x%29%3D%28x%5E2-3x%2B1%29%2F%28x%2B1%29%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E5%9B%9E%E7%AD%94+%E5%85%B11%E6%9D%A1+f%28x%29%3D%28x%26sup2%3B-3x-4%2B5%29%2F%28x%2B1%29%3D%5B%28x%2B1%29%28x-4%29%2B5%5D%2F%28x%2B1%29%3Dx-4%2B5%2F%28x%2B1%29%3D%28x%2B1%29%2B5%2F%28x%2B1%29-5x%3E-1x%2B1%3E1%E6%89%80%E4%BB%A5f%28x%29%3E%3D2%E2%88%9A%5B%28x%2B1%29%2A5%2F%28x%2B1%29%5D-5%3D2%E2%88%9A5-5%E6%89%80%E4%BB%A5%E6%9C%80%E5%B0%8F%E5%80%BC%E6%98%AF2%E2%88%9A5-5+%E7%94%B1f%28x%29%3D%28x%2B1%29%2B5%2F%28x%2B1%29-5%E5%8F%98%E4%B8%BAf%28x%29%3E%3D2)
已知x>-1,则f(x)=(x^2-3x+1)/(x+1)的最小值回答 共1条 f(x)=(x²-3x-4+5)/(x+1)=[(x+1)(x-4)+5]/(x+1)=x-4+5/(x+1)=(x+1)+5/(x+1)-5x>-1x+1>1所以f(x)>=2√[(x+1)*5/(x+1)]-5=2√5-5所以最小值是2√5-5 由f(x)=(x+1)+5/(x+1)-5变为f(x)>=2
已知x>-1,则f(x)=(x^2-3x+1)/(x+1)的最小值
回答 共1条
f(x)=(x²-3x-4+5)/(x+1)
=[(x+1)(x-4)+5]/(x+1)
=x-4+5/(x+1)
=(x+1)+5/(x+1)-5
x>-1
x+1>1
所以f(x)>=2√[(x+1)*5/(x+1)]-5=2√5-5
所以最小值是2√5-5
由f(x)=(x+1)+5/(x+1)-5变为f(x)>=2√[(x+1)*5/(x+1)]-5,是怎么来的?
已知x>-1,则f(x)=(x^2-3x+1)/(x+1)的最小值回答 共1条 f(x)=(x²-3x-4+5)/(x+1)=[(x+1)(x-4)+5]/(x+1)=x-4+5/(x+1)=(x+1)+5/(x+1)-5x>-1x+1>1所以f(x)>=2√[(x+1)*5/(x+1)]-5=2√5-5所以最小值是2√5-5 由f(x)=(x+1)+5/(x+1)-5变为f(x)>=2
f(x)=(x²-3x-4+5)/(x+1)
=[(x+1)(x-4)+5]/(x+1)
=x-4+5/(x+1)
=(x+1)+5/(x+1)-5
x>-1
x+1>1
所以f(x)>=2√[(x+1)*5/(x+1)]-5=2√5-5
所以最小值是2√5-5
f(x)=(x²-3x-4+5)/(x+1)
=[(x+1)(x-4)+5]/(x+1)
=x-4+5/(x+1)
=(x+1)+5/(x+1)-5
2√5-5
ww