Question

高中数学基本不等式急!已知x≥2/5,f(x）=x^2-4x+5/2x-4有最小值是多少已知a.b,c∈R,a+b+c=

Answer 1

⑴ f(x)=(x^2-4x+5)/(2x-4) =((x-2)^2+1)/(2(x-2)) =(x-2)/2+1/(2(x-2)) 当 (x-2)/2=1/(2(x-2))时有最小值即当x=3时,有最小值=2*√((x-2)/2*1/(2(x-2))) =2*1/2=1

⑵ 1/a+1/b+1/c

=(a+b+c)/a+(a+b+c)/b+(a+b+c)/c

=1+(b+c)/a+1(a+c)/b+1(a+b)/c

=3+b/c+c/b+a/c+c/a+a/b+b/a

>=3+2+2+2

=9