Question

 

Answer 1

Đặt A=B

=>\(\frac{2016x+2688}{x^2+1}=B\)

=>\(B\left(x^2+1\right)=2016x+2688\)

=>\(x^2\cdot B-2016x+B-2688=0\) (1)

\(\Delta=\left(-2016\right)^2-4B\left(B-2688\right)\)

\(=4064256-4B^2+10752B=-4\left(B+336\right)\left(B-3024\right)\)

Để phương trình (1) có nghiệm thì Δ>=0

=>-4(B+336)(B-3024)>=0

=>(B+336)(B-3024)<=0

=>-336<=B<=3024

=>-336<=A<=3024

\(A_{\min}=-336\) khi \(-336\left(x^2+1\right)=2016x+2688\)

=>\(x^2+1=-6x-8\)

=>\(x^2+6x+9=0\)

=>\(\left(x+3\right)^2=0\)

=>x+3=0

=>x=-3

\(A_{\max}=3024\) khi \(3024\left(x^2+1\right)=2016x+2688\)

=>\(3\left(x^2+1\right)=2x+\frac83\)

=>\(3x^2+3-2x-\frac83=0\)

=>\(3x^2-2x+\frac13=0\)

=>\(9x^2-6x+1=0\)

=>\(\left(3x-1\right)^2=0\)

=>3x-1=0

=>3x=1

=>\(x=\frac13\)