Ta có: \(B=\frac{2x^2-2x+5}{x^2+2}\)
\(=\frac{2x^2+4-2x+1}{x^2+2}=2+\frac{-2x+1}{x^2+2}\)
Đặt \(A=\frac{-2x+1}{x^2+2}\)
=>\(A\left(x^2+2\right)=-2x+1\)
=>\(A\cdot x^2+2A+2x-1=0\)
=>\(A\cdot x^2+2x+2A-1=0\) (1)
\(\Delta=2^2-4A\left(2A-1\right)=4-8A^2+4A=-4\left(2A^2-A-1\right)\)
=-4(A-1)(2A+1)
Để (1) có nghiệm thì -4(A-1)(2A+1)>=0
=>(A-1)(2A+1)<=0
=>\(-\frac12\le A\le1\)
=>\(A_{\max}=1\)
=>\(\frac{-2x+1}{x^2+2}\le1\forall x\) thỏa mãn ĐKXĐ
=>\(\frac{-2x+1}{x^2+2}+2\le3\forall x\) thỏa mãn ĐKXĐ
=>\(B\le3\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi \(\frac{-2x+1}{x^2+2}=1\)
=>\(x^2+2=-2x+1\)
=>\(x^2+2x+1=0\)
=>\(\left(x+1\right)^2=0\)
=>x+1=0
=>x=-1
