Sửa đề: Tìm x để B đạt GTLN
\(B=\dfrac{4}{x^2-2x+2}\)
\(=\dfrac{4}{x^2-2x+1+1}\)
\(=\dfrac{4}{\left(x-1\right)^2+1}\)
\(\left(x-1\right)^2>=0\forall x\)
=>\(\left(x-1\right)^2+1>=1\forall x\)
=>\(B=\dfrac{4}{\left(x-1\right)^2+1}< =\dfrac{4}{1}=4\forall x\)
Dấu '=' xảy ra khi x-1=0
=>x=1
Vậy: \(B_{max}=4\) khi x=1