Question

giúp mik bài 3,4 vs ạ ( mik cảm ơn ) mik cần gấp ạ 

Answer 1

Bài 4:
\(B=\dfrac{1}{x^2-4x+9}\)
\(=\dfrac{1}{\left(x^2-4x+4\right)+5}\)
\(=\dfrac{1}{\left(x-2\right)^2+5}\)
Do \(\left(x-2\right)^2\ge0\forall x\) nên \(B\ge\dfrac{1}{0+5}=\dfrac{1}{5}\)
Dấu "=" xảy ra khi \(\left(x-2\right)^2=0\)
\(\Rightarrow x=2\)
Vậy \(minB=\dfrac{1}{5}\) khi \(x=2\)

Answer 2