\(B=\frac{x^2-6x+14}{x^2-6x+12}\)
\(B=\frac{x^2-6x+12+2}{x^2-6x+12}\)
\(B=1+\frac{2}{\left(x-3\right)^2+3}\le1+\frac{2}{3}\)
\(B=1+\frac{2}{\left(x-3\right)^2+3}\le\frac{5}{3}\)
Dấu " = " xảy ra \(\Leftrightarrow x=3\)
B=\(\frac{x^2-6x+14}{x^2-6x+12}\)
=\(\frac{x^2-6x+9+3+2}{x^2-6x+9+3}\)
=\(\frac{\left(x^2-6x+9\right)+3+2}{\left(x^2-6x+9\right)+3}\)
=\(\frac{\left(x-3\right)^2+3+2}{\left(x-3\right)^2+3}\)
=\(\frac{\left(x-3\right)^2+3}{\left(x-3\right)^2+3}+\frac{2}{\left(x-3\right)^2+3}\)
=1+\(\frac{2}{\left(x-3\right)^2+3}\)
*Ta có:(x-3)2 \(\ge\) 0;với mọi x;cộng 3 vào 2 vế
\(\Rightarrow\)(x-3)2+3 \(\ge\) 0+3;với mọi x
\(\Rightarrow\)(x-3)2+3 \(\ge\) 3;với mọi x
\(\Rightarrow\)\(\frac{2}{\left(x-3\right)^2+3}\) \(\ge\) 0;với mọi x;lấy hai vế cộng cho1
\(\Rightarrow\)\(1+\frac{2}{\left(x-3\right)^2+3}\) \(\ge\)1+0;với mọi x
Vậy .................................
