Question

Tìm giá trị lớn nhất của biểu thức

\(A=\frac{2x^2-4x+7}{x^2-2x+2}\)

Answer 1

\(A=\frac{2x^2-4x+7}{x^2-2x+2}=\frac{2.\left(x^2-2x+2\right)+3}{x^2-2x+2}=2+\frac{3}{x^2-2x+1+1}=2+\frac{3}{\left(x-1\right)^2+1}\)

\(\text{Để A max}\Leftrightarrow\left(\frac{3}{\left(x-1\right)^2+1}\right)max\Leftrightarrow\left[\left(x-1\right)^2+1\right]min\)vì (x-1)²+1 > 0

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

Vậy Max A=5 <=> x=1

Answer 2

\(A=\frac{2x^2-4x+7}{x^2-2x+2}\)

\(A=\frac{2\left(x^2-2x+2\right)+3}{x^2-2x+2}\)

\(A=\frac{2\left(x^2-2x+2\right)}{x^2-2x+2}+\frac{3}{x^2-2x+2}\)

\(A=2+\frac{3}{x^2-2x+1+1}\)

\(A=2+\frac{3}{\left(x-1\right)^2+1}\le2+\frac{3}{0+1}=2+3=5\)

Dấu "=" xảy ra \(\Leftrightarrow x-1=0\Leftrightarrow x=1\)