Question

Tìm giá trị nhỏ nhất của biểu thức Q =\(\frac{2x^2+2}{\left(x+1\right)^2}\)

Answer 1

\(Q=\frac{2x^2+2}{\left(x+1\right)^2}\)

\(Q=\frac{x^2+2x+1+x^2-2x+1}{\left(x+1\right)^2}\)

\(Q=\frac{\left(x+1\right)^2+\left(x-1\right)^2}{\left(x+1\right)^2}\)

\(Q=\frac{\left(x+1\right)^2}{\left(x+1\right)^2}+\frac{\left(x-1\right)^2}{\left(x+1\right)^2}\)

\(Q=1+\frac{\left(x-1\right)^2}{\left(x+1\right)^2}\)

Vì \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow Q\ge1+0=1\)

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

Vậy........