\(A=x\left(x+2\right)+2\left(x-\frac{3}{2}\right)\)
\(=x^2+2x+2x-3\)
\(=x^2+4x-3\)
\(=x^2+4x+4-7\)
\(=\left(x+2\right)^2-7\ge-7\)
Dấu ' = ' \(\Leftrightarrow x+2=0\Rightarrow x=-2\)
\(A=x^2+2x+2x-3=x^2+4x-3.\)
\(A=x^2+4x+4-4-3=\left(x+2\right)^2-7\ge-7\)
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