Ta có: A = (x + 2)(x - 3) + x(x - 1) - 4 = x2 - 3x + 2x - 6 + x2 - x - 4 = 2x2 - 2x - 10 = 2(x2 - x + 1/4) - 21/2 = 2(x - 1/2)2 - 21/2
Ta luôn có: 2(x - 1/2)2 \(\ge\)0 \(\forall\)x
=> 2(x - 1/2)2 - 21/2 \(\ge\)-21/2 \(\forall\)x
hay A \(\ge\)-21/2 \(\forall\)x
Dấu "=" xảy ra <=> x -1/2 = 0 <=> x = 1/2
vậy Min của A = -21/2 tại x = 1/2
\(A=\left(x+2\right)\left(x-3\right)+x\left(x-1\right)-4\)
\(\Leftrightarrow A=x^2-x-6+x^2-x-4\)
\(\Leftrightarrow A=2x^2-2x-10\)
\(\Leftrightarrow A=2\left(x^2-x-5\right)\)
\(\Leftrightarrow A=2\left(x^2-2.\frac{1}{2}x+\frac{1}{4}-\frac{21}{4}\right)\)
\(\Leftrightarrow A=2\left[\left(x-\frac{1}{2}\right)^2-\frac{21}{4}\right]\)
\(\Leftrightarrow A=2\left[\left(x-\frac{1}{2}\right)^2\right]-\frac{42}{4}\ge-\frac{42}{4}\)
Vậy \(A_{min}=\frac{-42}{4}\Leftrightarrow x=\frac{1}{2}\)
