\(A=4x^2-12x+9-\left(x^2+6x+5\right)+2\)
\(=3x^2-18x+6\)
\(=3\left(x^2-6x+9\right)-21\)
\(=3\left(x-3\right)^2-21\ge-21\)
\(A_{min}=-21\) khi \(x=3\)
Đúng 3
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Ta có: \(A=\left(2x-3\right)^2-\left(x+1\right)\left(x+5\right)+2\)
\(=4x^2-12x+9-x^2-6x-5+2\)
\(=3x^2-18x+6\)
\(=3\left(x^2-9x+2\right)\)
\(=3\left(x^2-2\cdot x\cdot\dfrac{9}{2}+\dfrac{81}{4}-\dfrac{77}{4}\right)\)
\(=3\left(x-\dfrac{9}{2}\right)^2-\dfrac{231}{4}\ge-\dfrac{231}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{9}{2}\)
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