3: ta có: \(C=2x^2-4x+1\)
\(=2\left(x^2-2x+\dfrac{1}{2}\right)\)
\(=2\left(x^2-2x+1-\dfrac{1}{2}\right)\)
\(=2\left(x-1\right)^2-1\ge-1\forall x\)
Dấu '=' xảy ra khi x=1
4: Ta có: \(D=3x^2+2x-5\)
\(=3\left(x^2+\dfrac{2}{3}x-\dfrac{5}{3}\right)\)
\(=3\left(x^2+2\cdot x\cdot\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{16}{9}\right)\)
\(=3\left(x+\dfrac{1}{3}\right)^2-\dfrac{16}{3}\ge-\dfrac{16}{3}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{1}{3}\)
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