a: Ta có: \(A=16x^2+8x+5\)
\(=16x^2+8x+1+4\)
\(=\left(4x+1\right)^2+4\ge4\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{1}{4}\)
b: Ta có: \(B=2x^2-5x\)
\(=2\left(x^2-\dfrac{5}{2}x\right)\)
\(=2\left(x^2-2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}-\dfrac{25}{16}\right)\)
\(=2\left(x-\dfrac{5}{4}\right)^2-\dfrac{25}{8}\ge-\dfrac{25}{8}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{5}{4}\)
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