Question

Tìm Min

D=\(2x^2-8x-10\)

không làm tắt ạ!!!!

Answer 1

Ta có: \(D=2x^2-8x-10\)

\(=2\left(x^2-4x-5\right)\)

\(=2\left(x^2-4x+4-9\right)\)

\(=2\left(x-2\right)^2-18\ge-18\forall x\)

Dấu '=' xảy ra khi x=2

Answer 2

Answer 3

\(D=2x^2-8x-10=2\left(x^2-4x+4\right)-8-10=2\left(x-2\right)^2-18\)

Do \(2\left(x-2\right)^2\ge0\)

\(\Rightarrow D=2\left(x-2\right)^2-18\ge-18\)

\(minD=-18\Leftrightarrow2\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)