Question

CMR: biểu thức luôn dương với mọi x:
A=3x^2-x+20

Answer 1

\(A=3x^2-x+20=3\left(x^2-\dfrac{1}{3}x+\dfrac{20}{3}\right)=3\left(x^2-2.\dfrac{1}{6}x+\dfrac{1}{36}+\dfrac{239}{36}\right)\)

\(A=3\left[\left(x+\dfrac{1}{6}\right)^2+\dfrac{239}{36}\right]=3\left(x+\dfrac{1}{6}\right)^2+\dfrac{239}{12}\ge\dfrac{239}{12}\)

\(=>A>0\left(\forall x\right)\)

Answer 2

Ta có:A=3x²-x+20=2(x²-2x+1)+\(\left(x^2+2.\dfrac{3}{2}x+\dfrac{9}{4}\right)+\dfrac{73}{4}\)

            =\(2\left(x-1\right)^2+\left(x+\dfrac{3}{2}\right)^2+\dfrac{73}{4}\ge0\)