\(Q=-3x^2+x-11\)
\(Q=-3\left(x^2-\frac{1}{3}x+\frac{11}{3}\right)\)
\(Q=-3\left(x^2-2\cdot x\cdot\frac{1}{6}+\frac{1}{36}+\frac{131}{36}\right)\)
\(Q=-3\left[\left(x-\frac{1}{6}\right)^2+\frac{131}{36}\right]\)
\(Q=-3\left(x-\frac{1}{6}\right)^2-\frac{131}{108}< 0\forall x\)
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\(A=x^2-5x+6\)
\(A=x^2-2\cdot x\cdot\frac{5}{2}+\frac{25}{4}-\frac{1}{4}\)
\(A=\left(x-\frac{5}{2}\right)^2-\frac{1}{4}\ge\frac{-1}{4}\forall x\)
Đến đây chưa kết luận được A âm hay dương nha bạn
Q=-3x2+x-11
=-3(x2-\(\frac{1}{3}\)x)-11
=-3(x2-2.\(\frac{2}{3}x+\frac{4}{9}\))-11-(\(-3.\frac{4}{9}\))
=-3(x-\(\frac{2}{3}\))2-\(\frac{29}{3}\)
Ta có : \(\left\{{}\begin{matrix}-3\left(x-\frac{2}{3}\right)^2\le0với\forall x\\-\frac{29}{3}< 0\end{matrix}\right.\)
\(\Rightarrow\)-3(x-\(\frac{2}{3}\))2-\(\frac{29}{3}< 0\)
vậy Q<0
