a, Thay x=-3 vào B ta có:B=-3+23-3+2=-1-7=17Vậy B=17 tại x=-3b, M=AB=1x+2-2x4-x2+3x-2.x+23x+2, đk: x≠±2; x≠-23=x-2+2x+3x+2x-2x+2.x+23x+2=x-2+2x+3x+6x-2.13x+2=6x+4x-2.13x+2=23x+2x-2.13x+2=2x-2Vậy M=2x-2 c, N=M.x3-x2-2x=2x-2.xx2-x-2=2x-2.xx-2x+1=2xx+1x-2=2x2+2xx-2=2x2-4x+6x-12+12x-2=2x+6+12x-2=2x-2+12x-2+10Ta có: 2x-2+12x-2≥22x-2.12x-2=46, theo Cosi=>N=2x-2+12x-2+10≥46+10 với mọi xDấu = xảy ra<=>x-2=12x-2<=>x-22=12<=>x-2=±23<=>x=2±23Vậy x=2±23 thì N có GTNN=46+10..
\(P=2x-2x^2-5\) \(=-2\left(x^2-x+\dfrac{5}{2}\right)\) \(=-2\left[\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{1}{4}+\dfrac{5}{2}\right]\) \(=-2\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\right]\) \(\)\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\) Ta có : \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\Rightarrow-2\left(x-\dfrac{1}{2}\right)^2\le0\) \(\Rightarrow-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\) hay P ≤ \(-\dfrac{9}{2}\) Dấu = xảy ra \(\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\) Vậy \(Max_P=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{1}{2}\)
Tìm giá trị lớn nhất của biểu thức A = 2x–x2 –4 Các câu hỏi tương tự |
