\(A=x^2+x+2=\left(x^2+x+\frac{1}{4}\right)+\frac{7}{4}=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}\ge0+\frac{7}{4}=\frac{7}{4}.\) Dâu bàng xay ra khi: \(x=\frac{-1}{2}\)
\(B=4x^2-4x-1=\left(4x^2-4x+1\right)-2=\left(2x-1\right)^2-2\ge0-2=-2\Rightarrow B_{min}=-2\) Dâu bàng xay ra: \(x=\frac{1}{2}\)
\(C=x^2+y^2+2x-4y+2=x^2+y^2+2x-4y+5-3=\left(x^2+2x+1\right)+\left(y^2-4y+4\right)-3=\left(x+1\right)^2+\left(y-2\right)^2-3\ge0+0-3=-3\) Dâu bàng xay ra\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\y-2=0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
\(A=1-x^2+2x-1=1-\left(x-1\right)^2\le1-0=1\Rightarrow A_{max}=1.\text{Dâu "=" xay ra}\Leftrightarrow x=1\) \(B=-\left(x^2-4x-4\right)-3=-\left(x-2\right)^2-3\le0-3=-3\Rightarrow B_{max}=-3.\text{Dâu "=" xay ra}\Leftrightarrow x=2\)
svtkvtm làm câu d) bài 1 nha:)
\(D=\left(x^2+2xy+y^2\right)-\left(2x+2y\right)+1+1022\)
\(=\left(x+y\right)^2-2\left(x+y\right).1+1+1022\)
\(=\left(x+y-1\right)^2+1022\ge1022\)
Đẳng thức xảy ra khi x = 1 - y
Bài 1:
\( A = {x^2} + x + 2\\ A = \left( {{x^2} + 2.x.\dfrac{1}{2} + \dfrac{1}{4}} \right) - \dfrac{1}{4} + 2\\ A = {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{7}{4} \ge \dfrac{7}{4} \)
Vậy \({\min _A} = \dfrac{7}{4} \Leftrightarrow x = - \dfrac{1}{2}\)
\( B = 4{x^2} - 4x - 1\\ B = 4\left( {{x^2} - x} \right) - 1\\ B = 4\left( {{x^2} - 2.x.\dfrac{1}{2} + \dfrac{1}{4}} \right) - \dfrac{1}{4} - 1\\ B = 4{\left( {x - \dfrac{1}{2}} \right)^2} - \dfrac{5}{4} \ge - \dfrac{5}{4} \)
Vậy \({\min _B} = - \dfrac{5}{4} \Leftrightarrow x = \dfrac{1}{2} \)