a) \(A=x^2+x+1\)
\(A=x^2+x+\frac{1}{4}+\frac{3}{4}\)
\(A=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Có: \(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu = xảy ra khi: \(\left(x+\frac{1}{2}\right)^2=0\Rightarrow x+\frac{1}{2}=0\Rightarrow x=-\frac{1}{2}\)
Vậy: \(Min_A=\frac{3}{4}\) tại \(x=-\frac{1}{2}\)
b) \(B=2+x-x^2\)
\(B=\frac{9}{4}-x^2+x-\frac{1}{4}\)
\(B=\frac{9}{4}-\left(x-\frac{1}{2}\right)^2\)
Có: \(\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow\frac{9}{4}-\left(x-\frac{1}{2}\right)^2\le\frac{9}{4}\)
Dấu = xảy ra khi: \(\left(x-\frac{1}{2}\right)^2=0\Rightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)
Vậy: \(Max_B=\frac{9}{4}\) tại \(x=\frac{1}{2}\)
c) \(C=x^2-4x+1\)
\(C=x^2-4x+4-3\)
\(C=\left(x-2\right)^2-3\)
Có: \(\left(x-2\right)^2\ge0\Rightarrow\left(x-2\right)^2-3\ge-3\)
Dấu = xảy ra khi: \(\left(x-2\right)^2=0\Rightarrow x-2=0\Rightarrow x=2\)
Vậy: \(Min_C=-3\) tại \(x=2\)
Mấy bài kia tương tự, riêng bài g
g) \(G=h\left(h+1\right)\left(h+2\right)\left(h+3\right)\)
\(G=\left(h^2+3h\right)\left(h^2+3h+2\right)\)
Đặt: \(t=h^2+3h+1\)
\(\Leftrightarrow\hept{\begin{cases}h^2+3h=t-1\\h^2+3h+2=t+1\end{cases}}\)
\(\Leftrightarrow\left(h^2+3h\right)\left(h^2+3h+2\right)=\left(t-1\right)\left(t+1\right)=t^2-1=\left(h^2+3h+1\right)^2-1\)
Có: \(\left(h^2+3h+1\right)^2\ge0\Rightarrow\left(h^2+3h+1\right)^2-1\ge-1\)
Dấu = xảy ra khi: \(\left(h^2+3h+1\right)^2=0\Rightarrow h^2+3h+1=0\Rightarrow\left(h+\frac{3}{2}\right)^2-\frac{5}{4}=0\Rightarrow\orbr{\begin{cases}h=-\frac{\sqrt{5}}{2}-\frac{3}{2}\\h=\frac{\sqrt{5}}{2}-\frac{3}{2}\end{cases}}\)
Vậy: \(Min_G=-1\) tại \(\orbr{\begin{cases}h=-\frac{\sqrt{5}}{2}-\frac{3}{2}\\h=\frac{\sqrt{5}}{2}-\frac{3}{2}\end{cases}}\)
