\(A=\left(x^2+2x+1\right)+4=\left(x+1\right)^2+4Ma:\hept{\begin{cases}\left(x+1\right)^2\ge0\\4>0\end{cases}\Rightarrow\left(x+1\right)^2+4>0}\)với mọi x
\(B=\left(x^2-2.3x+9\right)+1=\left(x-3\right)^2+1\\ Ma:\hept{\begin{cases}\left(x-3\right)^2\ge0\\1>0\end{cases}\Rightarrow\left(x-3\right)^2+1}\)(dấu phái sau là do lỗi nha )
\(C=\left(x^2-2.\frac{3}{2}x+\frac{9}{4}\right)+\frac{11}{4}=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\)
\(Ma:\hept{\begin{cases}\left(x-\frac{3}{2}\right)^2\ge0\\\frac{11}{4}>0\end{cases}\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{11}{4}>0}\)
\(D=2\left(x^2-\frac{5}{2}x+7\right)=2\left(x^2-2.\frac{5}{4}x+\frac{25}{16}+\frac{87}{16}\right)=2\left(x-\frac{5}{4}\right)^2+\frac{87}{8}\)
\(Ma:\hept{\begin{cases}\left(x-\frac{5}{4}\right)^2\ge0\\\frac{87}{4}>0\end{cases}\Rightarrow2\left(x-\frac{5}{4}\right)^2+\frac{87}{8}>0}\)
Học tốt nha bạn
1 T I C K nha
