Viết lại đề câu a)
Câu b)
\(A=4x^2+4x+15\)
\(=\left(2x+1\right)^2+14\ge14\)
Dấu "=" xảy ra \(\Leftrightarrow x=-\frac{1}{2}\)
Vậy : Min \(A=14\Leftrightarrow x=-\frac{1}{2}\)
\(x^2-3x+7=\left(x-\frac{3}{2}\right)^2+\frac{19}{4}>0\)
Ta có \(A=4x^2+4x+15=\left(2x+1\right)^2+14\ge14\)
Dấu "=" xảy ra khi \(x=\frac{-1}{2}\)
Vậy Min \(A=14\Leftrightarrow x=\frac{-1}{2}\)
a, Ta có : \(x^2-3x+7\)
\(=x^2-2.x.\frac{3}{2}+\frac{9}{4}+\frac{19}{4}\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{19}{4}\)
Ta thấy \(\left(x-\frac{3}{2}\right)^2\ge0\)
=> \(\left(x-\frac{3}{2}\right)^2+\frac{19}{4}\ge\frac{19}{4}\)
Mà \(\frac{19}{4}>0\)
=> \(\left(x-\frac{3}{2}\right)^2+\frac{19}{4}>0\) với mọi x .
=> \(x^2-3x+7>0\forall x\)
b, Ta có : \(A=4x^2+4x+15\)
=> \(A=\left(2x+1\right)^2+14\)
Ta thấy : \(\left(2x+1\right)^2\ge0\)
=> \(\left(2x+1\right)^2+14\ge14\)
Vậy MinA = 14 khi 2x + 1 = 0 <=> \(x=-\frac{1}{2}\)
