\(a,4x^2+5x+3\)
\(=\left(4x^2+5x+\dfrac{25}{16}\right)+\dfrac{23}{16}\)
\(=\left(2x+\dfrac{5}{4}\right)^2+\dfrac{23}{16}\)
Với mọi giá trị của x ta có:
\(\left(2x+\dfrac{5}{4}\right)^2\ge0\Rightarrow\left(2x+\dfrac{5}{4}\right)^2+\dfrac{23}{16}\ge0\)
=>đpcm
b, \(7x^2-x+8=7\left(x^2-\dfrac{1}{7}x+\dfrac{1}{196}\right)+\dfrac{223}{28}\)
\(=\left(x-\dfrac{1}{14}\right)^2+\dfrac{223}{28}\)
Với mọi giá trị của x ta có:
\(\left(x-\dfrac{1}{14}\right)^2\ge0\Rightarrow\left(x-\dfrac{1}{14}\right)^2+\dfrac{223}{28}\ge0\)
=> đpcm
\(c,25x^2+8x+2017=25\left(x^2+\dfrac{8}{25}x+\dfrac{16}{625}\right)+\dfrac{50409}{25}\)\(=\left(x+\dfrac{4}{25}\right)^2+\dfrac{50409}{25}\)
Với mọi giá trị của x ta có:
\(\left(x+\dfrac{4}{25}\right)^2\ge0\Rightarrow\left(x+\dfrac{4}{25}\right)^2+\dfrac{50409}{25}\ge0\)=>đpcm
