\(A=5x^2+10x-3=5\left(x^2+2x+1\right)-8=5\left(x+1\right)^2-8\ge-8\)
Vậy \(MinA=-8\Leftrightarrow5\left(x+1\right)^2=0\Leftrightarrow x=-1\)
\(B=3x^2+3x-5=3\left(x^2+x+\frac{1}{4}\right)-5,75=3\left(x+\frac{1}{2}\right)^2-5,75\ge-5,75\)
Vậy \(MinB=-5,75\Leftrightarrow3\left(x+\frac{1}{2}\right)^2=0\Leftrightarrow x=-\frac{1}{2}\)
\(C=2x^2-3x-1=2\left(x^2-\frac{3}{2}x+\frac{9}{16}\right)-\frac{17}{8}=2\left(x-\frac{3}{4}\right)^2-\frac{17}{8}\ge-\frac{17}{8}\)
Vậy \(MinC=-\frac{17}{8}\Leftrightarrow2\left(x-\frac{3}{4}\right)^2=0\Leftrightarrow x=\frac{3}{4}\)
Bùi Hoàng Linh Chi
Min là giá trị nhỏ nhất
A=\(5x^2+10x-3\)
=\(5x^2+10x+5-8\)
=\(5.\left(x^2+2x+1\right)-8\)
=\(5.\left(x+1\right)^2-8\)
Vì \(\left(x+1\right)^2\ge0\forall x\)nên \(5.\left(x+1\right)^2\ge0\)\(\Rightarrow5.\left(x+1\right)^2-8\ge-8\forall x\)
Do đó min của A = -8 \(\Leftrightarrow\left(x+1^{ }\right)^2=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Vậy min của A=-8\(\Leftrightarrow x=-1\)
