Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
C = x2 - 6x + 11 = x2 - 6x + 9 + 2 = (x - 3)2 + 2 > 2
Vậy Min C = 2 <=> x = 3
\(P=a^2+a+1\)
\(=a^2+\frac{1}{2}\cdot2\cdot a+\frac{1}{4}+\frac{3}{4}\)
\(=\left(a+\frac{1}{2}\right)^2+\frac{3}{4}\)
\(\left(a+\frac{1}{2}\right)^2\ge0\Rightarrow\left(a+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
\(\Rightarrow P\ge\frac{3}{4}\)
dấu "=" xảy ra khi :
\(\left(a+\frac{1}{2}\right)^2=0\Rightarrow a+\frac{1}{2}=0\Rightarrow a=-\frac{1}{2}\)
vậy
Bài 1:
\(x^2-x+1=x^2-x+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu "=" khi \(x=\frac{1}{2}\)
Vậy \(Min=\frac{3}{4}\) khi \(x=\frac{1}{2}\)
Bài 2:
\(x^2+10x+2041=x^2+10x+25+2016\)
\(=\left(x^2+10x+25\right)+2016\)
\(=\left(x+5\right)^2+2016\ge2016\)
Dấu "=" khi \(x=-5\)
Vậy \(Min=2016\) khi \(x=-5\)
(a + 1)(a + 3)(a + 5)(a + 7)+15 = (a + 1)(a + 7)(a + 5)(a + 3)+15 = (a2 + 8a +7)(a2 + 8a +15) + 15
Đặt a2 + 8a + 11 = x => (a2 + 8a +7)(a2 + 8a +15) + 15 = (x-4)(x+4) +15 = x2 - 4 + 15 = x2 +11 \(\ge\)11
GTNN (a + 1)(a + 3)(a + 5)(a + 7) + 15 = 11 \(\Leftrightarrow\)x =0 \(\Leftrightarrow\)a2 + 8a + 11 = 0 \(\Leftrightarrow\)(a + 4)2 -5 = 0 \(\Leftrightarrow\)(a + 4 +\(\sqrt{5}\))(a + 4 -\(\sqrt{5}\))=0 \(\Leftrightarrow\)a = -4+ \(\sqrt{5}\)hoặc a = -4 - \(\sqrt{5}\)