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\(P=a^2+a+1\)
\(=a^2+\frac{1}{2}\cdot2\cdot a+\frac{1}{4}+\frac{3}{4}\)
\(=\left(a+\frac{1}{2}\right)^2+\frac{3}{4}\)
\(\left(a+\frac{1}{2}\right)^2\ge0\Rightarrow\left(a+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
\(\Rightarrow P\ge\frac{3}{4}\)
dấu "=" xảy ra khi :
\(\left(a+\frac{1}{2}\right)^2=0\Rightarrow a+\frac{1}{2}=0\Rightarrow a=-\frac{1}{2}\)
vậy
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Ta có:
M=/x+3/+/x-5/>=/x+3-x+5/
>=8
Vậy Min M=8 với mọi x
A=/x+13/+64
vì /x+13/>=0
=> /x+13/+64>=64
Vậy MinA=64 khi x=-13
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![](https://rs.olm.vn/images/avt/0.png?1311)
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\(A=x^2-x=x\left(x-1\right)\)
Với \(x\ge0\)
\(\Rightarrow A\ge0\)
Với\(x< 0\)
\(\Rightarrow x^2-x< 0\)
Vậy GTNN A là A < 0 <=> x < 0
Ta có :
B= x-x2
= -(x-x2)
= 1/4-(x2-x+1/4)
= 1/4-(x-1/2)2 < hoặc = 1/4
Vậy Bmax= 1/4 <=> x=1/2
GTNN 15
(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 - \(\sqrt{5}\)