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KT
2 tháng 3 2020
\(P=\frac{1}{16}\left(4x^2+4x+7\right)\left(2x+1\right)^2+\frac{9}{16}\ge\frac{9}{16}\)
"=" \(\Leftrightarrow x=-\frac{1}{2}\)
2 tháng 3 2020
P=x4+2x3+3x2+2x+1
= x^4 + 2x^2(x + 1) + x^2 + 2x + 1
= x^4 + 2x^2(x + 1) + (x + 1)^2
= (x^2 + x + 1)^2 > 0
xét P = 0 khi x^2 + x + 1 = 0
<=> (x+1/2)^2 + 3/4 = 0
<=> (x+1/2)^2 = -3/4
=> x thuộc tập hợp rỗng
S
0
LT
2
JY
1
M
24 tháng 12 2019
Bài giải
a, Ta có : \(A=\frac{x^2-2+1995}{x^2}=\frac{x^2}{x^2}-\frac{2+1995}{x^2}=1-\frac{1997}{x^2}\)
\(A\text{ đạt GTNN khi }\frac{1997}{x^2}\text{ đạt GTLN}\)
\(\Rightarrow\text{ }x^2\text{ nhỏ nhất }\left(x\ne0\right)\) Mà \(x^2\ge0\text{ }\Rightarrow\text{ }x^2=1\text{ }\Rightarrow\text{ }x\in\left\{\pm1\right\}\)
\(\Rightarrow\text{ Min A }=1-\frac{1997}{1}=1-1997=-1996\)
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