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1.
$x(x+2)(x+4)(x+6)+8$
$=x(x+6)(x+2)(x+4)+8=(x^2+6x)(x^2+6x+8)+8$
$=a(a+8)+8$ (đặt $x^2+6x=a$)
$=a^2+8a+8=(a+4)^2-8=(x^2+6x+4)^2-8\geq -8$
Vậy $A_{\min}=-8$ khi $x^2+6x+4=0\Leftrightarrow x=-3\pm \sqrt{5}$
2.
$B=5+(1-x)(x+2)(x+3)(x+6)=5-(x-1)(x+6)(x+2)(x+3)$
$=5-(x^2+5x-6)(x^2+5x+6)$
$=5-[(x^2+5x)^2-6^2]$
$=41-(x^2+5x)^2\leq 41$
Vậy $B_{\max}=41$. Giá trị này đạt tại $x^2+5x=0\Leftrightarrow x=0$ hoặc $x=-5$
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x^2+x+1/4+3/4
=(x+1/2)^2+3/4
=> A min=3/4
Câu kia tương tự .......
\(A=x^2+x+1=x^2+2x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge0,x\in R\)
nên \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4},x\in R\)
Vậy \(Min_A=\frac{3}{4}\)khi \(x+\frac{1}{2}=0\Rightarrow x=-\frac{1}{2}\)
\(B=\left(x+2\right)^2+\left(x-3\right)^2=x^2+2x+1+x^2-6x+9=2x^2-4x+10=2\left(x^2-2x+5\right)\)
\(B=2\left(x^2-2x+1+4\right)=2\left(x-1\right)^2+4\)
Vì \(2\left(x-1\right)^2\ge0,x\in R\)
nên \(2\left(x-1\right)^2+4\ge4,x\in R\)
Vậy \(Min_B=4\)khi \(x-1=0\Rightarrow x=1\)
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a.
\(A=\dfrac{2013}{x^2}-\dfrac{2}{x}+1=2013\left(\dfrac{1}{x}-\dfrac{1}{2013}\right)^2+\dfrac{2012}{2013}\ge\dfrac{2012}{2013}\)
Dấu "=" xảy ra khi \(x=2013\)
b.
\(B=\dfrac{4x^2+2-4x^2+4x-1}{4x^2+2}=1-\dfrac{\left(2x-1\right)^2}{4x^2+2}\le1\)
\(B_{max}=1\) khi \(x=\dfrac{1}{2}\)
\(B=\dfrac{-2x^2-1+2x^2+4x+2}{4x^2+2}=-\dfrac{1}{2}+\dfrac{\left(x+1\right)^2}{2x^2+1}\ge-\dfrac{1}{2}\)
\(B_{max}=-\dfrac{1}{2}\) khi \(x=-1\)
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x+y=2=> y=2-x {cach co ban nhat}
y^2=(2-x)^2=4-4x+x^2
N=x^2+y^2=x^2+(4-4x+x^2)
N=2x^2-4x+4
N=2(x^2-2x+2)
{tach ghep BP}
\(N=2.\left[\left(x^2-2.x+1^2\right)+2-1\right]\)
\(N=2\left(x-1\right)^2+2\left(2-1\right)\)
N\(\ge\)2(2-1)=2
dang thuc xay ra khi x=1; y=1
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\(A=\dfrac{x^2}{x-2}=\dfrac{x^2-4+4}{x-2}=\dfrac{\left(x-2\right)\left(x+2\right)+4}{x-2}\)
\(=x+2+\dfrac{4}{x-2}=x-2+\dfrac{4}{x-2}+4\)
mà \(x-2+\dfrac{4}{x-2}\ge2.\sqrt[]{x-2.\dfrac{4}{x-2}}=2.2=4\) Bất đẳng thức Cauchy)
\(\Rightarrow A=x-2+\dfrac{4}{x-2}+4\ge8\)
\(\Rightarrow GTNN\left(A\right)=8\)
\(x^2+x-2\)
\(=x^2+x+\dfrac{1}{4}-\dfrac{9}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2-\dfrac{9}{4}>=-\dfrac{9}{4}\forall x\)
Dấu '=' xảy ra khi \(x+\dfrac{1}{2}=0\)
=>\(x=-\dfrac{1}{2}\)