\(A=\frac{x^2-2x+2014}{x^2}=1-\frac{2}{x}+\frac{2014}{x^2}\)

Đặt \(\frac{1}{x}=a\)

=> \(A=1-2a+2014a^2\)

<=>\(A=2014\left(a^2-\frac{1}{1007}a+\frac{1}{2014}\right)\)

<=>\(A=2014\left(a^2-2\times a\times\frac{1}{2014}+\frac{1}{2014^2}-\frac{1}{2014^2}+\frac{1}{2014}\right)\)

<=>\(A=2014\left[\left(a-\frac{1}{2014}\right)^2+\left(\frac{1}{2014}-\frac{1}{2014^2}\right)\right]\)

<=>\(A=2014\left(a-\frac{1}{2014}\right)^2+2014\left(\frac{1}{2014}-\frac{1}{2014^2}\right)\)

<=>\(A=2014\left(a-\frac{1}{2014}\right)^2+1-\frac{1}{2014}\)

<=>\(A=2014\left(a-\frac{1}{2014}^2\right)+\frac{2013}{2014}\ge\frac{2013}{2014}\)

Vậy A đạt GTNN <=> \(A=\frac{2013}{2014}<=>a=\frac{1}{x}=\frac{1}{2014}<=>x=2014\)

Amin = 0 khi và chỉ khi x = 0

nhầm Amin = \(\frac{2013}{2014}\) khi và chỉ khi x = 2014 mình làm theo miền giá trị

\(D=\frac{x^2-2x+2014}{x^2}\)

\(D=\frac{x^2}{x^2}-\frac{2x}{x^2}+\frac{2014}{x^2}\)

\(D=1-\frac{2}{x}+\frac{2014}{x^2}\)

\(D=2014\cdot\frac{1}{x^2}-2\cdot\frac{1}{x}+1\)

Đặt \(\frac{1}{x}=a\)

\(D=2014a^2-2a+1\)

\(D=2014\left(a^2-a\cdot\frac{1}{1007}+\frac{1}{2014}\right)\)

\(D=2014\left(a^2-2\cdot a\cdot\frac{1}{2014}+\frac{1}{2014^2}+\frac{2013}{2014^2}\right)\)

\(D=2014\left[\left(a-\frac{1}{2014}\right)^2+\frac{2013}{2014^2}\right]\)

\(D=2014\left(a-\frac{1}{2014}\right)^2+\frac{2013}{2014}\ge\frac{2013}{2014}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow a=\frac{1}{2014}\Leftrightarrow\frac{1}{x}=\frac{1}{2014}\Leftrightarrow x=2014\)

Vậy....

\(A=\frac{1}{2017}-\frac{2}{2017x}+\frac{1}{x^2}=\left(\frac{1}{2017}-\frac{1}{x}\right)^2+\frac{1}{2017}-\frac{1}{2017^2}=\left(\frac{1}{2017}-\frac{1}{x}\right)^2+\frac{2016}{2017^2}\)

\(\Rightarrow A\ge\frac{2016}{2017^2}\)Dấu "=" xảy ra khi \(\left(\frac{1}{2017}-\frac{1}{x}\right)^2=0\Rightarrow x=2017\)

Vây ......

\(A=\frac{x^2-2x+2014}{x^2}\)

Ta có :

\(\frac{x^2-2x+2014}{x^2}-\frac{2013}{2014}=\frac{2014x^2-2.2014.x+2014^2-2013x^2}{2014x^2}=\frac{x^2-2.2004.x+2014^2}{2014x^2}=\frac{\left(x-2014\right)^2}{2014x^2}\ge\frac{2013}{2014}\)

\(\Rightarrow A\ge\frac{2013}{2014}\)

Dấu " = " xảy ra khi và chỉ khi \(x=2014\)

Vậy \(Min_A=\frac{2013}{2014}\Leftrightarrow x=2014\)

1.(√x -2)^2 ≥ 0 --> x -4√x +4 ≥ 0 --> x+16 ≥ 12 +4√x --> (x+16)/(3+√x) ≥4 
--> Pmin=4 khi x=4

2. Đặt \(\sqrt{x^2-4x+5}=t\ge1\)1

=> M=2x²-8x+\(\sqrt{x^2-4x+5}\)+6=2(t²-5)+t+6

<=> M=2t²+t-4\(\ge\)2.1²+1-4=-1

M_min=-1 khi t=1 hay x=2

\(A-\frac{2013}{2014}=\frac{x^2-2x+2014}{x^2}-\frac{2013}{2014}=\frac{2014x^2-2.2014.x+2014^2-2013x^2}{2014x^2}\)

\(=\frac{x^2-2.x.2014+2014^2}{2014x^2}=\frac{\left(x-2014\right)^2}{2014x^2}\ge0\)

=>\(A\ge\frac{2013}{2014}\)

Dấu "=" xảy ra khi x=2014

Vậy minA=2013/2014 khi x=2014

A=\(\frac{2014x^2-2.2014x-2014^2}{2014x^2}\)=\(\frac{2013x^2+\left(x^2-2.2014x-2014^2\right)}{2014x^2}\)=\(\frac{2013x^2+\left(x-2014\right)^2}{2014x^2}\)=\(\frac{2013}{2014}+\frac{\left(x-2014\right)^2}{2014x^2}\ge\frac{2013}{2014}\)

vậy minA=\(\frac{2013}{2014}\)dấu bằng xảy ra khi x=2014

Tớ có cách khác hay nè :

\(A=\frac{x^2-2x+2014}{x^2}\)

\(A=\frac{x^2}{x^2}-\frac{2x}{x^2}+\frac{2014}{x^2}\)

\(A=1-\frac{2}{x}+\frac{2014}{x^2}\)

\(A=1-2\cdot\frac{1}{x}+2014\cdot\left(\frac{1}{x}\right)^2\)

Đặt \(a=\frac{1}{x}\)ta có :

\(A=1-2a+2014a^2\)

\(A=2014\cdot\left(a^2-\frac{1}{1007}a+\frac{1}{2014}\right)\)

\(A=2014\cdot\left[a^2-2\cdot a\cdot\frac{1}{2014}+\left(\frac{1}{2014}\right)^2+\frac{2013}{2014^2}\right]\)

\(A=2014\cdot\left[\left(a-\frac{1}{2014}\right)^2+\frac{2013}{2014^2}\right]\)

\(A=2014\cdot\left(a-\frac{1}{2014}\right)^2+\frac{2013}{2014}\ge\frac{2013}{2014}\forall a\)

Dấu "=" xảy ra \(\Leftrightarrow a=\frac{1}{2014}\)

\(\Leftrightarrow\frac{1}{x}=\frac{1}{2014}\)

\(\Leftrightarrow x=2014\)

Vậy Amin = \(\frac{2013}{2014}\)khi và chỉ khi x = 2014