\(Q=\frac{2010,2011}{2014,2015}+\frac{2012,2013}{2014,2015}.\frac{1}{x^2}-\frac{2}{2014,2015}.\frac{1}{x}\)

\(=\frac{2010,2011}{2014,2015}+\frac{2012,2013}{2014,2015}\left(\frac{1}{x^2}-2.\frac{1}{x}.\frac{1}{2012,2013}+\frac{1}{\left(2012,2013\right)^2}\right)-\frac{1}{2012,2013}.\frac{1}{2014,2015}\)

\(=\frac{2010,2011}{2014,2015}-\frac{1}{2012,2013}.\frac{1}{2014,2015}+\frac{2012,2013}{2014,2015}\left(\frac{1}{x}-\frac{1}{2012,2013}\right)^2\)

\(\ge\frac{2010,2011}{2014,2015}-\frac{1}{2012,2013}.\frac{1}{2014,2015}\)

Dấu "=" xảy ra khi và chỉ khi \(x=2012,2013\)

Vậy GTNN của biểu thức là .....

1.

Vì x>0 nên \(A=\frac{16x+4+\frac{1}{x}}{2}\)

Áp dụng bất đẳng thức Côsi cho 2 số dương

\(16x+\frac{1}{x}\ge2\sqrt{16x.\frac{1}{x}}=2.4=8\). Dấu "=" khi \(16x=\frac{1}{x}\Rightarrow x^2=\frac{1}{16}\Rightarrow x=\frac{1}{4}\)

\(A=\frac{16x+4+\frac{1}{x}}{2}\ge\frac{8+4}{2}=6\)

Vậy GTNN của A là 6 khi \(x=\frac{1}{4}\)

2.

\(B=\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}=\frac{10}{ab}\)

Ta có: \(10=a+b\ge2\sqrt{ab}\Rightarrow\sqrt{ab}\le5\Rightarrow ab\le25\). Dấu "=" khi a = b = 5

\(\Rightarrow B=\frac{10}{ab}\ge\frac{10}{25}=\frac{2}{5}\)

Vậy GTNN của B là \(\frac{2}{5}\)khi a = b = 5

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

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

\(A=\frac{6x^2-2x+1}{x^2}=6-\frac{2}{x}+\frac{1}{x^2}\)

Đặt \(\frac{1}{x}=a\)ta có

\(A=6-2a+a^2=a^2-2a+1+5=\left(a-1\right)^2+5\)

Vì\(\left(a+1\right)^2\ge0\forall a\)

\(\Rightarrow A\ge5\forall a\)

GTNN của A=5 <=>a+1=0 <=>a=-1 =>x=-1

1.  x≥1 <=> \(\frac{1}{x}\le1\Leftrightarrow\frac{1}{x}+1\le2\Leftrightarrow A\le2\Rightarrow MaxA=2\Leftrightarrow x=1\)

2. Áp dụng bđt cosi cho x>0. ta có: \(x+\frac{1}{x}\ge2\sqrt{x.\frac{1}{x}}=2\Leftrightarrow P\ge2\Rightarrow MinP=2\Leftrightarrow x=\frac{1}{x}\Leftrightarrow x=1\)

3: \(A=\frac{x^2+x+4}{x+1}=\frac{\left(x^2+2x+1\right)-\left(x+1\right)+4}{x+1}=x+1-1+\frac{4}{x+1}\)

áp dụng cosi cho 2 số dương ta có: \(x+1+\frac{4}{x+1}\ge2\sqrt{x+1.\frac{4}{x+1}}=2\Leftrightarrow A+1\ge2\Rightarrow A\ge3\Rightarrow MinA=3\Leftrightarrow x+1=\frac{4}{x+1}\Leftrightarrow x=1\)

x=1 nhe nhap minh di ma ket ban voi minh nhe

\(A=\frac{3}{2+\sqrt{-x^2+2x+7}}=\frac{3}{2+\sqrt{8-\left(x^2-2x+1\right)}}=\frac{3}{2+\sqrt{8-\left(x-1\right)^2}}\)

Vì \(\left(x-1\right)^2\le0\Rightarrow8-\left(x-1\right)^2\le8\Rightarrow\sqrt{8-\left(x-1\right)^2}\le\sqrt{8}\)

\(\Rightarrow2+\sqrt{8-\left(x-1\right)^2}\le2+\sqrt{8}\)=>\(A=\frac{3}{2+\sqrt{8-\left(x-1\right)^2}}\ge\frac{3}{2+\sqrt{8}}\)

Dấu "=" xảy ra khi x=1

Vậy minA=\(\frac{3}{2+\sqrt{8}}\) khi x=1

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

Ta có: (x+1)² \(\ge0\)với \(\forall x\)Dấu "=" xảy ra khi x= -1

2017x² \(\ge0\)với \(\forall x\)Dấu "=" xảy ra khi x = 0

Suy ra \(\frac{\left(x+1\right)^2}{2017x^2}\)\(\ge\)0 với \(\forall x\)

<=> \(\frac{\left(x+1\right)^2+2016}{2017x^2}\)\(\ge\)2016 với \(\forall x\)

Mình nghĩ thế! 

\(A\ge\frac{2016}{2017^2}\)đẳng thức khi \(x=-4034\)