Áp dụng BĐT AM-GM ta có:

\(A=\frac{2011x+2012\sqrt{1-x^2}+2013}{\sqrt{1-x^2}}\)\(=\frac{2011x+2013}{\sqrt{1-x^2}}+2012\)

\(=\frac{2012\left(x+1\right)+\left(1-x\right)}{\sqrt{1-x^2}}+2012\)\(\ge\frac{2\sqrt{2012\left(x+1\right)\left(1-x\right)}}{\sqrt{1-x^2}}+2012\)

\(\ge\frac{2\sqrt{2012\left(1-x^2\right)}}{\sqrt{1-x^2}}+2012=2\sqrt{2012}+2012\)

\(ĐKXĐ:x\ge0\)

Ta có : \(D=\frac{2011x-2\sqrt{x}+1}{\sqrt{x}}=2011\sqrt{x}+\frac{1}{\sqrt{x}}-2\)

Theo BĐT AM - GM ta có :

\(2011\sqrt{x}+\frac{1}{\sqrt{x}}\ge2\sqrt{2011\sqrt{x}\cdot\frac{1}{\sqrt{x}}}=2\sqrt{2011}\)

\(\Rightarrow2011\sqrt{x}+\frac{1}{\sqrt{x}}-2\ge2\left(\sqrt{2011}-1\right)\)

Dấu "=" xảy ra \(\Leftrightarrow x=\frac{1}{2011}\)

Vậy \(D_{min}=2\left(\sqrt{2011}-1\right)\) tại \(x=\frac{1}{2011}\)

ĐKXĐ : \(x\ge0\)

Đặt \(A=\frac{2011x-2\sqrt{x}+1}{\sqrt{x}}\)

\(=2011\sqrt{x}-2+\frac{1}{\sqrt{x}}\)

\(=2011\sqrt{x}+\frac{1}{\sqrt{x}}-2\)

Áp dụng BĐT AM - GM cho hai số dương ta có :

\(2011\sqrt{x}+\frac{1}{\sqrt{x}}\ge2\sqrt{2011\sqrt{x}\cdot\frac{1}{\sqrt{x}}}=2\sqrt{2011}\)

Do đó : \(A\ge2\left(\sqrt{2011}-1\right)\)

Dấu "=" xảy ra \(\Leftrightarrow x=\frac{1}{2011}\)

Vậy \(A_{min}=2\left(\sqrt{2011}-1\right)\) khi \(x=\frac{1}{2011}\)

\(ĐK:x>0\)

Xét biểu thức\(\frac{2011x-2\sqrt{x}+1}{\sqrt{x}}=\frac{2011x-2\sqrt{x}+1}{\sqrt{x}}-2\left(\sqrt{2011}-1\right)+2\left(\sqrt{2011}-1\right)\)\(=\frac{2011x-2\sqrt{x}+1-2\sqrt{2011x}+2\sqrt{x}}{\sqrt{x}}+2\left(\sqrt{2011}-1\right)\)\(=\frac{\left(\sqrt{2011x}-1\right)^2}{\sqrt{x}}+2\left(\sqrt{2011}-1\right)\ge2\left(\sqrt{2011}-1\right)\)

\(\Rightarrow\frac{2011x-2\sqrt{x}+1}{\sqrt{x}}\ge2\left(\sqrt{2011}-1\right)\)

Đẳng thức xảy ra khi \(\sqrt{2011x}=1\Leftrightarrow2011x=1\Leftrightarrow x=\frac{1}{2011}\)

Vậy giá trị nhỏ nhất của biểu thức là \(2\left(\sqrt{2011}-1\right)\), đạt được khi \(x=\frac{1}{2011}\)

\(b,ĐKXĐ:x>0\)

\(D=2011\sqrt{x}-2+\frac{1}{\sqrt{x}}\)\(=2011\sqrt{x}+\frac{1}{\sqrt{x}}-2\)

Áp dụng bđt Cauchy cho 2 số dương \(2011\sqrt{x}\)và\(\frac{1}{\sqrt{x}}\)ta được:

\(2011\sqrt{x}+\frac{1}{\sqrt{x}}\ge2\sqrt{2011\sqrt{x}.\frac{1}{\sqrt{x}}}\)

\(\Leftrightarrow2011\sqrt{x}+\frac{1}{\sqrt{x}}-2\ge2\sqrt{2011}-2\)

\(\Leftrightarrow D\ge2\sqrt{2011}-2\)

Dấu "=" xảy ra \(\Leftrightarrow2011\sqrt{x}=\frac{1}{\sqrt{x}}\Leftrightarrow x=\frac{1}{2011}\left(TMĐK\right)\)

\(B=2x+3\sqrt{x}-28\)

Ta có điều kiện: \(x\ge0\)

Do đó \(B\ge2\cdot0+3\cdot0-28=-28\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\)

\(C=\frac{2011x-2\sqrt{x}+1}{\sqrt{x}}\)

\(C=2011\sqrt{x}-2+\frac{1}{\sqrt{x}}\)

Áp dụng bất đẳng thức Cô-si :

\(C\ge2\sqrt{\frac{2011\sqrt{x}}{\sqrt{x}}}-2=2\sqrt{2011}-2\)

Dấu "=" xảy ra \(\Leftrightarrow2011\sqrt{x}=\frac{1}{\sqrt{x}}\Leftrightarrow x=\frac{1}{2011}\)

Đặt \(\sqrt{2x-1}=a\ge0\)

Ta có \(2011x^2-a^2=2010xa\)

\(\Leftrightarrow\left(2010x^2-2010xa\right)+\left(x^2-a^2\right)=0\)

\(\Leftrightarrow\left(x-a\right)\left(2010x+x+a\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=a\\2011x=-a\left(loai\right)\end{cases}}\)

\(\Leftrightarrow x=1\)