\(A=0,6+\left|\dfrac{1}{2}-x\right|\\ Vì:\left|\dfrac{1}{2}-x\right|\ge\forall0x\in R\\ Nên:A=0,6+\left|\dfrac{1}{2}-x\right|\ge0,6\forall x\in R\\ Vậy:min_A=0,6\Leftrightarrow\left(\dfrac{1}{2}-x\right)=0\Leftrightarrow x=\dfrac{1}{2}\)

\(B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\\ Vì:\left|2x+\dfrac{2}{3}\right|\ge0\forall x\in R\\ Nên:B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\le\dfrac{2}{3}\forall x\in R\\ Vậy:max_B=\dfrac{2}{3}\Leftrightarrow\left|2x+\dfrac{2}{3}\right|=0\Leftrightarrow x=-\dfrac{1}{3}\)

a) ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne-2\end{cases}}\)

\(N=\frac{\left(x+2\right)^2}{x}.\left(1-\frac{x^2}{x+2}\right)-\frac{x^2+6x+4}{x}\)

\(N=\frac{\left(x+2\right)^2}{x}.\frac{x+2-x^2}{x+2}-\frac{x^2+6x+4}{x}\)

\(N=\frac{\left(x+2\right)\left(x+2-x^2\right)-x^2-6x-4}{x}\)

\(N=\frac{x^2+2x-x^3+2x+4-2x^2-x^2-6x-4}{x}\)

\(N=\frac{-x^3-2x^2-2x}{x}\)

\(N=\frac{-x\left(x^2+2x+2\right)}{x}\)

\(N=-\left(x^2+2x+2\right)\)

b) \(N=-\left(x^2+2x+2\right)\)

\(\Leftrightarrow N=-\left(x^2+2x+1+1\right)\)

\(\Leftrightarrow N=-\left(x+1\right)^2-1\le-1\)

Max N = -1 \(\Leftrightarrow x=-1\)

Vậy .......................

2. \(P=x^2-x\sqrt{3}+1=\left(x^2-x\sqrt{3}+\frac{3}{4}\right)+\frac{1}{4}=\left(x-\frac{\sqrt{3}}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\)

Dấu '=' xảy ra khi \(x=\frac{\sqrt{3}}{2}\)

Vây \(P_{min}=\frac{1}{4}\)khi \(x=\frac{\sqrt{3}}{2}\)

3. \(Y=\frac{x}{\left(x+2011\right)^2}\le\frac{x}{4x.2011}=\frac{1}{8044}\)

Dấu '=' xảy ra khi \(x=2011\)

Vây \(Y_{max}=\frac{1}{8044}\)khi \(x=2011\)

4. \(Q=\frac{1}{x-\sqrt{x}+2}=\frac{1}{\left(x-\sqrt{x}+\frac{1}{4}\right)+\frac{7}{4}}=\frac{1}{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{7}{4}}\le\frac{4}{7}\)

Dấu '=' xảy ra khi \(x=\frac{1}{4}\) 

Vậy \(Q_{max}=\frac{4}{7}\)khi \(x=\frac{1}{4}\)

Làm như thế nào ra \(\frac{x}{4x.2011}\)vậy bạn?

Dễ thấy: \(f\left(x\right)=\left(x+m-1\right)^2-m^2+5m-6\ge-m^2+5m-6\)

Giá trị nhỏ nhất của f(x) đạt lớn nhất tức \(-m^2+5m-6\) đạt lớn nhất

Mà \(g\left(m\right)=-m^2+5m-6=-\left(m-\dfrac{5}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)

g(m) đạt lớn nhất khi m=5/2

m cần tìm là 5/2