a) Ta có: 

\(Q=\sqrt{\left(1-3x\right)\left(x+\dfrac{1}{2}\right)}\) Q có nghĩa khi:

\(\left(1-3x\right)\left(x+\dfrac{1}{2}\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}1-3x\ge0\\x+\dfrac{1}{2}\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}1-3x\le0\\x+\dfrac{1}{2}\le\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x\le1\\x\ge-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}3x\ge1\\x\le-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\x\ge-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\x\le-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{1}{2}\le x\le\dfrac{1}{3}\\x\in\varnothing\end{matrix}\right.\)

\(\Leftrightarrow-\dfrac{1}{2}\le x\le\dfrac{1}{3}\)

b) Ta có: \(Q=\sqrt{\left(1-3x\right)\left(x+\dfrac{1}{2}\right)}\)

\(Q=\sqrt{x+\dfrac{1}{2}-3x^2-\dfrac{3}{2}x}\)

\(Q=\sqrt{-\left(3x^2+\dfrac{1}{2}x-\dfrac{1}{2}\right)}\)

\(Q=\sqrt{-3\left(x^2+\dfrac{1}{6}x-\dfrac{1}{6}\right)}\)

\(Q=\sqrt{-3\left(x^2+2\cdot\dfrac{1}{12}\cdot x+\dfrac{1}{144}-\dfrac{25}{144}\right)}\)

\(Q=\sqrt{-3\left(x+\dfrac{1}{12}\right)^2+\dfrac{25}{144}}\)

Mà: \(Q=\sqrt{-3\left(x+\dfrac{1}{12}\right)^2+\dfrac{25}{144}}\le\sqrt{\dfrac{25}{144}}=\dfrac{5}{12}\)

Dấu "=" xảy ra khi:

\(\Leftrightarrow-3\left(x+\dfrac{1}{12}\right)^2=0\)

\(\Leftrightarrow x+\dfrac{1}{12}=0\)

\(\Leftrightarrow x=-\dfrac{1}{12}\)

Vậy: \(Q_{max}=\dfrac{5}{12}.khi.x=-\dfrac{1}{12}\)

\(A=1-\sqrt{1-6x+9x^2}+\left(3x-1\right)^2\)

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

\(A=1-\left(3x-1\right)+\left(3x-1\right)^2\)

\(A=1-3x+1+9x^2-6x+1\)

\(A=9x^2-9x+3\)

\(A=\left(3x\right)^2-2.3x.\frac{9}{6}+\frac{81}{36}-\frac{27}{36}\)

\(A=\left(3x-\frac{9}{6}\right)^2-\frac{27}{36}\)

\(A=\left(3x-\frac{9}{6}\right)^2-\frac{3}{4}\ge0\forall x\)

Dấu = xảy ra khi:

\(3x-\frac{9}{6}=0\Leftrightarrow3x=\frac{9}{6}\Leftrightarrow x=0,5\)

Vậy A_min = -3/4 tại x = 0,5

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

A=1-/3x-1/+(3x-1)^2

đặt t=/3x-1/ với t>=0

khi đó A=t^2-t+1

A=t^2-t+1/4+3/4

A=(t-1/2)^2+3/4

khi đó A>=3/4

dấu bằng xảy ra khi t=1/2 hay x=1/2

Chúc bạn học tốt!

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

\(A=\left(3x-1\right)^2-\left|3x-1\right|+1\)

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

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=\frac{1}{2}\) ( tm ) 

+) Với \(x< \frac{1}{3}\)\(\Rightarrow\)\(A=\left(3x-1\right)^2+\left(3x-1\right)+\frac{1}{4}+\frac{3}{4}=\left(3x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=\frac{1}{6}\) ( tm ) 

Vậy GTNN của \(A=\frac{3}{4}\) khi \(\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{6}\end{cases}}\)

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?

BĐT \(\left(x+y\right)^2\ge4xy\)nhe bạn

a: \(C=\dfrac{3x+3\sqrt{x}-3-x+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\dfrac{1-1+\sqrt{x}}{1-\sqrt{x}}\)

\(=\dfrac{2x+3\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)

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

b: Để \(C=\sqrt{x}\) thì \(x-\sqrt{x}=\sqrt{x}+1\)

=>\(x-2\sqrt{x}-1=0\)

=>\(\Leftrightarrow x=3+2\sqrt{2}\)

c: |2x-5|=3

=>2x-5=3 hoặc 2x-5=-3

=>2x=2 hoặc 2x=8

=>x=4(nhận) hoặc x=1(loại)

Khi x=4 thì \(C=\dfrac{2+1}{2-1}=3\)