\(A=\sqrt{1-6x+9x^2}+\sqrt{9x^2-12x+4}\)

\(A=\sqrt{1^2-2\cdot3x\cdot1+\left(3x\right)^2}+\sqrt{\left(3x\right)^2-2\cdot2\cdot3x+2^2}\)

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

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

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

\(A=\left|-1\right|=1\)

Dấu "=" xảy ra \(\left(1-3x\right)\left(3x-2\right)\ge0\)

\(\Rightarrow\dfrac{1}{3}\le x\le\dfrac{2}{3}\)

Vậy: \(A_{min}=1\) khi \(\dfrac{1}{3}\le x\le\dfrac{2}{3}\)

\(Q=\sqrt{9x^2-6x+1}+\sqrt{25-30+9x^2}+2011\)

\(Q=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(5-3x\right)^2}+2011\)

\(Q=\left|3x-1\right|+\left|5-3x\right|+2011\)

Đặt \(Q'=\left|3x-1\right|+\left|5-3x\right|\ge\left|3x-1+5-3x\right|=4\)

Đẳng thức xảy ra \(\Leftrightarrow\left(3x-1\right)\left(5-3x\right)\ge0\)

\(\Leftrightarrow\frac{1}{3}\le x\le\frac{5}{3}\)

\(\Rightarrow Min_Q=Min_{Q'}+2011=4+2011=2015\)

Q = \(\sqrt{9x^2-6x+1}+\sqrt{25-30x+9x^2}+2011\)

Q = \(\sqrt{\left(3x-1\right)^2}+\sqrt{\left(3x-5\right)^2}+2011\)

Q = \(3x-1+3x-5+2011\)

Q = \(6x+2005\)

\(Q=\sqrt{9x^2-6x+1}+\sqrt{25-30x+9x^2}+2011\)

\(=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(3x-5\right)^2}+2011\)

\(=\left|3x-1\right|+\left|3x-5\right|+2011\)

Áp dụng BĐT \(\left|x\right|+\left|y\right|\ge\left|x+y\right|\)

\(\left|3x-1\right|+\left|3x-5\right|\ge\left|\left(3x-1\right)+\left(5-3x\right)\right|=4\)

(Dấu "="\(\Leftrightarrow\left(3x-1\right)\left(5-3x\right)\ge0\)

\(TH1:\hept{\begin{cases}3x-1\ge0\\5-3x\ge0\end{cases}}\Leftrightarrow\frac{1}{3}\le x\le\frac{5}{3}\)

​\(TH2:\hept{\begin{cases}3x-1\le0\\5-3x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le\frac{1}{3}\\x\ge\frac{3}{5}\end{cases}}\left(L\right)\)​)

\(\Rightarrow Q\ge2015\)

(Dấu "="\(\Leftrightarrow\frac{1}{3}\le x\le\frac{5}{3}\))

Vậy \(Q_{min}=2015\Leftrightarrow\frac{1}{3}\le x\le\frac{5}{3}\)

a) \(\sqrt{x}-x=-\left(x-\sqrt{x}\right)\)

\(=-\left[\left(\sqrt{x}\right)^2-2.\frac{1}{2}\sqrt{x}+\frac{1}{4}\right]+\frac{1}{4}\)

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

Vậy GTLN của bt là \(\frac{1}{4}\Leftrightarrow\sqrt{x}-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}\)

\(A=\left(x+3\right)^2+2\ge2\\ A_{min}=2\Leftrightarrow x=-3\\ B=\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{29}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{29}{4}\ge-\dfrac{29}{4}\\ B_{min}=-\dfrac{29}{4}\Leftrightarrow x=-\dfrac{3}{2}\\ C=\left(9x^2-12x+4\right)+2017=\left(3x-2\right)^2+2017\ge2017\\ C_{min}=2017\Leftrightarrow x=\dfrac{2}{3}\)

a: Sửa đề: \(M=3x-\sqrt[3]{27x^3+27x^2+9x+1}\)

\(=3x-\sqrt[3]{\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2+1^3}\)

\(=3x-\sqrt[3]{\left(3x+1\right)^3}\)

\(=3x-3x-1=-1\)

b: \(N=\sqrt[3]{8x^3+12x^2+6x+1}-\sqrt[3]{x^3}\)

\(=\sqrt[3]{\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2+1^3}-x\)

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

=2x+1-x

=x+1

a) \(\sqrt{1-6x+9x^2}=9\)

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

\(\Leftrightarrow\left|1-3x\right|=9\)

\(\Leftrightarrow\left[{}\begin{matrix}1-3x=9\\1-3x=-9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=1-9\\3x=1+9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=-8\\3x=10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{8}{3}\\x=\dfrac{10}{3}\end{matrix}\right.\)

b) \(\sqrt{2x-3}-\sqrt{x+1}=0\) (\(x\ge\dfrac{3}{2}\))

\(\Leftrightarrow\sqrt{2x-3}=\sqrt{x+1}\)

\(\Leftrightarrow2x-3=x+1\)

\(\Leftrightarrow2x-x=1+3\)

\(\Leftrightarrow x=4\left(tm\right)\)

c) \(\sqrt{9x^2+12+4}-2=3x\)

\(\Leftrightarrow\sqrt{\left(3x+2\right)^2}=3x+2\)

\(\Leftrightarrow\left|3x+2\right|=3x+2\)

\(\Leftrightarrow3x+2\ge0\)

\(\Leftrightarrow3x\ge-2\)

\(\Leftrightarrow x\ge-\dfrac{2}{3}\)

a: =>|3x-1|=9

=>3x-1=9 hoặc 3x-1=-9

=>x=-8/3 hoặc x=10/3

b: =>căn 2x-3=căn x+1

=>2x-3=x+1

=>x=4

c: =>|3x+2|=3x+2

=>3x+2>=0

=>x>=-2/3