\(P=\sqrt[]{9x^2-6x+1}+\sqrt[]{25-30x+9x^2}\)

\(\Leftrightarrow P=\sqrt[]{\left(3x-1\right)^2}+\sqrt[]{\left(5-3x\right)^2}\)

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

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

Vậy \(GTNN\left(P\right)=4\)

P = 4

Ta có \(9x^2-6x+1=\left(3x-1\right)^2,25-30x+9x^2=\left(5-3x\right)^2.\)

Suy ra \(B=\left|3x-1\right|+\left|5-3x\right|\ge\left|3x-1+5-3x\right|=4.\) (Ở đây ta sử dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|,\) với dấu bằng xảy ra khi và chỉ khi \(ab\ge0\)).

Mà khi \(x=\frac{1}{3}\) thì \(B=4.\) Vậy giá trị nhỏ nhất của B là 4.

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

Dấu '=' xảy ra khi (3x-1)(3x-5)<=0

=>1/3<=x<=5/3

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

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

=|3x-1|+|5-3x| ≥ |3x-1+5-3x|

<=> |3x-1|+|5-3x| ≥ |4|

=> Min A =4 khi (3x-1)(5-3x) ≥ 0

ta có bảng

=> x ≤ 1/3 hoặc x ≥ 5/3

vậy .....

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

\(A=3x-1+5-3x=4\)

\(A\)có giá trị ko phụ thuộc vào biến x

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

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

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

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

\(9x^2-6x+2=\left(3x-1\right)^2+1=t\ge1\)

\(Pt\Rightarrow\sqrt{t}+\sqrt{5t-1}=\sqrt{10-t}\)

\(\Leftrightarrow5t-1=10-t+t-2\sqrt{t\left(10t-1\right)}\)

\(\Leftrightarrow2\sqrt{t\left(10t-1\right)}+5t=11\)

\(\Rightarrow VT\ge VP\left(t\ge1\right)\Rightarrow t=1\Rightarrow x=\frac{1}{3}\)