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

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

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

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

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

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

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 \(\dfrac{1}{3}\le x\le\dfrac{2}{3}\)

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

Dấu '=' xảy ra khi (3x-1)(3x-5)<=0

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

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

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

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

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

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

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

=|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) \(A=\sqrt{4x^2+4x+2}=\sqrt{4x^2+4x+1+1}=\sqrt{\left(2x+1\right)^2+1}\)

Vì \(\left(2x+1\right)^2\ge0\forall x\)\(\Rightarrow\left(2x+1\right)^2+1\ge1\forall x\)

\(\Rightarrow A\ge\sqrt{1}=1\)

Dấu " = " xảy ra \(\Leftrightarrow2x+1=0\)\(\Leftrightarrow2x=-1\)\(\Leftrightarrow x=\frac{-1}{2}\)

Vậy \(minA=1\Leftrightarrow x=\frac{-1}{2}\)

b) \(B=\sqrt{2x^2-4x+5+1}=\sqrt{2x^2-4x+2+3+1}=\sqrt{2\left(x^2-2x+1\right)+4}\)

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

Vì \(\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow2\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow2\left(x-1\right)^2+4\ge4\forall x\)

\(\Rightarrow B\ge\sqrt{4}=2\)

Dấu " = " xảy ra \(\Leftrightarrow x-1=0\)\(\Leftrightarrow x=1\)

Vậy \(minB=2\Leftrightarrow x=1\)

Mơn bạn nha

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

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

\(\Rightarrow minA=\frac{3}{4}\Leftrightarrow x=\frac{1}{2}\)hoặc \(x=\frac{1}{6}\)

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

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

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

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

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.