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

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

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

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

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

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

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

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

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 \(\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{6}\end{cases}}\)

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

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

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

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

Đặt \(\left|3x-1\right|=a\ge0\)

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

\(\Rightarrow A_{min}=\frac{3}{4}\) khi \(a=\frac{1}{2}\Leftrightarrow\left|3x-1\right|=\frac{1}{2}\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{1}{6}\end{matrix}\right.\)

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

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

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

Dấu '=' xảy ra khi 3x-1=1/2 hoặc 3x-1=-1/2

=>3x=3/2 hoặc 3x=1/2

=>x=1/6 hoặc x=1/2

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

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

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

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

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

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

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 \(\dfrac{1}{3}\le x\le\dfrac{2}{3}\)

a/ \(A=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=\left[\left(x+1\right)\left(x-6\right)\right].\left[\left(x-2\right)\left(x-3\right)\right]\)

\(=\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\ge-36\)

Suy ra Min A = -36 <=> \(x^2-5x=0\Leftrightarrow x\left(x-5\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=5\end{array}\right.\)

b/ \(B=19-6x-9x^2=-9\left(x-\frac{1}{3}\right)^2+20\le20\)

Suy ra Min B = 20 <=> x = 1/3

a) \(A=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)\)

\(=\left[\left(x+1\right)\left(x-6\right)\right]\left[\left(x-2\right)\left(x-3\right)\right]\)

\(\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\)

Vì \(\left(x^2-5x\right)^2\ge0\)

=> \(\left(x^2-5x\right)^2-36\ge-36\)

Vậy GTNN của A là -36 khi \(x^2-5x=0\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=5\end{array}\right.\)

b) \(B=19-6x-9x^2=-\left(9x^2+6x+1\right)+20=-\left(3x+1\right)^2+20\)

Vì \(-\left(3x+1\right)^2\le0\)

=> \(-\left(3x+1\right)+20\le20\)

Vậy GTLN của B là 20 khi \(x=-\frac{1}{3}\)

B = 19 - 6x - 9x²

= - (9x² + 6x + 1 - 20)

= - [(3x + 1)² - 20]

(3x + 1)² lớn hơn hoặc bằng 0

(3x + 1)² + 20 lớn hơn hoặc bằng 20

- [(3x + 1)² + 20] nhỏ hơn hoặc bằng - 20

Vậy Max B = - 20 khi x = -1/3

Lời giải:

$G=1-\sqrt{(3x-1)^2}+(3x-1)^2=1-|3x-1|+|3x-1|^2$

Đặt $|3x-1|=a$ với $a\geq 0$

Ta cần tìm GTNN của $G=1-a+a^2$

Có: $G=(a-\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}$ với mọi $a\geq 0$

Do đó gtnn của $G$ là $\frac{3}{4}$