Ta có: \(A=\frac{3-2x}{3x-1}\)

=> \(3A=\frac{-\left(6x-9\right)}{3x-1}=\frac{-2\left(3x-1\right)+7}{3x-1}=-2+\frac{7}{3x-1}\)

Để A đạt GTLN <=> 3A đạt GTLN

<=> \(\frac{7}{3x-1}\)đạt GTLN 

<=> \(3x-1\)đạt GTNN

Do x \(\in\)Z <=> 3x - 1 = 2  <=> 3x = 3 <=> x = 1

Thay x = 1 vào 3A, ta có: \(-2+\frac{7}{3.1-1}=-2+7=5\)

        => \(A=5:3=\frac{5}{3}\)

Vậy x = 1 (x \(\in\)Z) thì A = \(\frac{3-2x}{3x-1}\)đạt Max

Cách khác nè. Ko bt đg ko

\(A=\frac{3-2x}{3x-1}\Leftrightarrow-A=\frac{2x-3}{3x-1}\Leftrightarrow-3A=\frac{6x-9}{3x-1}\)

\(\Leftrightarrow-3A=\frac{2\left(3x-1\right)-7}{3x-1}=2-\frac{7}{3x-1}\)

-3A đạt GTNN\(\Leftrightarrow\frac{7}{3x-1}\)đạt GTLN\(\Leftrightarrow3x-1\)đạt GTNN\(\Leftrightarrow3x-1=2\Leftrightarrow x=1\)(Vì x thuộc Z)

Khi đó \(\Leftrightarrow-3A==2-\frac{7}{3.1-1}=\frac{-3}{2}\)

Vậy GTLN của A là \(\frac{1}{2}\Leftrightarrow x=1\)

Sai thì mấy anh chị góp ý

\(ĐKXĐ:x\ge0\)

\(\sqrt{x}+6\)luôn lớn hơn 4 thỏa mãn ĐKXĐ:

Vậy \(x\ge0\) là giá trị cần tìm

\(\sqrt{x}+6>4\Rightarrow\sqrt{x}>-2\)

Luôn đúng với mọi giá trị của x 

Ta có \(VT=\frac{a^2}{a\sqrt{b}}+\frac{b^2}{b\sqrt{c}}+\frac{c^2}{c\sqrt{a}}\ge\frac{\left(a+b+c\right)^2}{a\sqrt{b}+b\sqrt{c}+c\sqrt{a}}\)

Mà \(a\sqrt{b}\le\frac{a^2+b}{2},b\sqrt{c}\le\frac{b^2+c}{2},c\sqrt{a}\le\frac{c^2+a}{2}\)

=> \(VT\ge\frac{2\left(a+b+c\right)^2}{a^2+b^2+c^2+a+b+c}=\frac{2\left(a+b+c\right)^2}{3+a+b+c}\)

Lại có \(a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\)

=> \(VT\ge\frac{2\left(a+b+c\right)^2}{3+3}=\frac{\left(a+b+c\right)^2}{3}\ge ab+bc+ac\)(ĐPCM)

Dấu bằng xảy ra khi a=b=c=1

\(D=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\)

\(=\frac{a-2\sqrt{ab}+b+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\frac{a+2\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\left(\sqrt{a}-\sqrt{b}\right)\)

\(=\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}\)

\(=2\sqrt{b}\)

\(D=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\)

\(D=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{-b+\sqrt{a}.\sqrt{b}}{\sqrt{b}}\)

\(D=\frac{\left[\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}\right].\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right).\sqrt{b}}-\frac{\left(\sqrt{a}.\sqrt{b}-b\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}.\left(\sqrt{a}+\sqrt{b}\right)}\)

\(D=\frac{\left[\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}\right]-\left(\sqrt{a}.\sqrt{b}-b\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}.\left(\sqrt{a}+\sqrt{b}\right)}\)

\(D=\frac{2b.\sqrt{a}+2b.\sqrt{b}}{\sqrt{b}.\left(\sqrt{a}+\sqrt{b}\right)}\)

\(D=\frac{2b.\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}\)

\(D=2\sqrt{b}\)