\(Q=x^2\left(4-3x\right)=\dfrac{4}{9}.\dfrac{3}{2}x.\dfrac{3}{2}x\left(4-3x\right)\)

\(Q\le\dfrac{1}{27}.\dfrac{4}{9}.\left(\dfrac{3x}{2}+\dfrac{3x}{2}+4-3x\right)^3=\dfrac{256}{243}\)

\(Q_{maxx}=\dfrac{256}{243}\) khi \(\dfrac{3x}{2}=4-3x\Leftrightarrow x=\dfrac{8}{9}\)

A = \(\frac{3x}{2}+\frac{2}{x-1}=3.\frac{x-1}{2}+\frac{2}{x-1}+\frac{3}{2}\)\(\ge2\sqrt{3}+\frac{3}{2}\)

\(\Rightarrow\)min A = \(2\sqrt{3}+\frac{3}{2}\Leftrightarrow x=\frac{2}{\sqrt{3}}+1\)(thỏa mãn)

B = \(x+\frac{3}{3x-1}=\frac{1}{3}\left(3x-1+\frac{9}{3x-1}+1\right)\)\(\ge\frac{1}{3}\left(2\sqrt{9}+1\right)=\frac{7}{3}\)

\(\Rightarrow\)min B = \(\frac{7}{3}\Leftrightarrow x=\frac{4}{3}\)

\(A\) \(=\) \(3x^2\left(8-x^2\right)\le3\frac{\left(x^2+8-x^2\right)^2}{4}=48\)

\(\Rightarrow\) maxA = 48 \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)(thỏa mãn)

\(B=\) \(4x\left(8-5x\right)\)\(=\frac{4}{5}.5x\left(8-5x\right)\le\frac{4}{5}.\frac{\left(5x+8-5x\right)^2}{4}=\frac{64}{5}\)

\(\Rightarrow\)max B = \(\frac{64}{5}\Leftrightarrow x=\frac{4}{5}\)(thỏa mãn)

C = \(4\left(x-1\right)\left(8-5x\right)=\frac{4}{5}.\left(5x-5\right)\left(8-5x\right)\)\(\le\frac{4}{5}.\frac{\left(5x-5+8-5x\right)^2}{4}=\frac{9}{5}\)

\(\Rightarrow\)max C = \(\frac{9}{5}\)\(\Leftrightarrow x=\frac{13}{10}\)(thỏa mãn)

D = \(x\left(3-\sqrt{3}\right)\)(quá dễ rồi)

Ai trả lời giùm em với

\(A=\frac{3}{4}.4.x^2\left(8-x^2\right)\le\frac{3}{4}\left(x^2+8-x^2\right)^2=48\)

\(A_{max}=48\) khi \(x^2=8-x^2\Rightarrow x=\pm2\)

\(B=\frac{1}{2}\left(2x-1\right)\left(6-2x\right)\le\frac{1}{8}\left(2x-1+6-2x\right)^2=\frac{25}{8}\)

\(B_{max}=\frac{25}{8}\) khi \(2x-1=6-2x\Rightarrow x=\frac{7}{4}\)

\(C=\frac{1}{\sqrt{3}}.\sqrt{3}x\left(3-\sqrt{3}x\right)\le\frac{1}{4\sqrt{3}}\left(\sqrt{3}x+3-\sqrt{3}x\right)^2=\frac{3\sqrt{3}}{4}\)

\(C_{max}=\frac{3\sqrt{3}}{4}\) khi \(\sqrt{3}x=3-\sqrt{3}x=\frac{\sqrt{3}}{2}\)

\(D=\frac{1}{20}.20x\left(32-20x\right)\le\frac{1}{80}\left(20x+32-20x\right)^2=\frac{64}{5}\)

\(D_{max}=\frac{64}{5}\) khi \(20x=32-20x\Rightarrow x=\frac{4}{5}\)

\(E=\frac{4}{5}\left(5x-5\right)\left(8-5x\right)\le\frac{1}{5}\left(5x-5+8-5x\right)=\frac{9}{5}\)

\(E_{max}=\frac{9}{5}\) khi \(5x-5=8-5x\Leftrightarrow x=\frac{13}{10}\)

Bài này chỉ tìm được Min thôi nhé:) 

Ta có: \(B=4x^2-3x^3=x^2\left(4-3x\right)\)

Vì \(0\le x< \frac{4}{3}\Rightarrow4-3x>0\)

\(\Rightarrow B\ge0\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(x^2=0\Rightarrow x=0\)

Vậy \(B_{Min}=0\Leftrightarrow x=0\)

Bài này mình tìm đc MAX bạn r bạn nhé
TH1: x=0 thì B=0 (1)
TH2; \(0< x< \frac{4}{3}\)

Suy ra: 4-3x >0

Ta có \(B=4x^2-3x^3 \)

<=> \(x.B=x.x.x.\left(4-3x\right)\) (do 0<x<4/3)

ÁP DỤNG BĐT CAUCHY cho các số dương ta đc 
\(x.B=x.x.x.\left(4-3x\right)\le\left(\frac{x+x+x+4-3x}{4}\right)^4=1\)

Suy ra \(B\le\frac{1}{x}\) (do 0<x<4/3)  (2)

Lại có \(\frac{1}{x}>0\) vói mọi 0<x<4/3 (3)

Nên từ (1), (2), (3) suy ra
\(MaxB=\frac{1}{x}\Leftrightarrow\hept{\begin{cases}x=x=x=4-3x\\0< x< \frac{4}{3}\end{cases} \Leftrightarrow\hept{\begin{cases}x=1\left(TMĐK\right)\\0< x< \frac{4}{3}\end{cases}}}\)
Khi đó Max B= 1

\(a,\dfrac{x^2+x+2}{\sqrt{x^2+x+1}}=\dfrac{x^2+x+1+1}{\sqrt{x^2+x+1}}=\sqrt{x^2+x+1}+\dfrac{1}{\sqrt{x^2+x+1}}\left(1\right)\)

Áp dụng BĐT cosi: \(\left(1\right)\ge2\sqrt{\sqrt{x^2+x+1}\cdot\dfrac{1}{\sqrt{x^2+x+1}}}=2\)

Dấu \("="\Leftrightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

đk câu a hình như sai đấy