Question

tìm GTLN

A=\(3x^2\left(8-x^2\right)\) với \(-2\sqrt{2}\le x\le2\sqrt{2}\)

B=4x(8-5x) với \(0\le x\le\frac{8}{5}\)

C=4(x-1)(8-5x) với \(1\le x\le\frac{8}{5}\)

D=x\(\left(3-\sqrt{3}\right)\) với \(0\le x\le\sqrt{3}\)

Tìm GTNN

A=\(\frac{3x}{2}+\frac{2}{x-1}\) với x>1

B=x+\(\frac{2}{3x-1}\) với x>1/3

Answer 1

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

Answer 2

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

\(\Rightarrow\) 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)

Answer 3

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)