Question

Tìm giá trị lớn nhất của các biểu thức :

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

b, B=(2x-1)(3-x) với 0,5\(\le x\le3\)

c, C=x(3-\(\sqrt{3}x\)) với 0\(\le x\le\sqrt{3}\)

d, D= 4x(8-5x) với 0\(\le x\le\frac{8}{5}̸\)

e, E= 4(x-1)(8-5x) với \(1\le x\le\frac{8}{5}\)

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Answer 1

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

\(A_{max}=48\) 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 \(5x-5=8-5x\Leftrightarrow x=\frac{13}{10}\)