Tìm GTLN của biểu thức :
\(Q=4x^2-3x^3\) với \(0\le x\le\dfrac{4}{3}\)
K
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0
8 tháng 1 2019
Câu a :
Ta có : \(\sqrt{5+3x}-\sqrt{5-3x}=a\)
\(\Leftrightarrow\left(\sqrt{5+3x}-\sqrt{5-3x}\right)^2=a^2\)
\(\Leftrightarrow5+3x-2\sqrt{\left(5+3x\right)\left(5-3x\right)}+5-3x=a^2\)
\(\Leftrightarrow10-2\sqrt{25-9x^2}=a^2\)
\(\Leftrightarrow2\sqrt{25-9x^2}=10-a^2\)
\(\Leftrightarrow\sqrt{25-9x^2}=\dfrac{10-a^2}{2}\)
\(\Leftrightarrow25-9x^2=\dfrac{\left(a^2-10\right)^2}{2}\)
\(\Leftrightarrow9x^2=25-\dfrac{\left(a^2-10\right)^2}{2}\)
\(\Leftrightarrow3x=\sqrt{\dfrac{50-\left(a^2-10\right)^2}{2}}\)
\(\Leftrightarrow x=\dfrac{\sqrt{50-\left(a^2-10\right)^2}}{3\sqrt{2}}\)
\(P=\dfrac{3\sqrt{2}.\sqrt{10+2\sqrt{\dfrac{10-a^2}{2}}}}{\sqrt{50-\left(a^2-10\right)^2}}\)
Bạn tự rút gọn nữa nhé :))
Câu b : \(M=\dfrac{2x+y+z-15}{x}+\dfrac{x+2y+z-15}{y}+\dfrac{x+y+2z-24}{z}\)
\(=\dfrac{x-3}{x}+\dfrac{y-3}{y}+\dfrac{z-12}{z}\)
\(=3-3\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{4}{z}\right)\le3-3\left[\dfrac{\left(1+1+2\right)^2}{12}\right]=-1\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 7 2020
\(A=\frac{3}{4}.4.x^2\left(8-x^2\right)\le\frac{3}{4}\left(x^2+8-x^2\right)^2=48\)
\(A_{max}=48\) 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 \(5x-5=8-5x\Leftrightarrow x=\frac{13}{10}\)
9 tháng 2 2020
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 = \(\frac{7}{3}\Leftrightarrow x=\frac{4}{3}\)
9 tháng 2 2020
\(A\) \(=\) \(3x^2\left(8-x^2\right)\le3\frac{\left(x^2+8-x^2\right)^2}{4}=48\)
\(\Rightarrow\) 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)
\(Q=x^2\left(4-3x\right)=\dfrac{4}{9}.\dfrac{3}{2}x.\dfrac{3}{2}x\left(4-3x\right)\)
\(Q\le\dfrac{1}{27}.\dfrac{4}{9}.\left(\dfrac{3x}{2}+\dfrac{3x}{2}+4-3x\right)^3=\dfrac{256}{243}\)
\(Q_{maxx}=\dfrac{256}{243}\) khi \(\dfrac{3x}{2}=4-3x\Leftrightarrow x=\dfrac{8}{9}\)