Áp dụng BĐT cosi:

\(A=\sqrt{\left(2x+1\right)\left(x+2\right)}+2\sqrt{x+3}-2x\\ A\le\dfrac{2x+1+x+2}{2}+\dfrac{4+x+3}{2}-2x\\ A\le\dfrac{3x+3}{2}+\dfrac{x+7}{2}-2x=\dfrac{3x+3+x+7-4x}{2}=5\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}2x+1=x+2\\4=x+3\end{matrix}\right.\Leftrightarrow x=1\)

\(P=\sqrt{\left(x+2\right)\left(2x+1\right)}+2\sqrt{x+3}-2x\)

\(P\le\dfrac{1}{2}\left(x+2+2x+1\right)+\dfrac{1}{2}\left(4+x+3\right)-2x=5\)

\(P_{max}=5\) khi \(x=1\)

\(A=\sqrt{2x^2+5x+2}+2\sqrt{x+3}-2x\)

\(2A=2\sqrt{2x^2+5x+2}+4\sqrt{x+3}-4x\)

\(2A=2\sqrt{\left(2x+1\right)\left(x+2\right)}+4\sqrt{x+3}-4x\)

\(\le2x+1+x+2+4+x+3-4x=10\)

=>2A\(\le10\Rightarrow A\le5\)

dấu bằng xảy ra \(\Leftrightarrow2x+1=x+2\)

và x+3=4

=>x=1

maxA=5 khi x=1

Khó vậy ta ????

\(5x^2+2xy+2y^2-\left(4x^2+4xy+y^2\right)=\left(x-y\right)^2\ge0\\ \Leftrightarrow5x^2+2xy+2y^2\ge4x^2+4xy+y^2=\left(2x+y\right)^2\)

\(\Leftrightarrow P\le\dfrac{1}{2x+y}+\dfrac{1}{2y+z}+\dfrac{1}{2z+x}=\dfrac{1}{9}\left(\dfrac{9}{x+x+y}+\dfrac{9}{y+y+z}+\dfrac{9}{z+z+x}\right)\\ \Leftrightarrow P\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{x}\right)\\ \Leftrightarrow P\le\dfrac{1}{9}\left(\dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}\right)=\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=1\)

Dấu \("="\Leftrightarrow x=y=z=1\)

\(\sqrt{5x^2+2xy+2y^2}=\sqrt{4x^2+2xy+y^2+x^2+y^2}\ge\sqrt{4x^2+2xy+y^2+2xy}=2x+y\)

\(\Rightarrow\dfrac{1}{\sqrt{5x^2+2xy+2y^2}}\le\dfrac{1}{2x+y}=\dfrac{1}{x+x+y}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}\right)=\dfrac{1}{9}\left(\dfrac{2}{x}+\dfrac{1}{y}\right)\)

Tương tự:

\(\dfrac{1}{\sqrt{5y^2+2yz+2z^2}}\le\dfrac{1}{9}\left(\dfrac{2}{y}+\dfrac{1}{z}\right)\) ; \(\dfrac{1}{\sqrt{5z^2+2zx+2x^2}}\le\dfrac{1}{9}\left(\dfrac{2}{z}+\dfrac{1}{x}\right)\)

Cộng vế:

\(P\le\dfrac{1}{9}\left(\dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}\right)=1\)

\(P_{max}=1\) khi \(x=y=z=1\)

nhờ casio và 1 số suy đoán  ta biết được max f(x) =7 khi x=0 ,giờ AM-GM ngược thôi :v

ta có: \(f\left(x\right)=\sqrt{\left(2x+3\right)\left(x+3\right)}+\sqrt{4\left(x+4\right)}-2x\)

Áp dụng bất đẳng thức cauchy :

\(\sqrt{\left(2x+3\right)\left(x+3\right)}\le\frac{1}{2}\left(3x+6\right)\)

\(\sqrt{4\left(x+4\right)}\le\frac{1}{2}\left(x+8\right)\)

\(\Rightarrow f\left(x\right)\le\frac{1}{2}\left(4x+14\right)-2x=2x+7-2x=7\)

đẳng thức xảy ra khi \(\hept{\begin{cases}2x+3=x+3\\4=x+4\end{cases}\Leftrightarrow x=0}\)

Còn ý liền trước nó nữa: 

Tìm tất cả các cặp số (x, y)  thỏa mãn \(2\left(x\sqrt{y-4}+y\sqrt{x-4}\right)=xy\)

LÀM GIÚP MK CÂU TÌM GTLN NHA

HELP ME, PLEASE!

a: ĐKXĐ: \(\left[{}\begin{matrix}x\ge3\\x\le2\end{matrix}\right.\)

b: ĐKXĐ: \(\left[{}\begin{matrix}x>\dfrac{2\sqrt{14}}{7}\\x< -\dfrac{2\sqrt{14}}{7}\end{matrix}\right.\)

c: ĐKXĐ: \(x=\dfrac{1}{3}\)

d: ĐKXĐ: \(-\dfrac{2}{3}< x\le\sqrt{3}\)