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Áp dụng bất đẳng thức Cosi, ta có : 

\(53=2x+3y\ge2\sqrt{2x.3y}=2\sqrt{6}.\sqrt{xy}\Rightarrow xy\le\left(\frac{53}{2\sqrt{6}}\right)^2\)

Do đó : \(P=\sqrt{xy+4}\le\sqrt{\left(\frac{53}{2\sqrt{6}}\right)^2+4}=\sqrt{\frac{2905}{24}}\)

Vậy : Max \(P=\sqrt{\frac{2905}{24}}\Leftrightarrow\left(x;y\right)=\left(\frac{53}{4};\frac{53}{6}\right)\)

Áp dụng bất đẳng thức Cosi ta có:

\(x+y\ge2\sqrt{xy}\)

\(\Rightarrow2\sqrt{S}\le12\Leftrightarrow\sqrt{S}\le6\Rightarrow S\le36\)

Dấu = xảy ra khi x=y=6

\(\sqrt{2x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\le\dfrac{1}{2}\left(x+y+x+z\right)=\dfrac{1}{2}\left(2x+y+z\right)\)

Tương tự: \(\sqrt{2y+xz}\le\dfrac{1}{2}\left(x+2y+z\right)\) ; \(\sqrt{2z+xy}\le\dfrac{1}{2}\left(x+y+2z\right)\)

Cộng vế:

\(P\le\dfrac{1}{2}\left(4x+4y+4z\right)=4\)

\(P_{max}=4\) khi \(x=y=z=\dfrac{2}{3}\)

P = \(1.\sqrt{2x+yz}+1.\sqrt{2y+xz}+1.\sqrt{2z+xy}\)

\(\le\sqrt{\left(1^2+1^2+1^2\right)\left(2x+yz+2y+xz+2z+xy\right)}\)

\(=\sqrt{3.\left(4+xy+yz+zx\right)}\)

Đã biết x² + y² + z² \(\ge\)xy + yz + zx

=> xy + yz + zx \(\le\dfrac{\left(x+y+z\right)^2}{3}\)

Khi đó \(P\le\sqrt{3\left(4+xy+yz+zx\right)}\le\sqrt{3\left[4+\dfrac{\left(x+y+z\right)^2}{3}\right]}\)

= 4 

Dấu "=" xảy ra <=> x = 2/3 

\(\sqrt{2x+yz}=\sqrt{\left(x+y+z\right)x+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\le\dfrac{x+2y+z}{2}\\ \Leftrightarrow P=\sum\sqrt{2x+yz}\le\dfrac{x+2y+z+2x+y+z+x+y+2z}{2}=\dfrac{4\left(x+y+z\right)}{2}=2\cdot2=4\)

Dấu \("="\Leftrightarrow x=y=z=\dfrac{2}{3}\)

P = \(1.\sqrt{2x+yz}+1.\sqrt{2y+xz}+1.\sqrt{2z+xy}\)

\(\le\sqrt{\left(1^2+1^2+1^2\right)\left(2x+yz+2y+xz+2z+xy\right)}\)(Bunyacovski)

\(=\sqrt{3\left[4+\left(xy+yz+zx\right)\right]}\)

\(\le\sqrt{3.\left[4+\dfrac{\left(x+y+z\right)^2}{3}\right]}=\sqrt{3.\left(4+\dfrac{4}{3}\right)}\) = 4

Dấu "=" xảy ra <=> x = y = z = 2/3 

Ta có: \(2x^2+\frac{y^2}{4}+\frac{1}{x^2}=4\)

=> \(\left(x^2+\frac{y^2}{4}\right)+\left(x^2+\frac{1}{x^2}\right)=4\)

Lại có: \(x^2+\frac{y^2}{4}\ge2.x.\frac{y}{2}=xy\) Và \(x^2+\frac{1}{x^2}\ge2.x.\frac{1}{x}=2\)

=> \(4\ge xy+2\)=> \(2\ge xy\)

=> \(A=2016+xy\le2016+2=2018\)

=> A_min=2018

\(\sqrt[]{\sqrt{ }\frac{ }{ }\sqrt[]{}3\hept{\begin{cases}\\\\\end{cases}}3\frac{ }{ }\sqrt{ }\cos\hept{\begin{cases}\\\\\end{cases}}\Omega3\cong}\)

oo

Đặt \(\left\{{}\begin{matrix}\sqrt{2x+3}=a\ge0\\\sqrt{y}=b\ge0\end{matrix}\right.\)

\(\Rightarrow b\left(b^2+1\right)-3a^2=\left(a^2+1\right)a-3b^2\)

\(\Rightarrow a^3-b^3+3a^2-3b^2+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2\right)+\left(a-b\right)\left(3a+3b\right)+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+3a+3b+1\right)=0\)

\(\Leftrightarrow a=b\Rightarrow\sqrt{2x+3}=\sqrt{y}\)

\(\Rightarrow y=2x+3\)

\(\Rightarrow M=x\left(2x+3\right)+3\left(2x+3\right)-4x^2-3\) tới đây chắc chỉ cần bấm máy