\(P=\sqrt{6\left(x+y\right)+9}+\sqrt{2}.\sqrt{51-6\left(x+y\right)}\)

\(P\le\sqrt{\left(1+2\right)\left[6\left(x+y\right)+9+51-6\left(x+y\right)\right]}=6\sqrt{5}\)

\(P_{max}=6\sqrt{5}\) khi \(x+y=\frac{11}{6}\)

Từ \(\left(x+\sqrt{1+y^2}\right)\left(y+\sqrt{1+x^2}\right)=1\)

\(\Rightarrow\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)

(Cách chứng minh tại đây):

Cho (x+\(\sqrt{y^2+1}\))(y+\(\sqrt{x^2+1}\))=1Tìm GTNN của P=2(x2+y2)+x+y  - Hoc24

\(\Rightarrow x+y=0\)

Do đó \(P=100\)

x,y thuộc N ôk

\(4=2^x+2^y\ge2\sqrt{2^{x+y}}\Rightarrow2^{x+y}\le4\Rightarrow x+y\le2\)

\(\Rightarrow xy\le1\)

\(P=4x^2y^2+2x^3+2y^3+10xy\)

\(P=4x^2y^2+10xy+2\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]\)

\(P\le4x^2y^2+10xy+4\left(4-3xy\right)=4x^2y^2-2xy+16\)

Đặt \(xy=t\Rightarrow0< t\le1\)

Xét hàm \(f\left(t\right)=4t^2-2t+16\) trên \((0;1]\)

\(\Rightarrow...\)

Ta có: \(\sqrt{x^2+y^2+4x-2y+5}+\sqrt{x^2+y^2-8x-14y+65}=6\sqrt{2}\)

\(\Leftrightarrow\sqrt{\left(x+2\right)^2+\left(y-1\right)^2}+\sqrt{\left(4-x\right)^2+\left(7-y\right)^2}=6\sqrt{2}\left(^∗\right)\)

Xét hai vectơ \(\overrightarrow{u}=\left(x+2;y-1\right)\)và \(\overrightarrow{v}=\left(4-x;7-y\right)\)

Ta có: \(\overrightarrow{u}+\overrightarrow{v}=\left(6;6\right)\Rightarrow\left|\overrightarrow{u}+\overrightarrow{v}\right|=\sqrt{6^2+6^2}=6\sqrt{2}\)

Do vậy \(\left(^∗\right)\)trở thành\(\overrightarrow{u}+\overrightarrow{v}=\left|\overrightarrow{u}+\overrightarrow{v}\right|\)

Điều này xảy ra khi và chỉ khi \(\overrightarrow{u}\)và \(\overrightarrow{v}\)cùng hướng

\(\Leftrightarrow\hept{\begin{cases}\left(x+2\right)\left(7-y\right)=\left(y-1\right)\left(4-x\right)\\\left(x+2\right)\left(4-x\right)\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=x+3\\-2\le x\le4\end{cases}}\)

Khi y = x + 3 thì \(x^2+y^2-2x+2y+2=2x^2+6x+17\)

Xét hàm số \(f\left(x\right)=2x^2+6x+17\)trên đoạn \(\left[-2;4\right]\)

Ta có: \(-\frac{6}{2.2}=\frac{-3}{2}\in\left[-2;4\right]\)và \(f\left(-2\right)=13;f\left(-\frac{3}{2}\right)=\frac{25}{2};f\left(4\right)=73\)

Suy ra \(|^{min}_{\left[-2;4\right]}f\left(x\right)=\frac{25}{2}\);\(|^{max}_{\left[-2;4\right]}f\left(x\right)=73\)

Do đó \(m=\frac{25}{2};M=73\)và \(n+M=\frac{171}{2}\)

Vậy \(n+M=\frac{171}{2}\)

Đặt \(\left\{{}\begin{matrix}x-4=a\\y-3=b\end{matrix}\right.\) \(\Rightarrow a^2+b^2=5\)

\(Q=\sqrt{\left(a+5\right)^2+b^2}+\sqrt{\left(a+3\right)^2+\left(b+4\right)^2}\)

\(\Rightarrow Q\le\sqrt{2\left[\left(a+5\right)^2+b^2+\left(a+3\right)^2+\left(b+4\right)^2\right]}\) (Bunhiacopxki)

\(\Rightarrow Q\le\sqrt{4\left(a^2+8a+b^2+4b+25\right)}\)

\(\Rightarrow Q\le\sqrt{4\left(a^2+2.4a+b^2+2.2b+25\right)}\)

\(\Rightarrow Q\le\sqrt{4\left(a^2+2\left(a^2+4\right)+b^2+2\left(b^2+1\right)+25\right)}\)

\(\Rightarrow Q\le\sqrt{4\left(3a^2+3b^2+35\right)}\le\sqrt{4\left(3.5+35\right)}=10\sqrt{2}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=6\\y=4\end{matrix}\right.\)

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