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\(P=\sqrt{6\left(x+y\right)+9}+\sqrt{2}.\sqrt{51-6\left(x+y\right)}\)
\(P\le\sqrt{\left(1+2\right)\left[6\left(x+y\right)+9+51-6\left(x+y\right)\right]}=6\sqrt{5}\)
\(P_{max}=6\sqrt{5}\) khi \(x+y=\frac{11}{6}\)
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 \(n+M=\frac{171}{2}\)
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2)ĐK:x\(\ge\frac{1}{2}\)
pt(2)\(\Leftrightarrow\left(y+1\right)^3\)+(y+1)=\(\left(2x\right)^3\)+2x
Xét hàm số: f(t)=\(t^3\)+t
f'(t)=3\(t^2\)+1>0,\(\forall\)t
\(\Rightarrow\)hàm số liên tục và đồng biến trên R
\(\Rightarrow\)y+1=2x
Thay y=2x-1 vào pt(1) ta đc:
\(x^2\)-2x=2\(\sqrt{2x-1}\)
\(\Leftrightarrow\left(x^2-4x+2\right)\left(1+\frac{4}{2x-2+2\sqrt{2x-1}}\right)=0\)
\(\Leftrightarrow x^2\)-4x+2=0(do(...)>0)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2+\sqrt{2}\Rightarrow y=3+2\sqrt{2}\\x=2-\sqrt{2}\Rightarrow y=3-2\sqrt{2}\end{array}\right.\)
4)ĐK:\(y\ge\frac{2}{3}\)
pt(1)\(\Leftrightarrow x-\sqrt{3y-2}=\sqrt{3y\left(3y-2\right)}-x\sqrt{x^2+2}\)
\(\Leftrightarrow x\left(\sqrt{x^2+2}+1\right)=\sqrt{3y-2}\left(\sqrt{3y}+1\right)\)
Xét hàm số:\(f\left(t\right)=t\left(\sqrt{t^2+2}+1\right)\)
\(\Rightarrow\)hàm số liên tục và đồng biến trên R
\(\Rightarrow x=\sqrt{3y-2}\)
Thay vào pt(2) ta đc:\(\sqrt{3y-2}+y+\sqrt{y+3}=4\)
\(\Leftrightarrow\sqrt{3y-2}-1+\sqrt{y+3}-2+y-1=0\)
\(\Leftrightarrow\left(y-1\right)\left(\frac{3}{\sqrt{3y-2}+1}+\frac{1}{\sqrt{y+3}+2}+1\right)=0\)
\(\Leftrightarrow y=1\Rightarrow x=1\)(do...)>0)
KL:...
\(\left(x^3+y^3\right)\left(x+y\right)=xy\left(1-x\right)\left(1-y\right)\Leftrightarrow\left(\frac{x^2}{y}+\frac{y^2}{x}\right)\left(x+y\right)=\left(1-x\right)\left(1-y\right)\left(1\right)\)
Ta có : \(\left(\frac{x^2}{y}+\frac{y^2}{x}\right)\left(x+y\right)\ge4xy\)
và \(\left(1-x\right)\left(1-y\right)=1-\left(x+y\right)+xy\le1-2\sqrt{xy}+xy\)
\(\Rightarrow1-2\sqrt{xy}+xy\ge4xy\Leftrightarrow0\) <\(xy\le\frac{1}{9}\)
Dễ chứng minh : \(\frac{1}{1+x^2}+\frac{1}{1+y^2}\le\frac{1}{1+xy};\left(x,y\in\left(0;1\right)\right)\)
\(\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}\le\sqrt{2\left(\frac{1}{1+x^2}+\frac{1}{1+y^2}\right)}\le\sqrt{2\left(\frac{2}{1+xy}\right)}=\frac{2}{\sqrt{1+xy}}\)
\(3xy-\left(x^2+y^2\right)=xy-\left(x-y\right)^2\le xy\)
\(\Rightarrow P\le\frac{2}{\sqrt{1+xy}}+xy=\frac{2}{\sqrt{1+t}}+t\), \(\left(t=xy\right)\), (0<\(t\le\frac{1}{9}\)
Xét hàm số :
\(f\left(t\right)=\frac{2}{\sqrt{t+1}}+t\) , (0<\(t\le\frac{1}{9}\)
\(\left(x+\sqrt{1+y^2}\right)\left(y+\sqrt{1+x^2}\right)=1\)
Nhân hai vế của pt với \(\left(x-\sqrt{1+y^2}\right)\left(y-\sqrt{1+x^2}\right)\)
\(\Leftrightarrow\left(x+\sqrt{1+y^2}\right)\left(x-\sqrt{1+y^2}\right)\left(y+\sqrt{1+x^2}\right)\left(y-\sqrt{1+x^2}\right)=\left(x-\sqrt{1+y^2}\right)\left(y-\sqrt{1+x^2}\right)\)
\(\Leftrightarrow\left(x^2-y^2-1\right)\left(y^2-x^2-1\right)=xy-x\sqrt{1+x^2}-y\sqrt{1+y^2}+\sqrt{\left(1+y^2\right)\left(1+x^2\right)}\)
\(\Leftrightarrow\left[-1+\left(x^2-y^2\right)\right]\left[-1-\left(x^2-y^2\right)\right]=2xy+2\sqrt{\left(1+x^2\right)\left(1+y^2\right)}-\left(xy+x\sqrt{1+y^2}+y\sqrt{1+x^2}+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\right)\)
\(\Leftrightarrow1^2-\left(x^2-y^2\right)^2=2xy+2\sqrt{\left(1+x^2\right)\left(1+y^2\right)}-\left(x+\sqrt{1+y^2}\right)\left(y+\sqrt{1+x^2}\right)\)
\(\Leftrightarrow1-\left(x^2-y^2\right)^2=2xy+2\sqrt{\left(1+x^2\right)\left(1+y^2\right)}-1\)
\(\Leftrightarrow2\left(1-xy\right)=\left(x^2-y^2\right)^2+2\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)(*)
Mặt khác : \(2\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2\sqrt{x^2+y^2+1+x^2y^2}\)
\(=2\sqrt{x^2+2xy+y^2+x^2y^2-2xy+1}\)
\(=2\sqrt{\left(x+y\right)^2+\left(xy-1\right)^2}\)
Vì \(\left(x^2-y^2\right)^2\ge0\forall x;y\) do đó theo (*) ta có :
\(2\left(1-xy\right)\ge2\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2\sqrt{\left(x+y\right)^2+\left(xy-1\right)^2}\)
\(\Leftrightarrow1-xy\ge\sqrt{\left(x+y\right)^2+\left(xy-1\right)^2}\ge\sqrt{\left(xy-1\right)^2}=\left|xy-1\right|\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-y^2\right)^2=0\\\left(x+y\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2-y^2=0\\x+y=0\end{matrix}\right.\)\(\Leftrightarrow x=-y\)
Thay vào P ta được :
\(P=x^7-x^7+2x^5-2x^5-3x^3+3x^3+4x-4x+100\)
\(P=0+0-0+0+100\)
\(P=100\)
Vậy...
p/s: mệt...