Đặt \(y=3-x\).Ta có:\(\hept{\begin{cases}x+y=3\\x^2+y^2\ge5\end{cases}\Leftrightarrow\hept{\begin{cases}x^2+y^2+2xy=9\\x^2+y^2\ge5\end{cases}}}\)

\(\Rightarrow x^2+y^2+4\left(x^2+y^2+2xy\right)\ge5+4.9=41\)

\(\Rightarrow5\left(x^2+y^2\right)+4\left(2xy\right)\ge41\)

Mặt khác \(16\left(x^2+y^2\right)^2+25\left(2xy\right)^2\ge40\left(x^2+y^2\right)\left(2xy\right)\left(1\right)\)

Cộng 2 vế của (1) với \(25\left(x^2+y^2\right)^2+16\left(2xy\right)^2\):

\(\Rightarrow41\left[\left(x^2+y^2\right)^2+\left(2xy\right)^2\right]\ge\left[5\left(x^2+y^2\right)+4\left(2xy\right)^2\right]\ge41\)

hay \(\left(x^2+y^2\right)^2+\left(2xy\right)^2\ge41\Leftrightarrow x^4+y^4+6x^2y^2\ge41\)

Vậy minP=41

You ơi , you thiếu điều kiện xảy ra dấu "="

Đặt \(t=x^2+\left(3-x\right)^2\Rightarrow t\ge5\)

Mặt khác: \(t=x^2+\left(3-x\right)^2=9-2x\left(3-x\right)\Rightarrow x\left(3-x\right)=\frac{9-t}{2}\)

Ta có: \(P=\left[x^2+\left(3-x\right)^2\right]^2+4x^2\left(3-x\right)^2=t^2+4\left(\frac{9-t}{2}\right)^2\)

\(=2t^2-18t+81=2\left(t-\frac{9}{2}\right)^2+\frac{81}{2}\)

Mà \(t\ge5\Rightarrow t-\frac{9}{2}\ge\frac{1}{2}\Rightarrow P\ge2.\left(\frac{1}{2}\right)^2+\frac{81}{2}=41\)

Đẳng thức xảy ra khi \(t=5\Leftrightarrow x^2+\left(3-x\right)^2=5\Leftrightarrow x^2-3x+2\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

Vậy \(MinP=41\), đạt được khi \(x\in\left\{1;2\right\}\)

phải là tìm giá trị lớn nhất chứ

Đặt \(x^2+\left(3-x\right)^2=a\ge5\)

Ta có: 

\(x\left(3-x\right)=-\frac{1}{2}\left(2x^2-6x\right)\)

\(=-\frac{1}{2}\left(x^2-6x+9+x^2-9\right)\)

\(=-\frac{1}{2}\left(x^2+\left(3-x\right)^2-9\right)=-\frac{1}{2}\left(a-9\right)\)

Áp dụng ta có: 

\(P=x^4+\left(3-x\right)^4+6x^2\left(3-x\right)^2=\left(x^2+\left(3-x\right)^2\right)^2+4x^2\left(3-x\right)^2\)

\(=a^2+\left(a-9\right)^2\)

\(=2a^2-18a+81=\left(2a^2-20a+50\right)+2a+31\)

\(=2\left(a-5\right)^2+2a+31\ge0+2.5+31=41\)

\(Q=x^2+\left(3-x\right)^2=\left[x+\left(3-x\right)\right]^2-2x\left(3-x\right)=3^2-2x\left(3-x\right)\)

đặt : t=2x.(3-x) => Q=9- t

\(Q\ge0\Rightarrow9-t\ge5\Rightarrow t\le4\)(*)

\(P=\left[x^2+\left(3-x\right)^2\right]^2+4x^2\left(3-x\right)^2=\left(9-t\right)^2+t^2\)

\(P=2t^2-18t+9^2=2\left(t^2-9.t\right)+9^2\)

\(P=2\left(t^2-2.\dfrac{9}{2}t+\dfrac{9^2}{4}\right)+9^2-\dfrac{9^2}{2}=2\left(t-\dfrac{9}{2}\right)^2+\dfrac{9^2}{2}\)

từ (*)

\(t\le4\Rightarrow\left(t-\dfrac{9}{2}\right)\le4-\dfrac{9}{2}=\dfrac{-1}{2}\Rightarrow\left(t-\dfrac{9}{2}\right)^2\ge\dfrac{1}{4}\)

\(P\ge2.\dfrac{1}{4}+\dfrac{9^2}{2}=\dfrac{1}{2}+\dfrac{81}{2}=\dfrac{82}{2}=41\)

đẳng thức khi t =4 <=> 2x(3-x) =4

<=>x^2 -3x =-2 <=>x^2 -3x+2=0 <=>x^2 -2x -(x-2)

<=>(x-1)(x-2) =0=>x={1;2}

Lời giải:

Đặt \(\left\{\begin{matrix} x=a\\ 3-x=b\end{matrix}\right.\). Theo điều kiện đb ta có: \(\left\{\begin{matrix} a+b=3\\ a^2+b^2\geq 5\end{matrix}\right.\)

\(\Rightarrow (a+b)^2-2ab\geq 5\Leftrightarrow 9-2ab\geq 5\)

\(\Leftrightarrow ab\leq 2\)

Ta có:

\(P=x^4+(3-x)^4+6x^2(3-x)^2\)

\(P=a^4+b^4+6a^2b^2=(a^2+b^2)^2+4a^2b^2\)

\(P=[(a+b)^2-2ab]^2+4a^2b^2=(9-2ab)^2+4a^2b^2\)

\(P=81+8a^2b^2-36ab=8(ab-2)^2-4ab+49\)

Ta có: \(\left\{\begin{matrix} (ab-2)^2\geq 0\\ ab\leq 2\end{matrix}\right.\) nên \(P\geq 0-4.2+49\Leftrightarrow P\geq 41\)

Vậy \(P_{\min}=41\)

Dấu bằng xảy ra khi \(ab=2\Leftrightarrow \text{x=2 or x=1}\)

\(A=\sqrt{x^4+4x^3+6x^2+4x+2}+\sqrt{y^4-8y^3+24y^2-32y+17}\)

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

Đặt \(\hept{\begin{cases}x+1=u\\y-2=v\end{cases}}\Rightarrow A=\sqrt{u^4+1}+\sqrt{v^4+1}\)(với \(u,v\inℝ\))

Điều kiện đã cho ban đầu trở thành \(\left(u+1\right)\left(v+1\right)=\frac{9}{4}\)

\(\Leftrightarrow uv+u+v+1=\frac{9}{4}\Leftrightarrow uv+u+v=\frac{5}{4}\)

Ta có: \(\hept{\begin{cases}\left(2u-1\right)^2\ge0\forall u\inℝ\\\left(2v-1\right)^2\ge0\forall v\inℝ\end{cases}}\Leftrightarrow\hept{\begin{cases}4u^2-4u+1\ge0\\4v^2-4v+1\ge0\end{cases}}\forall u,v\inℝ\)

\(\Rightarrow\hept{\begin{cases}4u^2+1\ge4u\\4v^2+1\ge4v\end{cases}}\Rightarrow u^2+v^2\ge u+v-\frac{1}{2}\forall u,v\inℝ\)(*)

và \(\left(u-v\right)^2\ge0\forall u,v\inℝ\Leftrightarrow u^2-2uv+v^2\ge0\forall u,v\inℝ\)

\(\Rightarrow u^2+v^2\ge2uv\forall u,v\inℝ\Leftrightarrow\frac{1}{2}\left(u^2+v^2\right)\ge uv\forall u,v\inℝ\)(**)

Cộng theo vế của (*) và (**), ta được: \(\frac{3}{2}\left(u^2+v^2\right)\ge uv+u+v-\frac{1}{2}=\frac{5}{4}-\frac{1}{2}=\frac{3}{4}\)

\(\Rightarrow u^2+v^2\ge\frac{1}{2}\)(**

Áp dụng bất đẳng thức Minkowski, ta được:

\(A=\sqrt{u^4+1}+\sqrt{v^4+1}\ge\sqrt{\left(u^2+v^2\right)^2+\left(1+1\right)^2}\)

\(=\sqrt{\left(u^2+v^2\right)^2+4}\ge\sqrt{\left(\frac{1}{2}\right)^2+4}=\sqrt{\frac{1}{4}+4}=\frac{\sqrt{17}}{2}\)

Đẳng thức xảy ra khi \(u=v=\frac{1}{2}\Leftrightarrow x=-\frac{1}{2};y=\frac{5}{2}\)

Vậy GTNN của A là \(\frac{\sqrt{17}}{2}\)đạt được khi \(x=-\frac{1}{2};y=\frac{5}{2}\)

Đặt \(a=2+x;b=y-1\) thì \(ab=\frac{9}{4}\)

Thì \(\sqrt{x^4+4x^3+6x^2+4x+2}=\sqrt{a^4-4a^3+6a^2-4a+2}\)

và \(\sqrt{y^4-8y^3+24y^2-32y+17}=\sqrt{b^4-4b^3+6b^2-4b+2}\) (cái này dùng phương pháp đồng nhất hệ số là xong)

Vậy ta tìm Min \(A=\sqrt{a^4-4a^3+6a^2-4a+2}+\sqrt{b^4-4b^3+6b^2-4b+2}\)

\(=\sqrt{\left(a^4-4a^3+4a^2\right)+2\left(a^2-2a+1\right)}+\sqrt{\left(b^4-4b^3+4b^2\right)+2\left(b^2-2b+1\right)}\)

\(=\sqrt{\left(a^2-2a\right)^2+\left[\sqrt{2}\left(a-1\right)\right]^2}+\sqrt{\left(b^2-2b\right)^2+\left[\sqrt{2}\left(b-1\right)\right]^2}\)

\(\ge\sqrt{\left(a^2+b^2-2a-2b\right)^2+2\left(a+b-2\right)^2}\)

\(\ge\sqrt{\left[\frac{\left(a+b\right)^2}{2}-2\left(a+b\right)\right]^2+2\left(a+b-2\right)^2}\)

\(=\sqrt{\left(\frac{t^2}{2}-2t\right)^2+2\left(t-2\right)^2}\left(t=a+b\ge2\sqrt{ab}=3\right)\)

\(=\sqrt{\frac{1}{4}\left(t-1\right)\left(t-3\right)\left(t^2-4t+5\right)+\frac{17}{4}}\ge\frac{\sqrt{17}}{2}\)

Trình bày hơi lủng củng, sr.