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

đúng đề nhá, ko có sai đâu

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 \(ab=2\Leftrightarrow \text{x=2 or x=1}\)

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