\(Q=x^2+\left(3-x\right)^2=\left[x+\left(3-x\right)\right]^2-2x\left(3-x\right)=3^2-2x\left(3-x\right)\)
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 \(ab=2\Leftrightarrow \text{x=2 or x=1}\)
