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\(Q=3xy\left(x+3y\right)-2xy\left(x+4y\right)-x^2\left(y-1\right)+y^2\left(1-x\right)+36\)\(\Leftrightarrow Q=3x^2y+9xy^2-2x^2y-8xy^2-x^2y+x^2+y^2-xy^2+36\)\(\Leftrightarrow Q=\left(3x^2y-2x^2y-x^2y\right)+\left(9xy^2-8xy^2-xy^2\right)+x^2+y^2+36\)\(\Leftrightarrow Q=x^2+y^2+36\ge36\forall x;y\)
Dấu " = " xảy ra
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=0\\y^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Vậy Min Q là : \(36\Leftrightarrow x=y=0\)
\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x-y\right)\left(x+y\right)-4x^2\\ P=\left(x-y-x-y\right)^2-4x^2\\ P=4y^2-4x^2=4\left(y-x\right)\left(x+y\right)\)
Đặt x = 4 - m; y = 4 + m
=> x2 + y2 = (4 - m)2 + (4 + m)2 = 16 - 8m + m2 + 16 + 8m + m2 = 32 + 2m2
Vì m2 >= 0 => 2m2 >= 0
=> 32 + 2m2 >= 32
Dấu bằng xảy ra khi: m2 = 0 => m = 0
Vậy x2 + y2min = 32 <=> x = y = 4
Ta có: \(x+y=4\) \(\Rightarrow\) \(y=4-x\)
Do đó: \(A=x^2+y^2=x^2+\left(4-x\right)^2=x^2+16-8x+x^2=2x^2-8x+16=2\left(x^2-4x+4\right)+8\)
\(A=2\left(x-2\right)^2+8\ge8\) với mọi \(x;y\)
Dấu \("="\) xảy ra \(\Leftrightarrow\) \(\left(x-2\right)^2=0\)
\(\Leftrightarrow\) \(x-2=0\)
\(\Leftrightarrow\) \(x=2\)
\(\Rightarrow\) \(y=2\) (do \(x+y=4\) )
Vậy, \(Min\) \(A=8\) \(\Leftrightarrow\) \(x=y=2\)
\(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(\Leftrightarrow A=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)
\(\Leftrightarrow A=\left(x^2-x+6x-6\right)\left(x^2+2x+3x+6\right)\)
\(\Leftrightarrow A=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(\Leftrightarrow A=\left(x^2+5x\right)^2-36\ge-36\forall x\)
Dấu " = " xảy ra
\(\Leftrightarrow x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy GTNN của A là : \(-36\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Áp dụng \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\)
Ta có \(P=\left(x^2\right)^3+\left(y^2\right)^3=\left(x^2+y^2\right)^3-3x^2y^2\left(x^2+y^2\right)\)
\(\Rightarrow P=1-3x^2y^2\ge1-3\dfrac{\left(x^2+y^2\right)^2}{4}=\dfrac{1}{4}\)
\(\Rightarrow P_{min}=\dfrac{1}{4}\) khi \(x^2=y^2=\dfrac{1}{2}\)