Ta có \(\left(x+10\right)^4+\left(x-3\right)^4=\left[\left(x+10\right)^2\right]^2+\left[\left(3-x\right)^2\right]^2\)
\(\ge\dfrac{\left[\left(x+10\right)^2+\left(3-x\right)^2\right]^2}{2}\) \(\ge\dfrac{\left[\dfrac{\left(x+10+3-x\right)^2}{2}\right]^2}{2}\) \(=\dfrac{\left(\dfrac{13^2}{2}\right)^2}{2}\)\(=\dfrac{28561}{8}\) (áp dụng 2 lần bất đẳng thức \(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}\))
Suy ra \(P\le2000-\dfrac{28561}{8}=-\dfrac{12561}{8}\).
Dấu "=" xảy ra \(\Leftrightarrow x+10=3-x\Leftrightarrow x=-\dfrac{7}{2}\)
Vậy \(maxP=-\dfrac{12561}{8}\), max xảy ra khi \(x=-\dfrac{7}{2}\)
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