Chủ đề này có 2 trả lời

[ThuThao36]

Cho hai số thực dương x và y thỏa mãn điều kiện $32x^{6}+4y^{3}=1$. Tìm giá trị lớn nhất của biểu thức:

$P=\frac{(2x^{2}+y+3)^{5}}{3(x^{2}+y^{2})-3(x+y)+2}$

[DOTOANNANG]

Từ $\it{:}$ $\it{32}\,\it{x}^{\,\it{6}}+ \it{4}\,\it{y}^{\,\it{3}}= \it{1}\,\,\Rightarrow \,\,\it{(}\,\,\it{y}- \it{2}\,\it{x}^{\,\it{2}}\,\,\it{)}\it{(}\,\,\it{x}- \frac{\it{1}}{\it{2}}\,\,\it{)}\leqq \it{0}\,\,,\,\,\it{(}\,\,\it{x}- \frac{\it{1}}{\it{2}}\,\,\it{)}\it{(}\,\,\it{y}- \frac{\it{1}}{\it{2}}\,\,\it{)}\leqq \it{0}\,\,,\,\,{\it{y}}'= -\,\frac{\it{16}\,\it{x}^{\,\it{5}}}{\it{y}^{\,\it{2}}}$ $\it{.}$ Khi đó $\it{:}$
${\it{P}}'\it{(}\,\,\it{x}\,\,\it{)}=$ $\frac{{\it{[}\,\,\it{(}\,\,\it{2}\,\it{x}^{\,\it{2}}+ \it{y}+ \it{3}\,\,\it{)}^{\,\it{5}}\,\,\it{]}}'\it{[}\,\,\it{3}\it{(}\,\,\it{x}^{\,\it{2}}+ \it{y}^{\,\it{2}}\,\,\it{)}- \it{3}\it{(}\,\,\it{x}+ \it{y}\,\,\it{)}+ \it{2}\,\,\it{]}- {\it{[}\,\,\it{3}\it{(}\,\,\it{x}^{\,\it{2}}+ \it{y}^{\,\it{2}}\,\,\it{)}- \it{3}\it{(}\,\,\it{x}+ \it{y}\,\,\it{)}+ \it{2}\,\,\it{]}}'\it{(}\,\,\it{2}\,\it{x}^{\,\it{2}}+ \it{y}+ \it{3}\,\,\it{)}^{\,\it{5}}}{\it{[}\,\,\it{3}\it{(}\,\,\it{x}^{\,\it{2}}+ \it{y}^{\,\it{2}}\,\,\it{)}- \it{3}\it{(}\,\,\it{x}+ \it{y}\,\,\it{)}+ \it{2}\,\,\it{]}^{\,\it{2}}}$ $= $

$=$ $\frac{\it{5}\it{(}\,\,\it{2}\,\it{x}^{\,\it{2}}+ \it{y}+ \it{3}\,\,\it{)}^{\,\it{4}}\left ( \it{4}\,\it{x}- \it{16}\,\frac{\it{x}^{\,\it{5}}}{\it{y}^{\,\it{2}}} \right )\it{[}\,\,\it{3}\it{(}\,\,\it{x}^{\,\it{2}}+ \it{y}^{\,\it{2}}\,\,\it{)}- \it{3}\it{(}\,\,\it{x}+ \it{y}\,\,\it{)}+ \it{2}\,\,\it{]}- \left ( \it{6}\,\it{x}- \it{3}- \it{96}\,\frac{\it{x}^{\,\it{5}}}{\it{y}}+ \it{48}\,\frac{\it{x}^{\,\it{5}}}{\it{y}^{\,\it{2}}} \right )\it{(}\,\,\it{2}\,\it{x}^{\,\it{2}}+ \it{y}+ \it{3}\,\,\it{)}^{\,\it{5}}}{\it{[}\,\,\it{3}\it{(}\,\,\it{x}^{\,\it{2}}+ \it{y}^{\,\it{2}}\,\,\it{)}- \it{3}\it{(}\,\,\it{x}+ \it{y}\,\,\it{)}+ \it{2}\,\,\it{]}^{\,\it{2}}}$ $= $

$=$ $\frac{\it{(}\,\,\it{2}\,\it{x}^{\,\it{2}}+ \it{y}+ \it{3}\,\,\it{)}^{\,\it{4}}\it{\{}\,\,\it{20}\,\it{x}\it{(}\,\,\it{y}- \it{2}\,\it{x}^{\,\it{2}}\,\,\it{)}\it{(}\,\,\it{y}+ \it{2}\,\it{x}^{\,\it{2}}\,\,\it{)}\it{[}\,\,\it{3}\it{(}\,\,\it{x}^{\,\it{2}}+ \it{y}^{\,\it{2}}\,\,\it{)}- \it{3}\it{(}\,\,\it{x}+ \it{y}\,\,\it{)}+ \it{2}\,\,\it{]}- \it{3}\it{[}\,\,\it{x}^{\,\it{5}}\it{(}\,\,\it{16}- \it{32}\,\it{y}\,\,\it{)}+ \it{y}^{\,\it{2}}\it{(}\,\,\it{2}\,\it{x}- \it{1}\,\,\it{)}\,\,\it{]}\,\,\it{\}}}{\it{y}^{\,\it{2}}\it{[}\,\,\it{3}\it{(}\,\,\it{x}^{\,\it{2}}+ \it{y}^{\,\it{2}}\,\,\it{)}- \it{3}\it{(}\,\,\it{x}+ \it{y}\,\,\it{)}+ \it{2}\,\,\it{]}^{\,\it{2}}}$

Dễ thấy ngay $\it{:}$ $\it{(}\,\,\it{x}- \frac{\it{1}}{\it{2}}\,\,\it{)}{\it{P}}'\it{(}\,\,\it{x}\,\,\it{)}\leqq \it{0}\,\,\Rightarrow \,\,\it{P}\it{(}\,\,\it{x}\,\,\it{)}\leqq \it{P}\it{(}\,\,\frac{\it{1}}{\it{2}}\,\,\it{)}= \it{2048}$

$\lceil$ Ý nghĩa hình học của đạo hàm $\it{(}$ $\it{!}$ $\it{)}$ $\rfloor$

By H-a-i-D-a-n-g-e-l $\it{(}$ $\it{D-...-A-N-G}$ $\it{)}$

Bài viết đã được chỉnh sửa nội dung bởi DOTOANNANG: 01-02-2019 - 15:10

[DOTOANNANG]

$\lceil$ https://math.stackex.../3275557/685500 $\rfloor$