Lời giải
giải phương trình:
a, $\frac{16}{\sqrt{x-3}}+\frac{4}{\sqrt{y-1}}+\frac{1225}{\sqrt{z-665}}=82-\sqrt{x-3}-\sqrt{y-1}-\sqrt{z-665}$
nghiệm đẹp!!
b, $\left\{\begin{matrix}\sqrt{1+x_{1}}+\sqrt{1+x_{2}}+...+\sqrt{1+x_{2000}}=2000\sqrt{\frac{2001}{2000}} & \\ \sqrt{1-x_{1}}+\sqrt{1-x_{2}}+...+\sqrt{1-x_{2000}}=2000\sqrt{\frac{1999}{2000}}& \end{matrix}\right.$
a)ĐK:...
$\frac{16}{\sqrt{x-3}}+\frac{4}{\sqrt{y-1}}+\frac{1225}{\sqrt{z-665}}=82-\sqrt{x-3}-\sqrt{y-1}-\sqrt{z-665}$
$\Leftrightarrow \frac{16}{\sqrt{x-3}}+\sqrt{x-3}+\frac{4}{\sqrt{y-1}}+\sqrt{y-1}+\frac{1225}{\sqrt{z-665}}+\sqrt{z-665}=82$
Áp dụng BĐT Cô-si ta có:
$\frac{16}{\sqrt{x-3}}+\sqrt{x-3}\geq 2\sqrt{\frac{16}{\sqrt{x-3}}.\sqrt{x-3}}=2\sqrt{16}=8$
$\frac{4}{\sqrt{y-1}}+\sqrt{y-1}\geq 2\sqrt{\frac{4}{\sqrt{y-1}}.\sqrt{y-1}}=2\sqrt{4}=4$
$\frac{1225}{\sqrt{z-665}}+\sqrt{z-665}\geq 2\sqrt{\frac{1225}{\sqrt{z-665}}.\sqrt{z-665}}=2\sqrt{1225}=70$
$\Rightarrow VT\geq 82$
Dấu '=' xảy ra khi:
$\left\{\begin{matrix} \frac{16}{\sqrt{x-3}}=\sqrt{x-3} & \\ \frac{4}{\sqrt{y-1}}=\sqrt{y-1} & \\\frac{1225}{\sqrt{z-665}}=\sqrt{z-665} & \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 16=x-3 & \\ 4=y-1 & \\1225=z-665 & \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=19(tm) & \\ y=5(tm) & \\z=1890(tm) & \end{matrix}\right.$
b)ĐK:...
PT(1) Áp dụng BĐT Bunhia ta có:
$(\sqrt{1+x_1}+\sqrt{1+x_2}+...+\sqrt{1+x_{2000}})^2 \leq 2000(2000+x_1+x_2+...+x_{2000})$
$\Leftrightarrow \left ( 2000.\sqrt{\frac{2001}{2000}} \right )^2 \leq 2000.(2000+x_1+x_2+...+x_{2000})$
$\Leftrightarrow x_1+x_2+...+x_{2000} \ge 1$
TT PT(2) ta có:
$\Leftrightarrow x_1+x_2+...+x_{2000} \leq 1$
Dấu '=' xảy ra khi: $x_1=x_2=x_3=...=x_{2000}=\frac{1}{2000}$
Bài viết đã được chỉnh sửa nội dung bởi minhhaiproh: 02-02-2024 - 00:01