Question

[Nguyễn Thị Thùy Linh]

1. CMR : \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}\)

2. Tìm x,y để \(C=-18-\left|2x-6\right|-\left|3y+9\right|\)đạt giá trị sớn nhất.

Answer 1

[Trần Nguyễn Bảo Quyên]

\(.1.\)

\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}\)

Ta có : \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}\)

\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}\)

....................

\(\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\)

____________

\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}.100=\sqrt{100}=10\)

Vậy : \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>0\)

Answer 2

[Lightning Farron]

Bài 2: Ta thấy:\(\left\{\begin{matrix}\left|2x-6\right|\ge0\\\left|3y+9\right|\ge0\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}-\left|2x-6\right|\le0\\-\left|3y+9\right|\le0\end{matrix}\right.\)

\(\Rightarrow-\left|2x-6\right|-\left|3y+9\right|\le0\)

\(\Rightarrow-18-\left|2x-6\right|-\left|3y+9\right|\le-18\)

\(\Rightarrow C\le-18\)

Dấu "=" xảy ra khi \(\left\{\begin{matrix}-\left|2x-6\right|=0\\-\left|3y+9\right|=0\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}x=3\\y=-3\end{matrix}\right.\)

Vậy với \(\left\{\begin{matrix}x=3\\y=-3\end{matrix}\right.\) thì C đạt GTLN là -18