1. lớp 9
2. ko bt làm
3. mới lớp sau lớp 9
4. ko bt làm
5. khao khát dc k
6. k cho mk ik~

Biến đi 

cục sì lầu bà tân vlog

nà ní???

What the fuck

ta có : \(x\sqrt{2017-y^2}\le\frac{x^2+2017-y^2}{2}\)

\(y\sqrt{2017-x^2}\le\frac{y^2+2017-x^2}{2}\)

Do đó \(x\sqrt{2017-y^2}+y\sqrt{2017-x^2}\le2017\)

dấu = xảy ra khi và chỉ khi :\(\hept{\begin{cases}x^2=2017-y^2\\y^2=2017-x^2\end{cases}}\)

\(\Leftrightarrow2\left(x^2+y^2\right)=2.2017\)(cộng vế với vế)

\(\Leftrightarrow x^2+y^2=2017\)

\(C=1-\frac{2}{2.3}+1-\frac{2}{3.4}+...+1-\frac{2}{2019.2020}\)

\(=2018-2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\right)\)

\(=2018-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)

\(=2018-2\left(\frac{1}{2}-\frac{1}{2020}\right)\)

\(=2018-2.\frac{1009}{2020}\)

\(=2018-\frac{1009}{1010}\)

\(=\frac{2037171}{1010}\)

\(x=\frac{1}{\sqrt[3]{4-\sqrt{15}}}+\sqrt[3]{4-\sqrt{15}}\)

<=> \(x^3=\frac{1}{4-\sqrt{15}}+3\left(\frac{1}{\sqrt[3]{4-\sqrt{15}}}+\sqrt[3]{4-\sqrt{15}}\right)\left(\frac{1}{\sqrt[3]{4-\sqrt{15}}}.\sqrt[3]{4-\sqrt{15}}\right)\)

                           \(+4-\sqrt{15}\)

<=> \(x^3=\frac{1}{4-\sqrt{15}}+4-\sqrt{15}+3x\)

<=> \(x^3-3x+2006=\frac{1}{4-\sqrt{15}}+4-\sqrt{15}+2006\)

<=> \(x^3-3x+2006=\frac{4+\sqrt{15}}{16-15}+4-\sqrt{15}+2006\)

<=> \(x^3-3x+2006=2014\)