\(\text{Ta có:}2;6;10;...;8010\text{ đều chia 4 dư 2}\)

\(\Rightarrow X\equiv2^2+3^2+4^2+....+2004^2\left(mod\text{ }10\right)\)

\(\text{ mà:}1^2+2^2+3^2+....+2004^2=\frac{2004.2005.4009}{6}=333.2005.4009\)

\(\Rightarrow X\equiv333.2005.4009-1\left(\text{mod 10}\right)\equiv3.5.9-1\equiv4\left(\text{mod 10}\right)\)

Vậy X có chữ số tận cùng là 4

\(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2^{10}-1}\)

\(< 1+\frac{1}{2}+\frac{1}{2}+\left(\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{2^2}\right)+..........\left(\frac{1}{2^9}+\frac{1}{2^9}+....+\frac{1}{2^9}\left(\text{512 số hạng }\frac{1}{2^9}\right)\right)\)

\(=1+1+1+1+1+1+1+1+1+1\)

\(=10\left(\text{điều phải chứng minh}\right)\)

\(\text{bài 2 câu b tương tự câu a}\)

câu 2 mình chưa hiểu lắm bạn có thể giải thích cho mình được không ạ? Mình cảm ơn ạ ^^

\(\frac{x^4}{a}=\frac{y^4}{b}=\frac{1}{a+b}=\frac{x^4+y^4}{a+b}\Rightarrow x^4+y^4=1.\)

Mà \(x^2+y^2=1\)=>\(x^4+y^4=x^2+y^2=1.\)

Nếu x =0 => y =1 => a =0 vô lí 

Xem lại đề  dc ko ( hay mình làm sai?)

đề đúng r bạn

a) \(\frac{5x-3}{5}+\frac{2x+1}{4}\le\frac{2-3x}{2}-5\)
\(\Leftrightarrow\frac{4\cdot\left(5x-3\right)}{20}+\frac{5\left(2x+1\right)}{20}\le\frac{10\left(2-3x\right)}{20}-\frac{20\cdot5}{20}\)
\(\Leftrightarrow20x-12+10x+5\le20-30x-100\)
\(\Leftrightarrow20x+10x+30x\le20-100+12-5\)
\(\Leftrightarrow60x\le-73\)
\(\Leftrightarrow x\le\frac{-73}{60}\)

Biến đổi tương đương thôi!

\(\frac{x^2-2x+2004}{x^2}\ge\frac{2003}{2004}\)

\(\Leftrightarrow2004x^2+2.2004.x+2004^2\ge2003x^2\)

\(\Leftrightarrow x^2+2.2004.x+2004^2\ge0\)

\(\Leftrightarrow\left(x+2004\right)^2\ge0\)(Luôn đúng)

ta có:

\(\frac{x+2}{2013}+\frac{x+5}{2010}>\frac{x+8}{2007}+\frac{x+11}{2004}\)

\(\Leftrightarrow\left(\frac{x+2}{2013}+1\right)+\left(\frac{x+5}{2010}+1\right)>\left(\frac{x+8}{2007}+1\right)+\left(\frac{x+11}{2004}+1\right)\)

\(\Leftrightarrow\frac{x+2015}{2013}+\frac{x+2015}{2010}>\frac{x+2015}{2007}+\frac{x+2015}{2004}\)

\(\Leftrightarrow\frac{x+2015}{2013}+\frac{x+2015}{2010}-\frac{x+2015}{2007}-\frac{x+2015}{2004}>0\)

\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}\right)>0\)

\(\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}x+2015>0\\\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}>0\end{cases}}\\\hept{\begin{cases}x+2015< 0\\\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}< 0\end{cases}}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}x+2015>0\\\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}>0\end{cases}}\\\hept{\begin{cases}x+2015< 0\\\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}< 0\end{cases}}\end{cases}}\)

tự làm nốt