Với \(a,b,c\ne0\); \(a+b+c\ne0\) , ta có:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\left(a+b+c\right)\left(ab+bc+ca\right)=abc\)

\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca\right)+c\left(ab+bc+ca\right)=abc\)

\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca\right)+abc+bc^2+c^2a=abc\)

\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca\right)+bc^2+c^2a=0\)

\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca\right)+c^2\left(a+b\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca+c^2\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left[b\left(a+c\right)+c\left(a+c\right)\right]=0\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b=0\\b+c=0\\c+a=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)

Không mất tính tổng quát, ta lấy \(a=-b\), ta có:

\(\frac{1}{a^{2005}}+\frac{1}{b^{2005}}+\frac{1}{c^{2005}}=\frac{1}{\left(-b\right)^{2005}}+\frac{1}{b^{2005}}+\frac{1}{c^{2005}}\)

\(=\frac{-1}{b^{2005}}+\frac{1}{b^{2005}}+\frac{1}{c^{2005}}=\frac{1}{c^{2005}}\) (1)

Ta có:\(\frac{1}{a^{2005}+b^{2005}+c^{2005}}=\frac{1}{\left(-b\right)^{2005}+b^{2005}+c^{2005}}\)

\(=\frac{1}{-b^{2005}+b^{2005}+c^{2005}}=\frac{1}{c^{2005}}\) (2)

Từ (1), (2), suy ra \(\frac{1}{a^{2005}}+\frac{1}{b^{2005}}+\frac{1}{c^{2005}}=\frac{1}{a^{2005}+b^{2005}+c^{2005}}\)

Cái chỗ không mất tính tổng quát đấy, là do a, b, c bình đẳng nhau.

\(a_n=\frac{1}{\sqrt{n}\sqrt{n+1}\left(\sqrt{n+1}+\sqrt{n}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

\(S_{2005}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{1+1}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2+1}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{3+1}}+...+\)

\(\frac{1}{\sqrt{2005}}-\frac{1}{\sqrt{2005+1}}\)

\(S_{2005}=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{2005}}-\frac{1}{\sqrt{2006}}\)

\(S_{2005}=1-\frac{1}{\sqrt{2006}}\)

PS : ko chắc :v 

mem nào k sai chỉ hộ t cái :v 

a)M=[(−4)3+43]:(1+3+5+...+2005)

M=\left[-64+64\right]\cdot(1+3+5+...+2005)M=[−64+64]⋅(1+3+5+...+2005)

M=0\cdot(1+3+5+...+2005)M=0⋅(1+3+5+...+2005)

M=0M=0

b, Như câu a

\(a)M=\left[(-4)^3+4^3\right]:(1+3+5+...+2005)\)

\(M=\left[-64+64\right]\cdot(1+3+5+...+2005)\)

\(M=0\cdot(1+3+5+...+2005)\)

\(M=0\)

b, Tương tự

Do a3+b3+c3=1;a+b+c=1→a3+b3+c3=a+b+c→3(a+b)(b+c)(c+a)=0→a=−b hoặc b=−c hoặc c=−aa3+b3+c3=1;a+b+c=1→a3+b3+c3=a+b+c→3(a+b)(b+c)(c+a)=0→a=−b hoặc b=−c hoặc c=−a
Nếu a=−ba=−b thì a2005+b2005+c2005=a2005−a2005+c2005=c2005=1 vì a-a+c=1a2005+b2005+c2005=a2005−a2005+c2005=c2005=1 vì a-a+c=1
Tương tự ta cũng được a2005+b2005+c2005=1a2005+b2005+c2005=1
Vậy với a+b+c=1;a3+b3+c3=1a+b+c=1;a3+b3+c3=1 thì a2005+b2005+c2005=1

do máy mình bị lỗi bàn phím nên giả sử a3 thì là a mũ 3 nha

cảm ơn

Đăng từng bài một rồi tui làm cho~

Nhìn như này hoa mắt lắm :(

làm hộ mình đi

Giải:

\(A=\dfrac{2005^2-2004}{2005^3+1}\)

\(\Leftrightarrow A=\dfrac{2005^2-2005+1}{\left(2005+1\right)\left(2005^2-2005+1\right)}\)

\(\Leftrightarrow A=\dfrac{1}{2005+1}\left(1\right)\)

\(B=\dfrac{2005^2+2006}{2005^3-1}\)

\(\Leftrightarrow B=\dfrac{2005^2+2005+1}{\left(2005-1\right)\left(2005^2+2005+1\right)}\)

\(\Leftrightarrow B=\dfrac{1}{2005-1}\left(2\right)\)

Ta có:

\(\left(1\right)< \left(2\right)\)

\(\Leftrightarrow A< B\)

Vậy ...