A= \(\left(1-\frac{1}{1+2}\right)\)\(\left(1-\frac{1}{1+2+3}\right)\) \(\left(1-\frac{1}{1+2+3+4}\right)\) .....\(\left(1-\frac{1}{1+2+3+...+2005+2006}\right)\)

A = \(\left(1-\frac{1}{3}\right)\) \(\left(1-\frac{1}{6}\right)\) \(\left(1-\frac{1}{10}\right)\) .... \(\left(1-\frac{1}{2013021}\right)\)

= \(\frac{2}{3}\) . \(\frac{5}{6}\) . \(\frac{9}{10}\) .....\(\frac{2013020}{2013021}\)

= \(\frac{4}{6}\).\(\frac{10}{12}\).\(\frac{18}{20}\)....\(\frac{4026040}{4026042}\)

= \(\frac{1.4}{2.3}\).\(\frac{2.5}{3.4}\).\(\frac{3.6}{4.5}\).\(\frac{2005.2008}{2006.2007}\)

= \(\frac{1.2.3.4...2005}{2.3.4.5...2006}\).\(\frac{4.5.6...2008}{3.4.5...2007}\)

= \(\frac{1}{2006}.\frac{2008}{3}=\frac{1004}{3009}\)

Đề bài là A = gì thế bạn?

Kết quả hơi lớn bạn nhé!

A=\(\frac{1}{31}\left[\frac{31}{5}\left(9-\frac{1}{2}\right)-\frac{17}{2}\left(4+\frac{1}{5}\right)+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{930}\right]\)

=\(\frac{1}{31}\left[\frac{31}{5}\left(\frac{18}{2}-\frac{1}{2}\right)-\frac{17}{2}\left(\frac{20}{5}+\frac{1}{5}\right)+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{30.31}\right]\)

=\(\frac{1}{31}\left[\frac{31}{5}.\frac{17}{2}-\frac{17}{2}.\frac{21}{5}+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{30}-\frac{1}{31}\right]\)

=\(\frac{1}{31}\left[\frac{17}{2}.\left(\frac{31}{5}-\frac{21}{5}\right)+1-\frac{1}{31}\right]\)

=\(\frac{1}{31}\left[\frac{17}{2}.\frac{10}{5}+\frac{31}{31}-\frac{1}{31}\right]\)

=\(\frac{1}{31}\left[\frac{17}{2}.2+\frac{30}{31}\right]\)

=\(\frac{1}{31}\left[17+\frac{30}{31}\right]\)

=\(\frac{1}{31}\left[\frac{527}{31}+\frac{30}{31}\right]\)

=\(\frac{1}{31}.\frac{557}{31}=\frac{557}{961}\)

A = bao nhiêu vậy bạn

\(\left(\frac{1}{24.25}+\frac{1}{25.26}+....+\frac{1}{29.30}\right).120+x=\frac{1}{3}\)

\(\left(\frac{1}{24}-\frac{1}{30}\right).120+x=\frac{1}{3}\)

120.\(\frac{1}{120}\)+x =\(\frac{1}{3}\)

1+x=\(\frac{1}{3}\)

\(\Rightarrow x=-\frac{2}{3}\)

\(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+x=\frac{1}{3}\)

\(\Rightarrow\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}+\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+x=\frac{1}{3}\)

\(\Rightarrow\left(\frac{1}{24}-\frac{1}{30}\right).120+x=\frac{1}{3}\)

\(\Rightarrow\frac{1}{120}.120+x=\frac{1}{3}\)

\(\Rightarrow1+x=\frac{1}{3}\)

\(\Rightarrow x=\frac{1}{3}-1\)

\(\Rightarrow x=-\frac{2}{3}\)

Vậy \(x=-\frac{2}{3}.\)

Chúc bạn học tốt!

Bổ sung đề:

Giải:

\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1\)

Vậy ...

Lời giải:

Biểu thức $P$ không đối xứng. Có lẽ đề bài đúng là:

\(P=\frac{1}{\sqrt{a^2-ab+3b^2+1}}+\frac{1}{\sqrt{b^2-bc+3c^2+1}}+\frac{1}{\sqrt{c^2-ab+3a^2+1}}\)

Áp dụng BĐT AM-GM ta có:

$a^2-ab+3b^2+1=(a^2+b^2)-ab+(b^2+1)+b^2\geq ab+2b+b^2$

$\Rightarrow \frac{1}{\sqrt{a^2-ab+3b^2+1}}\leq \frac{1}{\sqrt{ab+2b+b^2}}$

Mà cũng theo BĐT AM-GM kết hợp BĐT Cauchy_Schwarz:

\(\frac{1}{\sqrt{ab+2b+b^2}}=\frac{1}{\sqrt{b(a+b+2)}}\leq \frac{1}{4b}+\frac{1}{a+b+2}\leq \frac{1}{4b}+\frac{1}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{1}+\frac{1}{1}\right)\)

\(=\frac{1}{16}.\frac{1}{a}+\frac{5}{16}.\frac{1}{b}+\frac{1}{8}\)

Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế thu được:

$P\leq \frac{3}{8}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac{3}{8}\leq \frac{3}{8}.3+\frac{3}{8}=\frac{3}{2}$

Vậy $P_{\max}=\frac{3}{2}$. Dấu "=" xảy ra khi $a=b=c=1$

Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)

\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)

\(2A=1-\frac{1}{3^{99}}\)

\(\Rightarrow A=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\left(đpcm\right)\)