\(B=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{81}\right).\left(1-\frac{1}{100}\right)\)

\(B=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{80}{81}.\frac{99}{100}\)

\(B=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{8.10}{9.9}.\frac{9.11}{10.10}\)

\(B=\frac{1.2.3...8.9}{2.3.4...9.10}.\frac{3.4.5...10.11}{2.3.4...9.10}\)

\(B=\frac{1}{10}.\frac{11}{2}\)

\(B=\frac{11}{20}>\frac{11}{21}\)

Ta có : \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)

               \(=\frac{1}{2}.\frac{2}{3}....\frac{18}{19}.\frac{19}{20}\)

               \(=\frac{1.2....18.19}{2.3...19.20}\)

               \(=\frac{1}{20}>\frac{1}{21}\)

Vậy A > 1/21

\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)......\left(1-\frac{1}{81}\right)\left(1-\frac{1}{100}\right)\)

= \(-\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.......\frac{80}{81}.\frac{99}{100}\)

=\(-\frac{1.3.2.4.3.5..............8.10.9.11}{2^2.3^2.4^2.......10^2}=-\frac{\left(1.2.3.....9\right)\left(3.4.5....11\right)}{2.3.4....10.2.3.4.....10}=-\frac{11}{20}\)

\(\left(-\frac{1}{27}\right)^{53}=\left[\left(-\frac{1}{3}\right)^3\right]^{53}=\left(-\frac{1}{3}\right)^{159}\)

\(\left(-\frac{1}{243}\right)^{23}=\left[\left(-\frac{1}{3}\right)^5\right]^{23}=\left(-\frac{1}{3}\right)^{115}\)

Vì\(\left(-\frac{1}{3}\right)^{159}< \left(-\frac{1}{3}\right)^{115}\)nên: \(\left(-\frac{1}{27}\right)^{53}< \left(-\frac{1}{243}\right)^{23}\)

Nhung ơi tớ câu c tớ làm giống cái cậu Triều nhưng ko có dấu trừ

câu a lớn hơn còn câu b bé hơn

\(-\frac{1}{27}=-\frac{1}{3^3}\) => \(\left(-\frac{1}{27}\right)^{53}=\left(\left(-\frac{1}{3}\right)^3\right)^{53}=\left(-\frac{1}{3}\right)^{159}\)

\(-\frac{1}{243}=-\frac{1}{3^5}\) => \(\left(-\frac{1}{243}\right)^{23}=\left(\left(-\frac{1}{3}\right)^5\right)^{23}=\left(-\frac{1}{3}\right)^{115}\)

vẬY \(\left(-\frac{1}{27}\right)^{53}< \left(-\frac{1}{243}\right)^{23}\)

\(\left(\frac{-1}{27}\right)^{53}\)=\(\left(\frac{-1}{3}\right)^{3X53}\)=\(\left(\frac{-1}{3}\right)^{159}\)

\(\left(\frac{-1}{243}\right)^{23}\)=\(\left(\frac{-1}{3}\right)^{5X23}\)=\(\left(\frac{-1}{3}\right)^{115}\)

=>\(\left(\frac{-1}{3}\right)^{159}\)>\(\left(\frac{-1}{3}\right)^{115}\)

=>\(\left(\frac{-1}{27}\right)^{53}\)>\(\left(\frac{-1}{243}\right)^{23}\)

\(\left(X+\frac{1}{1.3}\right)+\left(X+\frac{1}{3.5}\right)+...+\left(X+\frac{1}{23.25}\right)=11.X+\)\(\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)

\(\Leftrightarrow12X+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)+11X\)\(+\frac{\left(1+\frac{1}{3}+...+\frac{1}{81}\right)-\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)}{2}\)

\(\Leftrightarrow X+\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\right)=\frac{242}{243}:2\)

\(\Leftrightarrow X+\frac{12}{25}=\frac{121}{243}\)

\(\Leftrightarrow X=\frac{109}{6075}\)

Vậy X=109/6075

Chắc Sai kết quả chứ công thức đúng nha!!!...

Fighting!!!...

Đặt: 

 \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{23.25}=\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{25-23}{23.25}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}=1-\frac{1}{25}=\frac{24}{25}\)

=> \(A=\frac{12}{25}\)

Đặt \(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)

   \(3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)

=> \(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)=1-\frac{1}{3^5}=\frac{242}{243}\)

=> \(2B=\frac{242}{243}\Rightarrow B=\frac{121}{243}\)

Giải phương trình:

\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}\right)+...+\left(x+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)\)

                        \(12x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{242}\right)\)

                                                                            \(12x+\frac{12}{25}=11x+\frac{121}{243}\)

                                                                             \(12x-11x=\frac{121}{243}-\frac{12}{25}\)

                                                                                                  \(x=\frac{109}{6075}\)

Ta có :

\(M=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{99}{100}=\frac{3.8.15.....99}{4.9.16.....100}=\frac{1.3.2.4.3.5.....9.11}{2.2.3.3.4.4.....10.10}\)\(=\frac{1.2.3...9}{2.3...10}.\frac{3.4...11}{2.3...10}=\frac{1}{10}.\frac{11}{2}=\frac{11}{20}< \frac{11}{19}\)

ta có M = (1- 1/4) (1- 1/9)... ( 1- 1/100)

             = 3/2^2.8/3^2 ... 99/10^2

             = 1.3/2^2 . 2.4/3^2 ... 9.11/10^ 2

             = 1.2.3...9/ 2.3.4...10 . 3.4.5... 11/ 2.3.4... 10

             = 1/10 . 11/2 = 11/20 < 11/19

              Vậy M < 11/19

\(\left(\frac{27}{64}\right)^{15}=\frac{\left(3^3\right)^{15}}{\left(2^6\right)^{15}}=\frac{3^{45}}{2^{90}}=\left(\frac{3}{2^2}\right)^{45}\)

\(\left(\frac{81}{256}\right)^{10}=\frac{\left(3^4\right)^{10}}{\left(2^8\right)^{10}}=\frac{3^{40}}{2^{80}}=\left(\frac{3}{2^2}\right)^{40}\)

Do \(\left(\frac{3}{2^2}\right)^{45}