x = từ 1 đến 10000....0

\(\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{15}\right)+....+\left(x+\frac{1}{575}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)

\(13x+\left(\frac{1}{1.3}+\frac{1}{3.5}+.....+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)

\(13x+\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)

\(2x+\frac{12}{25}=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)

Đặt \(A=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)

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

\(3A-A=1-\frac{1}{3^5}=\frac{242}{243}=2A\)

=> \(A=\frac{121}{243}\)

=> \(2x+\frac{12}{25}=\frac{121}{243}\)

=> \(2x=\frac{121}{243}-\frac{12}{25}=\frac{109}{6075}\)

=> x = ......

1, \(\frac{1}{2}-\left(6\frac{5}{9}+x-\frac{117}{8}\right):\left(12\frac{1}{9}\right)=0\)

   \(\left(\frac{6.9+5}{9}+x-\frac{117}{8}\right):\frac{12.9+1}{9}=\frac{1}{2}\)

 ( . là nhân nha) 

    \(\left(\frac{59}{9}-\frac{117}{8}+x\right):\frac{109}{9}=\frac{1}{2}\)

    \(\frac{59}{9}-\frac{117}{8}+x=\frac{1}{2}\cdot\frac{109}{9}\)

    \(\frac{59}{9}-\frac{117}{8}+x=\frac{109}{18}\)

   \(x=\frac{109}{18}-\frac{59}{9}+\frac{117}{8}\)

\(x=\frac{113}{8}\)

( \(\left(y+\frac{1}{3}\right)+\left(y+\frac{2}{9}\right)+\left(y+\frac{1}{27}\right)+\left(y+\frac{1}{81}\right)=\frac{56}{81}\)

   \(y+\frac{1}{3}+y+\frac{2}{9}+y+\frac{1}{27}+y+\frac{1}{81}=\frac{56}{81}\)

\(4y+\frac{1}{3}+\frac{2}{9}+\frac{1}{27}+\frac{1}{81}=\frac{56}{81}\)

\(4y+\frac{49}{81}=\frac{56}{81}\)

\(4y=\frac{7}{81}\)

y      =  7/81:4

y       = 7/324

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

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

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

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

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

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:

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

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

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

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

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

Chỉ biết \(x\) = \(\frac{109}{6075}\) thôi

Nhân 2 cả 2 vế lên:

\(\left(2x+\frac{2}{1x3}\right)+...+\left(2x+\frac{2}{23x25}\right)=22x+\frac{2}{3}+\frac{2}{9}+\frac{2}{81}+\frac{2}{243}\)2/243

\(24x+\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{23}-\frac{1}{25}\right)=22x+\frac{162+54+6+2}{243}\)

\(24x+\frac{24}{25}=22x+\frac{224}{243}\)

\(2x=\frac{224}{243}-\frac{24}{25}\)

\(2x=-\frac{232}{6025}\)

\(x=\frac{-116}{6075}\)

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

\(12x+\left[\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}\right)\right]=11.x+\left(\frac{81}{243}+\frac{27}{243}+\frac{3}{243}+\frac{1}{243}\right)\)

\(12x+\left[\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{25}\right)\right]=11.x+\frac{112}{243}\)

\(12x+\left(\frac{1}{2}.\frac{24}{25}\right)=11.x+\frac{112}{243}\)

\(12x+\frac{12}{25}=11x+\frac{112}{243}\)

\(11x-12x=\frac{112}{243}-\frac{12}{25}\)

\(-1x=-\frac{116}{6075}\)

\(x=-\frac{116}{6075}\div\left(-1\right)\)

\(x=\frac{116}{6075}\)

a)\(\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{7}\right)^7\)

\(x=\left(\frac{3}{7}\right)^7\div\left(\frac{3}{7}\right)^5\)

\(x=\left(\frac{3}{7}\right)^2\)

\(x=\frac{9}{49}\)

Vậy...

b)\(\left(-\frac{1}{3}\right)^3.x=\left(\frac{1}{3}\right)^4\)

\(\left(-\frac{1}{3}\right)^3.x=\left(-\frac{1}{3}\right)^4\)

\(x=\left(-\frac{1}{3}\right)^4\div\left(\frac{-1}{3}\right)^3\)

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

Vậy...

c)\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)

=>\(x-\frac{1}{2}=\frac{1}{3}\)

\(x=\frac{1}{3}+\frac{1}{2}\)

\(x=\frac{5}{6}\)

Vậy...

d)\(\left(x+\frac{1}{4}\right)^4=\left(\frac{2}{3}\right)^4\)

=>\(x+\frac{1}{4}=\frac{2}{3}\)

\(x=\frac{2}{3}-\frac{1}{4}\)

\(x=\frac{5}{12}\)

Vậy...

Phù, mãi mới xong, tk cho mk nha bn

a) x=1

b) x=1

c) x= -(245/81)

d) x= 1/27

e) x=3

g) x=4

cần gấp