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

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

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

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

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

Vậy X=109/6075

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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:

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

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

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

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

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

a) \(2^x+5=21\)

\(\Rightarrow2^x=21-5=16\Rightarrow2^x=2^4\)

Vậy x  = 4

b) \(2^x-1+3^2=5^2+2.5\)

\(\Rightarrow2^x-1+9=35\)

\(\Rightarrow2^x=35-9+1=27\)

Vậy x không có giá trị

c;d;e;f làm tương tự

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

a, Ta có \(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}\)

<=> \(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}-\frac{x-4}{2008}=0\)

<=> \(\left(\frac{x-1}{2011}-1\right)+\left(\frac{x-2}{2010}-1\right)-\left(\frac{x-3}{2009}-1\right)-\left(\frac{x-4}{2008}-1\right)=0\)

<=>\(\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\) 

<=> \(\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)

Mà \(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)

=> \(x-2012=0=>x=2012\)

b, \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{\left(2x-1\right)\left(2x+1\right)}=\frac{49}{99}\)

=>\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2x-1\right)\left(2x+1\right)}=2\cdot\frac{49}{99}\)

=>\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2x-1}-\frac{1}{2x+1}=\frac{98}{99}\)

=>\(1-\frac{1}{2x+1}=\frac{98}{99}\)

=>\(\frac{2x}{2x+1}=\frac{98}{99}\)

=>2x = 98

=>x = 49

a) \(2^x+5=21\)

\(2^x=21-5\)

\(2^x=16\)

\(2^x=2^4\)

\(\Rightarrow x=4\)

Vậy \(x=4\)

a ) \(2^x+5=21\)

 \(2^x=21-5\)

\(2^x=16\)

\(2^x=2^4\)

\(\Rightarrow x=4\)

câu 2) 2.5 là 2 phẩy 5 hay 2 nhân 5 vậy?

1) ( 2x -15 )⁵ = ( 2x - 15 )³

( 2x -15 )⁵ - ( 2x - 15 )³ = 0

( 2x - 15 )³ .  [ ( 2x - 15 )² - 1 ] = 0

\(\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}}\)

\(\orbr{\begin{cases}2x-15=0\\2x-15=1\end{cases}}\)

\(\orbr{\begin{cases}2x=15\\2x=16\end{cases}}\)

\(\orbr{\begin{cases}x=\frac{15}{2}\\x=8\end{cases}}\)

a.15-(5-2x)=-4

\(\Leftrightarrow\)15-5+2x=-4

\(\Leftrightarrow\)2x=-4-15+5

\(\Leftrightarrow\)2x=-14

\(\Leftrightarrow\)x=-7

b.TH1:x-3\(\ge\)0 \(\Rightarrow\)x\(\ge\)3

Ta có \(|\)x-3\(|\)=x-3

PT trên\(\Leftrightarrow\)x-3+1=4

             \(\Leftrightarrow\)x=4+3-1

            \(\Leftrightarrow\)x=6(nhận)

TH2:x-3<0\(\Leftrightarrow\)x<3

Ta có:\(|\)x-3\(|\)=-x+3

PT trên\(\Leftrightarrow\)-x+3+1=4

                       -x=4-3-1

                       x=0(nhận)

Vậy S={0;6}

chỗ 100000 là 1000000000000000000.mười tám chữ số 0

/ / LÀ DẤU GIÁ TRỊ TUYỆT ĐỐI NHÉ!

\(\left(\frac{3}{1.3}+\frac{3}{3.5}+.......+\frac{3}{97.99}\right).\left(2x+1\right)=x+\frac{1}{33}\)

\(\Rightarrow[\frac{3}{2}.(\frac{2}{1.3}+\frac{2}{3.5}+.......+\frac{2}{97.99})].\left(2x+1\right)=x+\frac{1}{33}\)

\(\Rightarrow[\frac{3}{2}.(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{97}-\frac{1}{99})].\left(2x+1\right)=x+\frac{1}{33}\)

\(\Rightarrow[\frac{3}{2}.(1-\frac{1}{99})].\left(2x+1\right)=x+\frac{1}{33}\)

\(\Rightarrow\left(\frac{3}{2}.\frac{98}{99}\right).\left(2x+1\right)=x+\frac{1}{33}\)

\(\Rightarrow\frac{49}{33}.\left(2x+1\right)=x+\frac{1}{33}\)

\(\Rightarrow\frac{49}{33}.2x+\frac{49}{33}=x+\frac{1}{33}\)

\(\Rightarrow\frac{98}{33}.x+\frac{49}{33}=x+\frac{1}{33}\)

\(\Rightarrow\frac{98}{33}.x-x=\frac{1}{33}-\frac{49}{33}\)

\(\Rightarrow\frac{65}{33}.x=\frac{-16}{11}\)

\(\Rightarrow x=\frac{-16}{11}:\frac{65}{33}\)

\(\Rightarrow x=\frac{-48}{65}\)

Vậy \(x=\frac{-48}{65}\)