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M = - ( 4/1.5 + 4/5.9 + ..................+ 4/(n-4).n
M = - (1-1/5 + 1/5 - 1/9 +..............+1/(n-4) - 1/n
M = -(1-1/n)
M = -1 + 1/n
M = -n + 1
Ta có : \(-\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-.....-\frac{4}{\left(n+4\right)n}\)
\(=-\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+......+\frac{4}{n\left(4+n\right)}\right)\)
\(=-\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+......+\frac{1}{n}-\frac{1}{n+4}\right)\)
\(=-\left(1-\frac{1}{n+4}\right)\)
\(=-\left(\frac{n+4}{n+4}-\frac{1}{n+4}\right)\)
\(=-\frac{n+3}{n+4}\)
\(S=\frac{-4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n-4\right).n}\)
\(=-\left(\frac{1}{1}-\frac{1}{5}\right)-\left(\frac{1}{5}-\frac{1}{9}\right)-\left(\frac{1}{9}-\frac{1}{13}\right)-...-\left(\frac{1}{n-4}-\frac{1}{n}\right)\)
\(=-\frac{1}{1}+\frac{1}{5}-\frac{1}{5}+\frac{1}{9}-\frac{1}{9}+\frac{1}{13}-...-\frac{1}{n-4}+\frac{1}{n}\)
\(=-\frac{1}{1}+\frac{1}{n}=\frac{1}{n}+1\)
M=\(\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n-4\right).n}\)
\(M=1-\frac{1}{5}-\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)
\(M=1-\frac{1}{5}-\frac{1}{5}+\frac{1}{n}\)
\(M=\frac{3}{5}+\frac{1}{n}\)
Mình chỉ giải đến đây thôi vì chẳng biết n bằng mấy cả
= - (1-1/5 +1/5 -1/9 +1/9 -1/13 +1/n + 1/n+4)
=-(1-1/n+4)
=-1+1/n+4
\(\text{Đề bài sai : }\frac{4}{\left(n-4\right)^n}->\frac{4}{\left(n-4\right)^n}\)
\(\text{Ta có :}\)
\(S=\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n-4\right)n}\)
\(=\left(\frac{1}{1}-\frac{1}{5}\right)-\left(\frac{1}{5}-\frac{1}{9}\right)-...-\left(\frac{1}{n-4}-\frac{1}{n}\right)\)
\(=\frac{1}{1}-\frac{1}{5}-\frac{1}{5}+\frac{1}{9}-...-\frac{1}{n-4}+\frac{1}{n}\)
\(=\frac{1}{1}-\frac{1}{5}-\frac{1}{5}+\frac{1}{n}\)
\(=\frac{3}{5}+\frac{1}{n}\)
\(=\frac{3}{5}+\frac{1}{n}\)
\(=\frac{3n+5}{5n}\)
\(\text{Vậy ...}\)
Ta có :
\(M=-\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n+4\right)n}\)
\(\Leftrightarrow\)\(M=-\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n+4}-\frac{1}{n}\right)\)
\(\Leftrightarrow\)\(M=-\left(1-\frac{1}{n}\right)\)
\(\Leftrightarrow\)\(M=-\frac{n}{n}+\frac{1}{n}\)
\(\Leftrightarrow\)\(M=\frac{-n+1}{n}\)
Vậy \(M=\frac{-n+1}{n}\)
\(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left(\frac{9}{45}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\Rightarrow\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}\)
\(\Rightarrow x=\frac{7}{\frac{21}{45}}=15\)
Vậy \(x=15\).
a) \(\frac{x-1}{21}=\frac{3}{x+1}\)( ĐKXĐ : x khác -1 )
<=> ( x - 1 )( x + 1 ) = 21.3
<=> x2 - 1 = 63
<=> x2 = 64
<=> x2 = ( ±8 )2
<=> x = ±8 ( tmđk )
b) \(\frac{7}{x}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{41\cdot45}=\frac{29}{45}\)( ĐKXĐ : x khác 0 )
<=> \(\frac{7}{x}+\left(\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{41\cdot45}\right)=\frac{29}{45}\)
<=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
<=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
<=> \(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
<=> \(\frac{7}{x}=\frac{7}{15}\)
<=> x = 15 ( tmđk )
a) \(\frac{x-1}{21}=\frac{3}{x+1}\Leftrightarrow\left(x-1\right)\left(x+1\right)=3.21\)
\(\Leftrightarrow x^2-1=63\Rightarrow x^2=63+1=64\Rightarrow x=\pm8\)
b) \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}=\frac{7}{15}\Rightarrow x=15\)
