Giải phương trình :\(\left(\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}\right).x=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)
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NT
13 tháng 5 2016
\(A=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103+...}+\frac{1}{10.110}\)
\(A=\frac{1}{100}(\frac{100}{1.101}+\frac{100}{2.102}+\frac{100}{3.103}+...+\frac{100}{10.110})\)
\(A=\frac{1}{100}(\frac{1}{1}-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110})\)
\(A=\frac{1}{100}((\frac{1}{1}+\frac{1}{2}+...+\frac{1}{10})-(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}))\) ok?
\(B=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)
\(B=\frac{1}{10}(\frac{10}{1.11}+\frac{10}{2.12}+...+\frac{10}{100.110})\)
\(B=\frac{1}{10}(\frac{1}{1}-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110})\)
\(B=\frac{1}{10}((\frac{1}{1}+\frac{1}{2}+...+\frac{1}{100})-(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{110}))\)=\(\frac{1}{10}((\frac{1}{1}+\frac{1}{2}+...+\frac{1}{10})-(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}))\)
B=10A
A.x=10A suy ra x=10
gõ xong mém xỉu. :)
1 tháng 3 2017
a) \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}\right).\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right).\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{15}\right).\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)
\(\Leftrightarrow\frac{7}{15}\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)
\(\Leftrightarrow\frac{7}{15}x-\frac{7}{15}=\frac{3}{5}x-\frac{7}{15}\)
\(\Leftrightarrow\frac{7}{15}x-\frac{7}{15}-\frac{3}{5}x+\frac{7}{15}=0\)
\(\Leftrightarrow\frac{8}{15}x=0\)
\(\Leftrightarrow x=0\)
CC
7 tháng 1 2017
$\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{103.105}\right).\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\\ \Leftrightarrow \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{103}-\frac{1}{105}\right).\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\\ \Leftrightarrow \frac{1}{2}.\left(1-\frac{1}{105}\right).\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\\ \Leftrightarrow \frac{52}{105}.\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\\ \Leftrightarrow \frac{52}{105}x-\frac{52}{105}=\frac{3}{5}x-\frac{7}{15}\\ \Leftrightarrow x=-\frac{3}{11}$
NN
8 tháng 1 2017
b) Đặt \(A=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}\)
A\(=\frac{1}{100}\left(\frac{100}{1.101}+\frac{100}{2.102}+\frac{1}{3.103}+...+\frac{100}{10.110}\right)\)
A\(=\frac{1}{100}\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)\)
A\(=\frac{1}{100}\left[\left(1+\frac{1}{2}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right]\)Đặt \(B=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{10}{100.110}\)
\(B=\frac{1}{10}\left(\frac{10}{1.11}+\frac{10}{2.12}+...+\frac{10}{100.110}\right)\)
\(B=\frac{1}{10}\left(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)
\(B=\frac{1}{10}\left[\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{110}\right)\right]\)\(=\frac{1}{10}\left[\left(1+\frac{1}{2}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right]\)\(B=10A\)
\(A.x=10A\)
\(=>x=10\)
Vì gõ trên Hoc24 khá lâu nên mình gửi hình ảnh cho lẹ
17 tháng 3 2016
khi ko mún tích thì tích 1 tích
khi mún tích thì tích 50 tích
14 tháng 3 2018
\(100E\)\(=\frac{100}{1.101}+\frac{100}{2.102}+..........+\frac{100}{10.110}\)
\(=1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+........+\frac{1}{10}-\frac{1}{110}\)
\(10F=\frac{10}{1.11}+\frac{10}{2.12}+......+\frac{10}{100.110}\)
\(=1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+......+\frac{1}{100}-\frac{1}{110}\)
\(=1+\frac{1}{2}+...+\frac{1}{10}+\frac{1}{11}+....+\frac{1}{100}-\frac{1}{11}-\frac{1}{12}-....-\frac{1}{100}-\frac{1}{101}-...-\frac{1}{110}\)
\(=1+\frac{1}{2}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\)\(=100E\)
\(\Rightarrow10F=100E\Rightarrow\frac{E}{F}=\frac{1}{10}\)
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