\(a,\left(\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{13.15}\right)\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)
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13 tháng 1 2019
\(\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{13.15}\right)\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)
\(\Leftrightarrow\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{13.15}\right)\left(x-1\right)=\frac{6}{5}x-\frac{14}{15}\)
\(\Leftrightarrow\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\right)\left(x-1\right)=\frac{6}{5}x-\frac{14}{15}\)
\(\Leftrightarrow\left(1-\frac{1}{15}\right)\left(x-1\right)=\frac{6}{5}x-\frac{14}{15}\)
\(\Leftrightarrow\frac{14}{15}\left(x-1\right)=\frac{6}{5}x-\frac{14}{15}\)
\(\Leftrightarrow\frac{14}{15}x-\frac{14}{15}=\frac{6}{5}x-\frac{14}{15}\)
\(\Leftrightarrow-\frac{4}{15}x=\frac{28}{15}\)
\(\Leftrightarrow x=7\)
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 \(B=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{10}{100.110}\)
\(B=\frac{1}{10}\left(\frac{10}{1.11}+\frac{10}{2.12}+...+\frac{10}{100.110}\right)\)
\(B=\frac{1}{10}\left(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)
\(B=\frac{1}{10}\left[\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{110}\right)\right]\)\(=\frac{1}{10}\left[\left(1+\frac{1}{2}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right]\)\(B=10A\)
\(A.x=10A\)
\(=>x=10\)
20 tháng 4 2019
a,A=\(\frac{1}{2}.\left(\frac{2.2}{1.3}.\frac{3.3}{2.4}......\frac{2016.2016}{2015.2017}\right)=\frac{1}{2}.\left(\frac{2.3.4...2016}{1.2....2015}.\frac{2.3.4...2016}{3.4....2017}\right)=\frac{1}{2}.\left(\frac{2016.2}{2017}\right)=\frac{4032}{4034}=\frac{2016}{2017}\)
Hok tốt
20 tháng 4 2019
\(\left|x\right|=\frac{1}{2}\Rightarrow x=\orbr{\begin{cases}\frac{1}{2}\\-\frac{1}{2}\end{cases}}\)
TH1:\(x=\frac{1}{2}\)
\(\Rightarrow\frac{1}{2}-\frac{3}{2}+5=4\)
TH2:\(x=\frac{-1}{2}\)
\(\Rightarrow\frac{1}{2}+\frac{3}{2}+5=7\)
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