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TA
19 tháng 1 2018
Dat A=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{13.15}\)
2A=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{13.15}\)
= 1-\(\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-....+\dfrac{1}{13}-\dfrac{1}{15}\)
= 1-\(\dfrac{1}{15}=\dfrac{14}{15}\)
=> A=\(\dfrac{7}{15}\)
Ta co : \(\dfrac{7}{15}\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
=> \(\dfrac{7}{15}x-\dfrac{7}{15}+\dfrac{7}{15}=\dfrac{3}{5}x\)
=> \(\dfrac{7}{15}x-\dfrac{3}{5}x=0\)
=> x\(\left(\dfrac{7}{15}-\dfrac{3}{5}\right)=0\)
=> x\(\left(-\dfrac{2}{15}\right)=0\)
=> x=0
19 tháng 1 2018
\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{13.15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
<=>\(\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{13.15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
<=>\(\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
<=>\(\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
<=> \(\dfrac{7}{15}\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
<=>\(\dfrac{7}{15}x-\dfrac{7}{15}=\dfrac{3}{5}x-\dfrac{7}{15}\)
<=>\(\dfrac{7}{15}x-\dfrac{3}{5}x=\dfrac{-7}{15}+\dfrac{7}{15}\)
<=> \(\dfrac{-2}{15}x=0\)
<=> \(x=0\)
Vậy: \(s=\left\{0\right\}.\)
13 tháng 1 2019
\(\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{13.15}\right)\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)
\(\Leftrightarrow\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{13.15}\right)\left(x-1\right)=\frac{6}{5}x-\frac{14}{15}\)
\(\Leftrightarrow\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\right)\left(x-1\right)=\frac{6}{5}x-\frac{14}{15}\)
\(\Leftrightarrow\left(1-\frac{1}{15}\right)\left(x-1\right)=\frac{6}{5}x-\frac{14}{15}\)
\(\Leftrightarrow\frac{14}{15}\left(x-1\right)=\frac{6}{5}x-\frac{14}{15}\)
\(\Leftrightarrow\frac{14}{15}x-\frac{14}{15}=\frac{6}{5}x-\frac{14}{15}\)
\(\Leftrightarrow-\frac{4}{15}x=\frac{28}{15}\)
\(\Leftrightarrow x=7\)
3 tháng 3 2019
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{15}\right)=\frac{7}{15}\)
Ta có: \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}\right)\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)
\(\Leftrightarrow\frac{7}{15}\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\Leftrightarrow\frac{7}{15}x=\frac{3}{5}x\)
\(\Leftrightarrow\frac{2}{15}x=0\Leftrightarrow x=0\)
Tập nghiệm: \(S=\left\{0\right\}\)
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 \(B=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{10}{100.110}\)
\(B=\frac{1}{10}\left(\frac{10}{1.11}+\frac{10}{2.12}+...+\frac{10}{100.110}\right)\)
\(B=\frac{1}{10}\left(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)
\(B=\frac{1}{10}\left[\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{110}\right)\right]\)\(=\frac{1}{10}\left[\left(1+\frac{1}{2}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right]\)\(B=10A\)
\(A.x=10A\)
\(=>x=10\)
Đề: \(\frac{\left(x-1\right)}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{3}{5x}-\frac{7}{15}\)
\(\Leftrightarrow\frac{\left(x-1\right)}{2}\left(1-\frac{1}{15}\right)=\frac{3}{5x}-\frac{7}{15}\)
\(\Leftrightarrow\frac{\left(x-1\right)}{2}.\frac{14}{15}=\frac{3}{5x}-\frac{7}{15}\Leftrightarrow\frac{7\left(x-1\right)+7}{3}=\frac{3}{x}\)
\(\Leftrightarrow\frac{7x}{3}=\frac{3}{x}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{7}}{3}\\x=-\frac{\sqrt{7}}{3}\end{cases}}\)