\(2\left(\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1}{9}\)
\(\Leftrightarrow2\left(\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1}{9}\)
\(\Leftrightarrow2\left(\frac{1}{9}-\frac{1}{x+1}\right)=\frac{1}{9}\)
\(\Leftrightarrow\frac{1}{9}-\frac{1}{x+1}=\frac{1}{9}\div2\)
\(\Leftrightarrow\frac{1}{9}-\frac{1}{x+1}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{9}-\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{18}\)
\(\Leftrightarrow x+1=18\)
\(\Leftrightarrow x=18-1\)
\(\Leftrightarrow x=17\)
\(\left|x\right|-\frac{3}{4}=\frac{5}{3}\)
\(\Leftrightarrow\left|x\right|=\frac{5}{3}+\frac{3}{4}\)
\(\Leftrightarrow\left|x\right|=\frac{20}{12}+\frac{9}{12}\)
\(\Leftrightarrow\left|x\right|=\frac{29}{12}\)
\(\Leftrightarrow x=\pm\frac{29}{12}\)
Bài 1 :
\(2.\left(\frac{1}{9.10}+\frac{1}{10.11}+...+\frac{1}{x.\left(x+1\right)}\right)\)= \(\frac{1}{9}\)
=> \(\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+1}\)= \(\frac{1}{18}\)
\(\frac{1}{9}-\frac{1}{x+1}=\frac{1}{18}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
=> x + 1 = 18
=> x = 17
vậy x = 17
\(2\left(\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1}{9}\)
<=> \(2\left(\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1}{9}\)
<=> \(2\cdot\left(\frac{1}{9}-\frac{1}{x+1}\right)=\frac{1}{9}\)
<=> \(\frac{1}{9}-\frac{1}{x+1}=\frac{1}{18}\)
<=> \(\frac{1}{x+1}=\frac{1}{18}\)
<=> \(x+1=18\)
<=> \(x=17\)
\(\left|x\right|-\frac{3}{4}=\frac{5}{3}\)
<=> \(\left|x\right|=\frac{29}{12}\)
<=> x = 29/12 hoặc x = -29/12
Bài 1 :
l x l - 3/4 = 5/3
=> l x l = 29/12
=> x = \(\pm\)29/12
Vậy x = \(\pm\)29/12
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