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#)Well ! Bài này cg dạng tầm cỡ vừa :v
Làm cho ai v ? mau vô nhận bài đê !
ờ tụi làm cho bạn trên lớp í mà
Mà tui ngu quá nhóm lun từ số 1 nhanh hơn ko
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{x\left(2x+1\right)}=\frac{1}{10}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{2x\left(2x+1\right)}=\frac{1}{10}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}\right)+2.\left(\frac{1}{3}-\frac{1}{4}\right)+2.\left(\frac{1}{4}-\frac{1}{5}\right)+2.\left(\frac{1}{5}-\frac{1}{6}\right)+...+2.\left(\frac{1}{2x}-\frac{1}{2x+1}\right)=\frac{1}{10}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2x}-\frac{1}{2x+1}\right)=\frac{1}{10}\)
\(2.\left(\frac{1}{2}-\frac{1}{2x+1}\right)=\frac{1}{10}\)
\(\frac{1}{2}-\frac{1}{2x+1}=\frac{1}{10}:2\)
\(\frac{1}{2}-\frac{1}{2x+1}=\frac{1}{20}\)
\(\frac{1}{2x+1}=\frac{1}{2}-\frac{1}{20}\)
\(\frac{1}{2x+1}=\frac{9}{20}\)
\(\Rightarrow2x+1=\frac{20}{9}\)
\(\Rightarrow2x=\frac{20}{9}-1\)
\(\Rightarrow2x=\frac{11}{9}\)
\(\Rightarrow x=\frac{11}{9}:2\)
\(\Rightarrow x=\frac{11}{18}\)
a) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.........+\frac{2}{x\left(x+1\right)}=\frac{1998}{2000}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.......+\frac{2}{x\left(x+1\right)}=\frac{1998}{2000}\)
\(\Leftrightarrow2.\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+......+\frac{1}{x\left(x+1\right)}\right]=\frac{1998}{2000}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{x\left(x+1\right)}=\frac{999}{2000}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+......+\frac{1}{x}-\frac{1}{x+1}=\frac{999}{2000}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{999}{2000}\)\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2000}\)
\(\Leftrightarrow x+1=2000\)\(\Leftrightarrow x=1999\)
Vậy \(x=1999\)
b) \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+......+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
\(\Leftrightarrow\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{15.2}{93}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+......+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Leftrightarrow\frac{1}{2x+3}=\frac{1}{93}\)\(\Leftrightarrow2x+3=93\)
\(\Leftrightarrow2x=90\)\(\Leftrightarrow x=45\)
Vậy \(x=45\)
\(=1-2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+......+\frac{1}{90}\right)=1-2\left(\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{9.10}\right)=1-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.......-\frac{1}{10}\right)=1-2\left(\frac{1}{2}-\frac{1}{10}\right)=1-\frac{2.4}{10}=1-\frac{4}{5}=\frac{1}{5}\)