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Sửa đề : \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Leftrightarrow\)\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Leftrightarrow\)\(\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}.\frac{1}{2}\)
\(\Leftrightarrow\)\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{18}\)
\(\Leftrightarrow\)\(x+1=18\)
\(\Leftrightarrow\)\(x=18-1\)
\(\Leftrightarrow\)\(x=17\)
Vậy \(x=17\)
Chúc bạn học tốt ~
\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+10\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+10.\frac{4}{55}=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+\frac{40}{55}=\frac{3}{11}\)
\(\Leftrightarrow\)\(x=\frac{3}{11}-\frac{40}{55}\)
\(\Leftrightarrow\)\(x=\frac{-5}{11}\)
Vậy \(x=\frac{-5}{11}\)
Chúc bạn học tốt ~
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+....+\frac{1}{97.100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{97}-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\frac{99}{100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{33}{100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow x=\frac{0,33\times100}{0,33}=100\)
\(a,\)\(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=-\frac{37}{45}\)
\(x+\left(\frac{9-5}{5.9}+\frac{13-9}{9.13}+\frac{17-13}{13.17}+...+\frac{45-41}{41.45}\right)=-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+....+\frac{1}{41}-\frac{1}{45}\right)-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(x+\frac{8}{45}=-\frac{37}{45}\)
\(x=-\frac{37}{45}-\frac{8}{45}\)
\(x=-1\)
Bài 1:
a)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(=1-\frac{1}{2017}\)
\(=\frac{2016}{2017}\)
b)\(=1008\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=1008\cdot\left(1-\frac{1}{2017}\right)\)
Bài 2:
a)\(A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\)
\(=\frac{1}{3}-\frac{1}{21}\)
\(=\frac{2}{7}\)
b)\(B=\frac{5}{28}+\frac{5}{70}+...+\frac{5}{700}\)
\(=\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{25.28}\)
\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(=\frac{5}{3}\cdot\frac{6}{28}\)
\(=\frac{15}{14}\)
Bài 3:
a)Đặt \(A=-\frac{20}{11.13}-\frac{20}{13.15}-...-\frac{20}{53.55}\)
\(=-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)\)
\(=-\left[10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)\right]\)
\(=-\left[10\left(\frac{1}{11}-\frac{1}{55}\right)\right]\)
\(=-\left[10\cdot\frac{4}{55}\right]\)
\(=-\frac{8}{11}\).Thay vào ta có: \(x-\frac{8}{11}=\frac{2}{9}\)
\(\Leftrightarrow x=\frac{94}{99}\)
b)\(\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
\(x+1=18\)
\(x=17\)
<=> x\(-10\left(\frac{1}{11x13}+\frac{1}{13x15}+...+\frac{1}{53x55}\right)\)) =\(\frac{3}{11}\)
x\(-10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
X-10\(\left(\frac{1}{11}-\frac{1}{55}\right)\)=\(\frac{3}{11}\)
X-\(\frac{40}{55}\)=\(\frac{3}{11}\)
X=\(\frac{3}{11}+\frac{40}{55}=\frac{15+40}{55}=\frac{55}{55}=1\)
Đặt \(A=\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{11}{13}}-x-\frac{1}{9}=\frac{2}{3}+\frac{2}{15}\frac{2}{35}+\frac{2}{63}\)
\(\Rightarrow A=\frac{0,55+\frac{11}{7}+\frac{11}{13}}{0,15+\frac{3}{7}+\frac{11}{13}}-x-\frac{1}{9}=\frac{42}{63}+\frac{14}{105}+\frac{6}{105}+\frac{2}{63}\)
\(\Rightarrow A=\frac{\frac{11}{20}+\frac{11}{7}+\frac{11}{13}}{\frac{3}{20}+\frac{3}{7}+\frac{11}{13}}-x-\frac{1}{9}=\frac{44}{63}+\frac{20}{105}\)
\(\Rightarrow A=\frac{\frac{1001}{1820}+\frac{2860}{1820}+\frac{1540}{1820}}{\frac{273}{1820}+\frac{780}{1820}+\frac{1540}{1820}}-x-\frac{1}{9}=\frac{44}{63}+\frac{4}{21}\)
\(\Rightarrow A=\frac{\frac{5401}{1820}}{\frac{2593}{1820}}-x-\frac{1}{9}=\frac{44}{63}+\frac{12}{63}\)
\(\Rightarrow A=\frac{5401}{1820}:\frac{2593}{1820}-x-\frac{1}{9}=\frac{56}{63}\)
\(\Rightarrow A=\frac{5401}{1820}.\frac{1820}{2593}-x-\frac{1}{9}=\frac{8}{9}\)
\(\Rightarrow A=\frac{5401}{2593}-x-\frac{1}{9}=\frac{8}{9}\)
\(\Rightarrow\frac{5401}{2593}-x=\frac{8}{9}+\frac{1}{9}\)
\(\Rightarrow\frac{5401}{2593}-x=1\)
\(\Rightarrow x=\frac{5401}{2593}-1\)
\(\Rightarrow x=\frac{2808}{2593}\)
Vậy \(x=\frac{2808}{2593}\)
Chúc bn học tốt