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HD: Vũ Phương Vy em chỉ cần đặt ts c rồi rút gọn
ko chép lại đề nha
=\(A=\frac{2\left(1-\frac{2}{19}+\frac{2}{23}\right)-\frac{1}{1010}}{3\left(1-\frac{1}{19}+\frac{1}{23}-\frac{1}{2020}\right)}\)\(.\frac{4\left(1-\frac{1}{29}+\frac{1}{41}\right)-\frac{1}{505}}{5\left(1-\frac{1}{29}+\frac{1}{41}\right)-\frac{1}{404}}\)
rồi em chỉ cần rút gọn tiếp
p/s đến đây thấy đề kì kì sao đó
em chek lại đề đc k
a: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}+\dfrac{-\dfrac{1}{4}\cdot\dfrac{-2}{3}-\dfrac{3}{4}:\dfrac{1}{6}}{\dfrac{3}{2}\cdot\left(\dfrac{-2}{3}-\dfrac{3}{4}\cdot\dfrac{-2}{3}\right)}\)
\(=\dfrac{3}{4}+\dfrac{\dfrac{2}{12}-\dfrac{9}{2}}{\dfrac{3}{2}\cdot\dfrac{-1}{6}}=\dfrac{3}{4}+\dfrac{-13}{3}:\dfrac{-3}{12}\)
\(=\dfrac{3}{4}+\dfrac{13}{3}\cdot\dfrac{12}{3}=\dfrac{3}{4}+\dfrac{156}{9}=\dfrac{217}{12}\)
b: \(A=158\left(\dfrac{12\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}{4\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}:\dfrac{5\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}{6\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}\right)\cdot\dfrac{50550505}{711711711}\)
\(=158\cdot\left(3\cdot\dfrac{6}{5}\right)\cdot\dfrac{50550505}{711711711}\)
\(\simeq40.39\)
\(A=\frac{3}{2}-\frac{5}{6}+\frac{13}{12}-\frac{19}{20}+\frac{31}{30}-\frac{41}{42}+\frac{57}{56}-\frac{71}{72}+\frac{91}{90}-\frac{109}{110}\)
\(\Rightarrow A=\left(1+\frac{1}{2}\right)-\left(1-\frac{1}{6}\right)+\cdot\cdot\cdot+\left(1+\frac{1}{90}\right)-\left(1-\frac{1}{110}\right)\)
\(\Rightarrow A=1+\frac{1}{2}-1+\frac{1}{6}+\cdot\cdot\cdot+1+\frac{1}{90}-1+\frac{1}{110}\)
\(\Rightarrow A=\left[\left(1-1\right)+\frac{1}{2}+\frac{1}{6}\right]+\cdot\cdot\cdot+\left[\left(1-1\right)+\frac{1}{90}+\frac{1}{110}\right]\)
\(\Rightarrow A=\frac{1}{2}+\frac{1}{6}+\cdot\cdot\cdot+\frac{1}{90}+\frac{1}{110}\)
\(\Rightarrow A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{10}-\frac{1}{11}\)
\(\Rightarrow A=1-\frac{1}{11}\)
\(\Rightarrow A=\frac{10}{11}\)
\(a,\)\(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=-\frac{37}{45}\)
\(x+\left(\frac{9-5}{5.9}+\frac{13-9}{9.13}+\frac{17-13}{13.17}+...+\frac{45-41}{41.45}\right)=-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+....+\frac{1}{41}-\frac{1}{45}\right)-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(x+\frac{8}{45}=-\frac{37}{45}\)
\(x=-\frac{37}{45}-\frac{8}{45}\)
\(x=-1\)