BÀi 2:
\(\left|x-\frac{2019}{2020}\right|\ge0\forall x\)
=>\(-\frac{2020}{2019}\left|x-\frac{2019}{2020}\right|\le0\forall x\)
=>\(-\frac{2020}{2019}\left|x-\frac{2019}{2020}\right|+\frac{2019}{2020}\le\frac{2019}{2020}\forall x\)
Dấu '=' xảy ra khi \(x-\frac{2019}{2020}=0\)
=>\(x=\frac{2019}{2020}\)
Bài 1:
a: \(A=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{402-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac35+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)
\(=\frac{5\left(31-\frac27-\frac{1}{11}+\frac{1}{23}\right)}{13\left(31-\frac27-\frac{1}{11}+\frac{1}{23}\right)}+\frac{3\left(\frac15+\frac{1}{13}-\frac{3}{10}\right)}{\frac15+\frac{1}{13}-\frac{3}{10}}=\frac{5}{13}+3=\frac{44}{13}\)
b: \(B=-\frac54+\frac35-1\frac{3}{14}:\left|-\frac{34}{21}\right|+\frac{-5}{17}:\left|-\frac{1}{34}\right|+\frac53\cdot\left(\frac12-\frac25\right)\)
\(=-\frac54+\frac35-\frac{17}{14}\cdot\frac{21}{34}+\frac{-5}{17}\cdot34+\frac53\cdot\frac{1}{10}\)
\(=-\frac54+\frac35-\frac{3}{2\cdot2}+\left(-10\right)+\frac{5}{30}=-\frac54+\frac35-\frac34+\left(-10\right)+\frac16\)
\(=-\frac{75}{60}+\frac{36}{60}-\frac{45}{60}+\frac{\left(-600\right)}{60}+\frac{10}{60}=\frac{-674}{60}=-\frac{337}{30}\)
c: \(C=1-\frac{1}{5\cdot10}-\frac{1}{10\cdot15}-\cdots-\frac{1}{95\cdot100}\)
\(=1-\frac15\left(\frac{5}{5\cdot10}+\frac{5}{10\cdot15}+\cdots+\frac{5}{95\cdot100}\right)\)
\(=1-\frac15\left(\frac15-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\cdots+\frac{1}{95}-\frac{1}{100}\right)\)
\(=1-\frac15\left(\frac15-\frac{1}{100}\right)=1-\frac15\cdot\frac{19}{100}=1-\frac{19}{500}=\frac{481}{500}\)
d: \(D=\frac17+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{775}+\frac{1}{1147}\)
\(=\frac{1}{1\cdot7}+\frac{1}{7\cdot13}+\cdots+\frac{1}{31\cdot37}\)
\(=\frac16\left(\frac{6}{1\cdot7}+\frac{6}{7\cdot13}+\cdots+\frac{6}{31\cdot37}\right)=\frac16\left(1-\frac17+\frac17-\frac{1}{13}+\cdots+\frac{1}{31}-\frac{1}{37}\right)\)
\(=\frac16\left(1-\frac{1}{37}\right)=\frac16\cdot\frac{36}{37}=\frac{6}{37}\)
