\(-\dfrac{25}{20}>0\)

\(\dfrac{20}{25}>0\)

\(\Rightarrow-\dfrac{25}{20}< \dfrac{20}{25}\)

\(\dfrac{x}{7}+\dfrac{2}{y}=\dfrac{-1}{15}\)

=>\(\dfrac{xy+14}{7y}=\dfrac{-1}{15}\)

=>\(15\left(xy+14\right)+7y=0\)

=>\(15xy+7y=-210\)

=>y(15x+7)=-210

mà 15x+7 chia 15 dư 7

nên \(\left(15x+7;y\right)\in\left\{\left(7;-30\right)\right\}\)

=>\(\left(x;y\right)\in\left(0;-30\right)\)

help me pls:(

a) \(x-\dfrac{3}{5}=\dfrac{2}{7}\)

\(\Rightarrow x=\dfrac{2}{7}+\dfrac{3}{5}\)

\(\Rightarrow x=\dfrac{10}{35}+\dfrac{21}{35}\)

\(\Rightarrow x=\dfrac{31}{35}\)

b) \(x+\dfrac{20}{11\cdot13}+\dfrac{20}{13\cdot15}+...+\dfrac{20}{53\cdot55}=\dfrac{3}{11}\)

\(\Rightarrow x+10\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{53\cdot55}\right)=\dfrac{3}{11}\)

\(\Rightarrow x+10\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)
\(\Rightarrow x+10\left(\dfrac{1}{11}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)

\(\Rightarrow x+10\cdot\dfrac{4}{55}=\dfrac{3}{11}\)

\(\Rightarrow x+\dfrac{40}{55}=\dfrac{3}{11}\)

\(\Rightarrow x=\dfrac{3}{11}-\dfrac{40}{55}\)

\(\Rightarrow x=\dfrac{-25}{55}=\dfrac{-5}{11}\)

a) 

\(\dfrac{4}{5}-\left(\dfrac{-2}{3}\right)-\dfrac{1}{10}-\dfrac{2}{3}\\ =\dfrac{4}{5}+\dfrac{2}{3}-\dfrac{1}{10}-\dfrac{2}{3}\\ =\left(\dfrac{4}{5}-\dfrac{1}{10}\right)+\left(\dfrac{2}{3}-\dfrac{2}{3}\right)\\ =\dfrac{7}{10}+0\\ =\dfrac{7}{10}\)  

b) 

\(\dfrac{1}{3}-\dfrac{-1}{2}+\dfrac{1}{13}-\dfrac{5}{6}\\ =\left(\dfrac{1}{2}+\dfrac{1}{3}\right)+\dfrac{1}{13}-\dfrac{5}{6}\\ =\dfrac{5}{6}+\dfrac{1}{13}-\dfrac{5}{6}\\ =\dfrac{1}{13}\) 

c) 

\(\dfrac{-5}{12}-\left(\dfrac{-5}{6}-\dfrac{5}{12}\right)\\ =\dfrac{-5}{12}+\dfrac{5}{6}+\dfrac{5}{12}\\ =\left(-\dfrac{5}{12}+\dfrac{5}{12}\right)+\dfrac{5}{6}\\ =\dfrac{5}{6}\)

\(\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{2008}-1\right).\left(\dfrac{1}{2009}-1\right)\)

\(=-\dfrac{1}{2}.-\dfrac{2}{3}.-\dfrac{3}{4}...-\dfrac{2007}{2008}.-\dfrac{2008}{2009}\)

\(=-\left(\dfrac{1.2.3...2007.2008}{2.3.4...2008.2009}\right)\)

\(=-\dfrac{1}{2009}\)

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Leftrightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)

\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)

\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\Leftrightarrow x=-2004\)

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

=>\(\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)

=>\(\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)

=>\(\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

=>x+2004=0

=>x=-2004

Người thợ săn trộn số lít sơn xanh và trắng là :
        3,8 + 0,7 = 4,5 ( l )
Người thợ sơn còn lại số lít sơn xanh nhạt là :
       4,5 - 1,25 = 3,25 ( l )
            Đáp số : 3,25l sơn xanh nhạt

a) \(\dfrac{-3}{5}+\dfrac{28}{5}.\left(\dfrac{13}{56}-\dfrac{5}{24}+\dfrac{1}{7}\right)\\ =-\dfrac{3}{5}+\dfrac{28}{5}.\dfrac{13}{56}-\dfrac{28}{5}.\dfrac{5}{24}+\dfrac{28}{5}.\dfrac{1}{7}\\ =-\dfrac{3}{5}+\dfrac{28.13}{5.2.28}-\dfrac{4.7.5}{5.4.6}+\dfrac{4.7.1}{5.7}\\ =-\dfrac{3}{5}+\dfrac{13}{10}-\dfrac{7}{6}+\dfrac{4}{5}\\ =\dfrac{1}{5}+\dfrac{13}{10}-\dfrac{7}{6}\\ =\dfrac{6}{30}+\dfrac{39}{30}-\dfrac{35}{30}=\dfrac{10}{30}=\dfrac{1}{3}\)

b) \(\dfrac{5}{7}.\dfrac{11}{18}+\dfrac{3}{7}.\dfrac{5}{8}+\dfrac{4}{9}\\ =\dfrac{5}{7}.\dfrac{11}{18}+\dfrac{5}{7}.\dfrac{3}{8}+\dfrac{4}{9}\\ =\dfrac{5}{7}.\left(\dfrac{11}{18}+\dfrac{3}{8}\right)+\dfrac{4}{9}\\ =\dfrac{5}{7}.\dfrac{7}{4}+\dfrac{4}{9}\\ =\dfrac{5}{4}+\dfrac{4}{9}=\dfrac{61}{36}\)

c) \(4\dfrac{5}{9}:\left(\dfrac{-5}{7}\right)+\dfrac{49}{9}:\left(\dfrac{-5}{7}\right)\\=\left(4\dfrac{5}{9}+\dfrac{49}{9}\right):\dfrac{-5}{7}\\ =\left(\dfrac{41}{9}+\dfrac{49}{9}\right).\dfrac{-7}{5}\\ =10.\dfrac{-7}{5}=-14\)

Mình sửa lại câu b nhé:

\(=\dfrac{5}{7}.\dfrac{11}{18}+\dfrac{5}{7}.\dfrac{3}{8}+\dfrac{4}{9}\\ =\dfrac{5}{7}.\left(\dfrac{11}{18}+\dfrac{3}{8}\right)+\dfrac{4}{9}\\ =\dfrac{5}{7}.\dfrac{71}{72}+\dfrac{4}{9}=\dfrac{355}{504}+\dfrac{4}{9}\\ =\dfrac{193}{168}\)

a)

b)Ta có: \(x-y=7\Rightarrow x=y+7\)

\(\dfrac{2x+7}{3x-y}=\dfrac{2\left(y+7\right)+7}{3\left(y+7\right)-y}=\dfrac{2y+21}{2y+21}=1\)

\(\dfrac{2y-7}{3y-x}=\dfrac{2y-7}{3y-\left(y+7\right)}=\dfrac{2y-7}{2y-7}=1\)

\(\Rightarrow C=\dfrac{2x+7}{3x-y}+\dfrac{3y-7}{3y-x}=1+1=2\)

\(\left(\dfrac{1}{4}-\dfrac{2}{3}\right):\dfrac{3x}{5}=\dfrac{5}{2}\Leftrightarrow\dfrac{-5}{12}:\dfrac{3x}{5}=\dfrac{5}{2}\Leftrightarrow\dfrac{3x}{5}=-\dfrac{1}{6}\Leftrightarrow x=-\dfrac{5}{18}\)

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