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Bài 1:
a) Ta có: \(\dfrac{2}{5}\cdot x+\dfrac{1}{3}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{2}{5}\cdot x=\dfrac{1}{5}-\dfrac{1}{3}=\dfrac{-2}{15}\)
\(\Leftrightarrow x=\dfrac{-2}{15}:\dfrac{2}{5}=\dfrac{-2}{15}\cdot\dfrac{5}{2}\)
hay \(x=-\dfrac{1}{3}\)
Vậy: \(x=-\dfrac{1}{3}\)
b) Ta có: \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}\)
\(\Leftrightarrow x=\dfrac{5}{3}:\dfrac{3}{10}=\dfrac{5}{3}\cdot\dfrac{10}{3}\)
hay \(x=\dfrac{50}{9}\)
Vậy: \(x=\dfrac{50}{9}\)
c) Ta có: \(\dfrac{4}{9}-\dfrac{5}{3}\cdot x=-2\)
\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{4}{9}+2=\dfrac{22}{9}\)
\(\Leftrightarrow x=\dfrac{22}{9}:\dfrac{5}{3}=\dfrac{22}{9}\cdot\dfrac{3}{5}\)
hay \(x=\dfrac{22}{15}\)
Vậy: \(x=\dfrac{22}{15}\)
d) Ta có: \(\dfrac{5}{7}:x-3=\dfrac{-2}{7}\)
\(\Leftrightarrow\dfrac{5}{7}:x=\dfrac{-2}{7}+3=\dfrac{19}{21}\)
\(\Leftrightarrow x=\dfrac{5}{7}:\dfrac{19}{21}=\dfrac{5}{7}\cdot\dfrac{21}{19}\)
hay \(x=\dfrac{15}{19}\)
Vậy:\(x=\dfrac{15}{19}\)
A = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + 10 - 11 - 12 + .........- 299 - 300 + 301 + 302
A = 1 + ( 2 - 3 - 4 + 5 ) + ( 6 - 7 -8 + 9 ) + ( 10 - 11 - 12 + 13 ) + ............- ( 298 - 299 - 300 + 301 ) + 302
A = 1 + 303
A = 303
A = 1+ 2 - 3 - 4 - 5 + 6 - 7 -8 + 9 +......+ 298 - 299 - 300 + 301 + 302
A = 1 + ( 2 - 3 - 4 + 5 ) + ( 6 - 7 - 8 + 9 ) + ......+ ( 298 - 299 - 300 + 301 ) + 302
A = 1 + ( 2 + 5 - 3 - 4 ) + ( 6 + 9 - 7 - 8 ) + .......+ ( 298 + 301 - 299 - 300) + 302
A = 1 + 0 + 0 + .......+ 0 + 302
A = 1 + 302
A = 303
\(C1:2\dfrac{4}{9}+1\dfrac{1}{6}=\dfrac{22}{9}+\dfrac{7}{6}=\dfrac{44}{18}+\dfrac{21}{18}=\dfrac{65}{18}\)
\(C2:2\dfrac{4}{9}+1\dfrac{1}{6}=\left(2+1\right)+\left(\dfrac{4}{9}+\dfrac{1}{6}\right)=3+\dfrac{11}{18}=\dfrac{54}{18}+\dfrac{11}{18}=\dfrac{65}{18}\)
\(2\dfrac{4}{9}+1\dfrac{1}{6}=\dfrac{22}{9}+\dfrac{7}{6}=\dfrac{44}{18}+\dfrac{21}{18}=\dfrac{65}{18}\)