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2 giờ trước (15:57)

a: Ta có: \(3x+\left(x-\frac{9}{20}\right)=-\frac{13}{40}\)

=>\(3x+x-\frac{9}{20}=-\frac{13}{40}\)

=>\(4x=-\frac{13}{40}+\frac{9}{20}=-\frac{13}{40}+\frac{18}{40}=\frac{5}{40}=\frac18\)

=>\(x=\frac18:4=\frac{1}{32}\)

b: \(x+\left(\frac14x-2,5\right)=-\frac{11}{20}\)

=>\(x+\frac14x-2,5=-\frac{11}{20}\)

=>\(1,25x=-0,55+2,5=1,95\)

=>\(x=\frac{1.95}{1.25}=\frac{195}{125}=\frac{39}{25}\)

c: \(\frac35x+\left(x+0,5\right)=-\frac{13}{15}\)

=>\(\frac35x+x+0,5=-\frac{13}{15}\)

=>\(\frac85x=-\frac{13}{15}-0,5=-\frac{26}{30}-\frac{15}{30}=-\frac{41}{30}\)

=>\(x=-\frac{41}{30}:\frac85=-\frac{41}{30}\cdot\frac58=\frac{-41}{6\cdot8}=-\frac{41}{48}\)

d: \(-\frac23x+\left(4x-\frac67\right)=\frac{9}{21}\)

=>\(-\frac23x+4x-\frac67=\frac37\)

=>\(\frac{10}{3}x=\frac37+\frac67=\frac97\)

=>\(x=\frac97:\frac{10}{3}=\frac97\cdot\frac{3}{10}=\frac{27}{70}\)

S
2 giờ trước (16:05)

bài 11: câu a:

\(3x+\left(x-\frac{9}{20}\right)=-\frac{13}{40}\)

\(3x+x-\frac{9}{20}=-\frac{13}{40}\)

\(4x=-\frac{13}{40}+\frac{9}{20}\)

\(4x=-\frac{13}{40}+\frac{18}{40}\)

\(4x=\frac{5}{40}\)

\(4x=\frac18\)

\(x=\frac18:4=\frac18\cdot\frac14=\frac{1}{32}\)

b. \(x+\left(\frac14x-2,5\right)=-\frac{11}{20}\)

\(x+\frac14x-2,5=-\frac{11}{20}\)

\(\frac54x-2,5=-\frac{11}{20}\)

\(\frac54x=-\frac{11}{20}+2,5\)

\(\frac54x=\frac{39}{20}\)

\(x=\frac{39}{20}:\frac54=\frac{39}{20}\cdot\frac45=\frac{39}{25}\)

c. \(\frac35x+\left(x+0,5\right)=-\frac{13}{15}\)

\(\frac35x+x+0,5=-\frac{13}{15}\)

\(\frac85x+\frac12=-\frac{13}{15}\)

\(\frac85x=-\frac{13}{15}-\frac12\)

\(\frac85x=-\frac{41}{30}\)

\(x=-\frac{41}{30}:\frac85=-\frac{41}{30}\cdot\frac58=-\frac{41}{48}\)

\(d.-\frac23x+\left(4x-\frac67\right)=\frac{9}{21}\)

\(-\frac23x+4x-\frac67=\frac{9}{21}\)

\(\frac{10}{3}x=\frac97\)

\(x=\frac97:\frac{10}{3}=\frac97\cdot\frac{3}{10}=\frac{27}{70}\)