Question

[Kuruishagi zero]

1.tính

a) \(\frac{0,75+0,6-\frac{3}{7}-\frac{3}{13}}{2,75+2,2-\frac{11}{17}-\frac{11}{13}}\)

b)\(\frac{2}{-5}\)+ \(\frac{-3}{7}\)+ \(\frac{-7}{10}\)+ \(\frac{3}{-8}\)

c) 1\(\frac{1}{8}\)- \(\frac{8}{4}\)+ \(\frac{2}{25}\)+\(\frac{1}{4}\)- \(\frac{5}{10}\)+ \(\frac{19}{25}\)- \(\frac{1}{9}\)+ \(\frac{2}{25}\)- \(\frac{1}{81}\)

2.tìm x

a) \(\frac{3}{5}\)+ \(\frac{3}{2}\). x = \(\frac{-5}{7}\)

b) (   4x - \(\frac{1}{3}\)) . (  \(\frac{3}{2}\). x + \(\frac{5}{6}\)) = 0

Answer 1

[Nobi Nobita]

1.a) Sửa lại đề: \(\frac{11}{17}\)ở mẫu chuyển thành \(\frac{11}{7}\)

\(\frac{0,75+0,6-\frac{3}{7}-\frac{3}{13}}{2,75+2,2-\frac{11}{7}-\frac{11}{13}}=\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{7}-\frac{3}{13}}{\frac{11}{4}+\frac{11}{5}-\frac{11}{7}-\frac{11}{13}}\)\(=\frac{3\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}{11\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}=\frac{3}{11}\)

( vì \(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\ne0\))

2.a) \(\frac{3}{5}+\frac{3}{2}.x=\frac{-5}{7}\)\(\Leftrightarrow\frac{3}{2}.x=\frac{-5}{7}-\frac{3}{5}\)

\(\Leftrightarrow\frac{3}{2}.x=\frac{-46}{35}\)\(\Leftrightarrow x=\frac{-46}{35}:\frac{3}{2}\)\(\Leftrightarrow x=\frac{-92}{105}\)

Vậy \(x=\frac{-92}{105}\)

b) \(\left(4x-\frac{1}{3}\right).\left(\frac{3}{2}x+\frac{5}{6}\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}4x-\frac{1}{3}=0\\\frac{3}{2}x+\frac{5}{6}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=\frac{1}{3}\\\frac{3}{2}x=\frac{-5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-5}{9}\end{cases}}\)

Vậy \(x=\frac{-5}{9}\)hoặc \(x=\frac{1}{12}\)