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a) \(\frac{-2}{5}+\frac{3}{10}+\frac{-3}{5}\)
\(=\left[\left(-\frac{2}{5}\right)+\left(-\frac{3}{5}\right)\right]+\frac{3}{10}\)
\(=\left(-1\right)+\frac{3}{10}\)
\(=-\frac{7}{10}.\)
c) \(15\frac{1}{4}:\frac{5}{7}-25\frac{1}{4}:\frac{5}{7}\)
\(=\frac{61}{4}:\frac{5}{7}-\frac{101}{4}:\frac{5}{7}\)
\(=\left(\frac{61}{4}-\frac{101}{4}\right):\frac{5}{7}\)
\(=\left(-10\right):\frac{5}{7}\)
\(=-14.\)
Chúc bạn học tốt!
giúp mik vs, mik bik các pạn giờ này đang ngủ rùi nhưng giúp mik lần này thui.yêu các pạn nhìu
\(5\frac{1}{2}+\left(-3\right)=\frac{11}{2}+\frac{-3}{1}\)\(=\frac{11}{2}+\frac{-6}{2}=\frac{5}{2}\)\(;\)
\(4\frac{9}{11}+\left(-2\frac{1}{11}\right)=\frac{53}{11}+\frac{-23}{11}\)\(=\frac{30}{11}\)\(;\)
\(2\frac{1}{2}+\left(-6\right)=\frac{5}{2}+\frac{-6}{1}\)\(=\frac{5}{2}+\frac{-12}{2}=\frac{-7}{2}\)\(;\)
\(\left(-\frac{4}{5}\right)+\frac{1}{2}=\frac{-4}{5}+\frac{1}{2}\)\(=\frac{-8}{10}+\frac{5}{10}=\frac{-3}{10}\)\(;\)
\(4,3-\left(-1,2\right)=4,3+1,2=5,5\)\(=\frac{55}{10}=\frac{11}{2}\)\(;\)
\(0-\left(-0,4\right)=0+0,4=0,4\)\(=\frac{4}{10}=\frac{2}{5}\)\(;\)
\(\frac{-2}{3}-\frac{-1}{3}=\frac{-2}{3}+\frac{1}{3}=\frac{-1}{3}\)\(;\)
\(\frac{-1}{2}-\frac{-1}{6}=\frac{-1}{2}+\frac{1}{6}\)\(=\frac{-3}{6}+\frac{1}{6}=\frac{-2}{6}=\frac{-1}{3}\)\(;\)
\(x+\frac{1}{3}=\frac{3}{4}\) \(;\) \(x-\frac{2}{5}=\frac{5}{7}\) \(;\)
\(x=\frac{3}{4}-\frac{1}{3}\) \(x=\frac{5}{7}+\frac{2}{5}\)
\(x=\frac{5}{12}\) \(x=\frac{39}{35}\)
\(-x-\frac{2}{3}=-\frac{6}{7}\) \(;\) \(\frac{4}{7}-x=\frac{1}{3}\)
\(\frac{6}{7}-\frac{2}{3}=x\) \(\frac{4}{7}-\frac{1}{3}=x\)
\(\frac{4}{21}=x\) \(\Leftrightarrow\)\(x=\frac{4}{21}\) \(\frac{5}{21}=x\)\(\Leftrightarrow\)\(x=\frac{5}{12}\)
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 \(x=\frac{1}{12}\)
a)\(\frac{1}{7}.\frac{1}{3}+\frac{1}{7}.\frac{1}{2}-\frac{1}{7}\)
\(=\frac{1}{7}.\left(\frac{1}{3}+\frac{1}{2}\right)-\frac{1}{7}\)
\(=\frac{1}{7}.\left(\frac{2}{6}+\frac{3}{6}\right)-\frac{1}{7}\)
\(=\frac{1}{7}.\frac{5}{6}-\frac{1}{7}\)
\(=\frac{5}{42}-\frac{1}{7}\)
\(=\frac{5}{42}-\frac{6}{42}=-\frac{1}{42}\)
\(3-\frac{2}{3}+\frac{3}{5}\cdot\left(-\frac{10}{9}-\frac{25}{3}\right)-\frac{5}{6}\)
\(=3-\frac{2}{3}+\frac{3}{5}\cdot\left(-\frac{10}{9}-\frac{75}{9}\right)-\frac{5}{6}\)
\(=3-\frac{2}{3}+\frac{3}{5}\cdot-\frac{85}{9}-\frac{5}{6}\)
\(=3-\frac{2}{3}+\left(-\frac{17}{3}\right)-\frac{5}{6}\)
\(=\frac{-25}{6}\)
\(3-\frac{2}{3}+\frac{3}{5}.\left(\frac{-10}{9}-\frac{25}{3}\right)-\frac{5}{6}\)
\(=3-\frac{2}{3}+\frac{3}{5}.\left(\frac{-10}{9}-\frac{75}{9}\right)-\frac{5}{6}\)
\(=3-\frac{2}{3}+\frac{3}{5}.\frac{-85}{9}-\frac{5}{6}\)
\(=3-\frac{2}{3}+\frac{3.\left(-85\right)}{5.9}-\frac{5}{6}\)
\(=3-\frac{2}{3}+\frac{1.\left(-17\right)}{1.3}-\frac{5}{6}\)
\(=3-\frac{2}{3}+\frac{-17}{3}-\frac{5}{6}\)
\(=\frac{3}{1}-\frac{2}{3}+\frac{-17}{3}-\frac{5}{6}\)
\(=\frac{18}{6}-\frac{4}{6}+\frac{-34}{6}-\frac{5}{6}\)
\(=\frac{18-4+\left(-34\right)-5}{6}\)
\(=\frac{-25}{6}\)