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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 \(x=\frac{1}{12}\)
e) \(\frac{1}{7}.\frac{-3}{8}+\frac{-13}{8}.\frac{1}{7}\)
\(=\frac{1}{7}.\left[\left(-\frac{3}{8}\right)+\left(-\frac{13}{8}\right)\right]\)
\(=\frac{1}{7}.\left(-2\right)\)
\(=-\frac{2}{7}.\)
Chúc bạn học tốt!
a)\(\frac{1}{7}.\frac{1}{3}+\frac{1}{7}.\frac{1}{2}-\frac{1}{7}\)
\(=\frac{1}{7}.\left(\frac{1}{3}+\frac{1}{2}\right)-\frac{1}{7}\)
\(=\frac{1}{7}.\left(\frac{2}{6}+\frac{3}{6}\right)-\frac{1}{7}\)
\(=\frac{1}{7}.\frac{5}{6}-\frac{1}{7}\)
\(=\frac{5}{42}-\frac{1}{7}\)
\(=\frac{5}{42}-\frac{6}{42}=-\frac{1}{42}\)
a) \(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-\frac{1}{11.15}-\frac{1}{15.19}-\frac{1}{19.23}-\frac{1}{23.27}\)
\(=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\frac{1}{15.19}+\frac{1}{19.23}+\frac{1}{23.27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{19}-\frac{1}{19}+\frac{1}{23}-\frac{1}{23}+\frac{1}{27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}.\frac{8}{27}\)
\(=\frac{1}{2}-\frac{2}{27}\)
\(=\frac{23}{54}\)
b) \(1-\frac{1}{5.10}-\frac{1}{10.15}-\frac{1}{15.20}-...-\frac{1}{95.100}\)
\(=1-\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{95.100}\right)\)
\(=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+\frac{1}{20}-...-\frac{1}{95}-\frac{1}{100}\right)\)
\(=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{100}\right)\)
\(=1-\frac{1}{5}.\frac{19}{100}\)
\(=1-\frac{19}{500}\)
\(=\frac{481}{500}\)
a/ \(\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-\frac{32}{17}+\frac{14}{21}=\left(\frac{15}{34}+\frac{19}{34}\right)+\left(\frac{7}{21}+\frac{14}{21}\right)-\frac{32}{17}=1+1-\frac{32}{17}=\frac{2}{17}\)
a, 6/7 + (2/11 - 6/7) - (13/11 + 1)
= 6/7 + 2/11 - 6/7 - 13/11 - 1
= (6/7 - 6/7) - (13/11 - 2/11) - 1
= 0 - 1 - 1
= -2
1.\(\left(\frac{-10}{15}+1\frac{1}{6}-0,5\right):\left(4\frac{4}{105}-5\frac{4}{81}\right)\)
=\(\left(\frac{-10}{15}+\frac{7}{6}-0,5\right):\left(\frac{425}{105}-\frac{409}{81}\right)\)
=\(0:\left(\frac{425}{105}-\frac{409}{81}\right)\)
=\(0\)
2. \(\frac{1}{2}\left(\frac{-11}{19}\right)-50\%.\left(\frac{-1}{19}\right)+\frac{10}{19}.\frac{1111}{2222}\)
=\(\frac{-11}{38}-\frac{-1}{38}+\frac{5}{19}\)
=\(\frac{-11}{38}+\frac{1}{38}+\frac{5}{19}\)
=\(0\)
k cho mik nha