Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(3\frac{1}{2}+4\frac{2}{5}=\left(3+4\right)+\left(\frac{1}{2}+\frac{2}{5}\right)=7+\frac{9}{10}=7\frac{9}{10}\)
nha....................................................
\(A=-7+\frac{3}{4}-\frac{1}{3}-6+\frac{5}{4}-\frac{4}{3}-3-\frac{7}{4}+\frac{5}{3}\)
\(A=\left(-7-6-3\right)+\left(\frac{3}{4}+\frac{5}{4}-\frac{7}{4}\right)+\left(\frac{5}{3}-\frac{1}{3}-\frac{4}{3}\right)\)
\(A=-16+\frac{1}{4}+0\)
\(A=-15\frac{3}{4}\)
\(A=\left(-7+\frac{3}{4}-\frac{1}{3}\right)-\left(6-\frac{5}{4}+\frac{4}{3}\right)-\left(3+\frac{7}{4}-\frac{5}{3}\right)\)
\(=-7+\frac{3}{4}-\frac{1}{3}-6+\frac{5}{4}-\frac{4}{3}-3-\frac{7}{4}+\frac{5}{3}\)
\(=\left(-7-6-3\right)+\left(\frac{3}{4}+\frac{5}{4}-\frac{7}{4}\right)+\left(\frac{-1}{3}-\frac{4}{3}+\frac{5}{3}\right)\)
\(=-16-\frac{1}{4}\)
1) \(\frac{4}{7}-\frac{1}{14}+\left|\frac{-5}{21}\right|\)
\(=\frac{4}{7}-\frac{1}{14}+\frac{5}{21}\)
\(=\frac{24}{42}-\frac{3}{42}+\frac{10}{42}\)
\(=\frac{31}{42}\)
2) \(\left|\frac{-2}{3}\right|-\frac{1}{2}+3\)
\(=\frac{2}{3}-\frac{1}{2}+\frac{3}{1}\)
\(=\frac{4}{6}-\frac{3}{6}+\frac{18}{6}\)
\(=\frac{19}{6}\)
3) \(\left|\frac{7}{-4}\right|-\frac{5}{8}+\frac{-2}{3}\)
\(=\frac{7}{4}-\frac{5}{8}+\frac{-2}{3}\)
\(=\frac{42}{24}-\frac{15}{24}+\frac{-16}{24}\)
\(=\frac{11}{24}\)
4) \(\frac{4}{5}+\left|\frac{-3}{2}\right|+\frac{1}{-4}\)
\(=\frac{4}{5}+\frac{3}{2}+\frac{-1}{4}\)
\(=\frac{16}{20}+\frac{30}{20}+\frac{-5}{20}\)
\(=\frac{41}{20}\)
5) \(\left|\frac{-1}{4}\right|-3+\frac{3}{4}\)
\(=\frac{1}{4}-\frac{3}{1}+\frac{3}{4}\)
\(=\frac{1}{4}-\frac{12}{4}+\frac{3}{4}\)
\(=-2\)
6) \(\left|\frac{-1}{3}\right|-\frac{5}{4}+\frac{1}{5}\)
\(=\frac{1}{3}-\frac{5}{4}+\frac{1}{5}\)
\(=\frac{20}{60}-\frac{75}{60}+\frac{12}{60}\)
\(=\frac{-43}{60}\)
Bài 1:
a) Ta có: \(\frac{5}{6}-\frac{2}{3}+\frac{1}{4}\)
\(=\frac{10}{12}-\frac{8}{12}+\frac{3}{12}\)
\(=\frac{2+3}{12}=\frac{5}{12}\)
b) Ta có: \(1\frac{11}{12}-\frac{5}{12}\cdot\left(\frac{4}{5}-\frac{1}{10}\right):\frac{-5}{12}\)
\(=\frac{23}{12}-\frac{5}{12}\cdot\left(\frac{8}{10}-\frac{1}{10}\right)\cdot\frac{-12}{5}\)
\(=\frac{23}{12}-\frac{5}{12}\cdot\frac{7}{10}\cdot\frac{-12}{5}\)
\(=\frac{23}{12}-\frac{-7}{10}\)
\(=\frac{115}{60}+\frac{42}{60}=\frac{157}{60}\)
Bài 2:
a) Ta có: \(\frac{1}{2}\cdot x-\frac{2}{5}=\frac{1}{5}\)
\(\Leftrightarrow\frac{1}{2}\cdot x=\frac{1}{5}+\frac{2}{5}=\frac{3}{5}\)
\(\Leftrightarrow x=\frac{3}{5}:\frac{1}{2}=\frac{3}{5}\cdot2=\frac{6}{5}\)
Vậy: \(x=\frac{6}{5}\)
b) Ta có: \(\left(1-2x\right)\cdot\frac{4}{3}=\left(-2\right)^3\)
\(\Leftrightarrow\left(1-2x\right)\cdot\frac{4}{3}=-8\)
\(\Leftrightarrow1-2x=-8:\frac{4}{3}=-8\cdot\frac{3}{4}=-6\)
\(\Leftrightarrow-2x=-6-1=-7\)
hay \(x=\frac{7}{2}\)
Vậy: \(x=\frac{7}{2}\)
kazuto kirigaya thật là bt làm ko đó ko bt thì nói đi còn bt thì làm đi
a/ \(4\frac{2}{5}:2=\frac{22}{5}.\frac{1}{2}=\frac{22}{10}=\frac{11}{5}\)
b/ \(4\frac{2}{5}:2=\left(4+\frac{2}{5}\right):2=2+\frac{1}{5}=\frac{11}{5}\)