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\(A=-1-4=-5\)
\(B=\frac{4}{3}.\frac{15}{7}-16\)
\(B=\frac{20}{7}-16\)
\(B=\frac{-92}{7}\)
\(C=\frac{28}{15}.0,25.3+\left(\frac{8}{15}-\frac{1}{4}\right)\div1\frac{23}{24}\)
\(C=1,4+\frac{17}{60}\div\frac{47}{24}\)
\(C=1,4+\frac{34}{235}\)
\(C=\frac{363}{235}\)
\(A=\frac{-15}{8}+\frac{7}{8}-4\)
\(=-1-4=-5\)
\(B=\left(4-2\frac{2}{3}\right).2\frac{1}{7}-1\frac{3}{5}:\frac{1}{10}\)
\(=\frac{4}{3}.\frac{15}{7}-\frac{8}{5}:\frac{1}{10}\)
\(=\frac{20}{7}-16=\frac{-92}{7}\)
\(C=1\frac{13}{15}.\left(0,5\right)^2.3+\left(\frac{8}{15}-25\%\right):1\frac{23}{24}\)
\(=\frac{28}{15}.\frac{1}{4}.3+\frac{17}{60}:\frac{47}{24}\)
\(=\frac{7}{15}.3+\frac{17}{60}:\frac{47}{24}\)
\(=\frac{7}{5}+\frac{34}{235}=\frac{363}{235}\)
Ta có : \(A=8\frac{2}{7}-\left(3\frac{4}{9}+4\frac{2}{7}\right)\)
\(\Rightarrow A=\frac{58}{7}-\left(\frac{31}{9}+\frac{30}{7}\right)\)
\(\Rightarrow A=\frac{58}{7}-\frac{487}{63}=\frac{5}{9}\)
P/s:Câu B tương tự nhé
Tiếp B của @Phạm Tuấn Đạt
\(B=\left(10\frac{2}{9}+2\frac{3}{5}\right)-6\frac{2}{9}\)
\(\Rightarrow B=\left(\frac{92}{9}+\frac{13}{5}\right)-\frac{56}{9}\)
\(B=\left(\frac{92}{9}-\frac{56}{9}\right)+\frac{13}{5}\)
\(B=\frac{36}{9}+\frac{13}{5}\)
\(B=4+\frac{13}{5}\)
\(B=\frac{20}{5}+\frac{13}{5}=\frac{33}{5}\)
B1
a) \(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):16\frac{2}{3}=0\)
\(1-\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=0\)
\(1-\left(x-\frac{11}{6}\right).\frac{3}{50}=0\)
\(\left(x-\frac{11}{6}\right).\frac{3}{50}=1-0\)
\(\left(x-\frac{11}{6}\right).\frac{3}{50}=1\)
\(x-\frac{11}{6}=1:\frac{3}{50}\)
\(x-\frac{11}{6}=\frac{50}{3}\)
\(x=\frac{50}{3}+\frac{11}{6}\)
\(x=\frac{37}{2}\)
b) \(\frac{3}{5}+\frac{5}{7}:x=\frac{1}{3}\)
\(\frac{5}{7}:x=\frac{1}{3}-\frac{3}{5}\)
\(\frac{5}{7}:x=-\frac{4}{15}\)
\(x=\frac{5}{7}:\left(-\frac{4}{15}\right)\)
\(x=-\frac{75}{28}\)
c) \(\left(4\frac{1}{2}-\frac{2}{5}.x\right):\frac{7}{4}=\frac{11}{9}\)
\(\left(\frac{9}{2}-\frac{2}{5}.x\right):\frac{7}{4}=\frac{11}{9}\)
\(\frac{9}{2}-\frac{2}{5}.x=\frac{11}{9}.\frac{7}{4}\)
\(\frac{9}{2}-\frac{2}{5}.x=\frac{11}{2}\)
\(\frac{2}{5}.x=\frac{9}{2}-\frac{11}{2}\)
\(\frac{2}{5}.x=-1\)
\(x=-1:\frac{2}{5}\)
\(x=-\frac{5}{2}\)
B2
a) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{2}{6}\right).24:5-\frac{9}{22}:\frac{15}{121}\)
\(=\left(\frac{3}{6}+\frac{2}{6}+\frac{2}{6}\right).24:5-\frac{9}{22}.\frac{121}{15}\)
\(=\frac{7}{6}.24:5-\frac{33}{10}\)
\(=28:5-\frac{33}{10}\)
\(=\frac{28}{5}-\frac{33}{10}\)
\(=\frac{56}{10}-\frac{33}{10}\)
\(=\frac{23}{10}\)
b) \(\frac{5}{14}+\frac{18}{35}+\left(1\frac{1}{4}-\frac{5}{4}\right):\left(\frac{5}{12}\right)^2\)
\(=\frac{25}{70}+\frac{36}{70}+\left(\frac{5}{4}-\frac{5}{4}\right):\frac{25}{144}\)
\(=\frac{61}{70}+0:\frac{25}{144}\)
\(=\frac{61}{70}+0\)
\(=\frac{61}{70}\)
A=\(7\frac{4}{15}-3\frac{2}{9}-2\frac{4}{15}\)
=\(7\frac{4}{15}-2\frac{4}{15}-3\frac{2}{9}\)
= \(5-3\frac{2}{9}\)
= \(5-\frac{29}{9}\)
= \(\frac{45}{9}-\frac{29}{9}\)
= \(\frac{11}{9}\)
=\(1\frac{2}{9}\)
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