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T
14 tháng 4 2019
\(A=\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(A=\left(\frac{3}{8}+\frac{-6}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(A=\left(\frac{-3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(A=\left(\frac{-36}{24}+\frac{56}{24}\right):\frac{5}{6}+\frac{1}{2}\)
\(A=\frac{5}{6}:\frac{5}{6}+\frac{1}{2}\)
\(A=\frac{5}{6}\times\frac{6}{5}+\frac{1}{2}\)
\(A=1+\frac{1}{2}\)
\(A=\frac{1}{1}+\frac{1}{2}=\frac{2}{2}+\frac{1}{2}\)
\(A=\frac{3}{2}\)
9 tháng 8 2019
\(bai1:a,\frac{3}{7}\cdot\frac{-5}{9}+\frac{4}{9}\cdot\frac{3}{7}-\frac{3}{7}\cdot\frac{8}{9}\)
\(< =>\frac{-15}{63}+\frac{12}{63}-\frac{24}{63}\)
\(< =>\frac{-15+12-24}{63}\)
\(< =>\frac{-3}{7}\)
\(b,1\frac{13}{15}\cdot0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
\(< =>\frac{28}{15}\cdot\frac{3}{4}-\left(\frac{11}{20}+\frac{1}{4}\right):\frac{7}{5}\)
\(< =>\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)
\(< =>\frac{7}{5}-\frac{4}{7}\)
\(< =>\frac{29}{35}\)
\(bai2:\)
\(a,\frac{-3}{4}\cdot x-\frac{4}{10}=\frac{1}{5}\)
\(< =>\frac{-3}{4}\cdot x=\frac{1}{5}+\frac{4}{10}\)
\(< =>\frac{-3}{4}\cdot x=\frac{3}{5}\)
\(< =>x=\frac{3}{5}:\frac{-3}{4}\)
\(< =>x=\frac{-4}{5}\)
\(b,3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
\(< =>3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)
\(< =>\left[3\left(x-\frac{1}{3}\right)\right]=\frac{1}{12}< =>x-\frac{1}{3}=\frac{1}{12}:3=\frac{1}{36}=>x=\frac{1}{36}+\frac{1}{3}=>x=\frac{13}{36}\)
\(< =>\left[\frac{1}{3}\cdot x\right]=\frac{1}{12}< =>x=\frac{1}{12}:\frac{1}{3}=>x=\frac{1}{4}\)
MM
9 tháng 8 2019
Bài 1:
a)\(\frac{3}{7}.\frac{-5}{9}+\frac{4}{9}.\frac{3}{7}-\frac{3}{7}.\frac{8}{9}\) b,\(1\frac{13}{15}.0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
\(=\frac{3}{7}.(\frac{-5}{9}+\frac{4}{9}-\frac{8}{9})\) \(=\frac{28}{15}.\frac{3}{4}-\left(\frac{11}{20}+\frac{5}{20}\right):\frac{7}{5}\)
\(=\frac{3}{7}.\frac{-9}{9}\) \(=\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)
\(=\frac{-3}{7}\) \(=\frac{7}{5}-\frac{4}{7}\)
\(=\frac{29}{35}\)
Bài 2:
a)\(\frac{-3}{4}x-\frac{4}{10}=\frac{1}{5}\) b,\(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
\(\frac{-3}{4}x\) \(=\frac{1}{5}+\frac{4}{10}\) \(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)
\(\frac{-3}{4}x\) \(=\frac{3}{5}\) \(\left(x.3-\frac{1}{3}.3\right)+\frac{1}{3}x=\frac{1}{12}\)
\(x\) \(=\frac{3}{5}:\frac{-3}{4}\) \(\left(x.3-1\right)+\frac{1}{3}x=\frac{1}{12}\)
\(x\) \(=\frac{4}{-5}\) \(x.\left(3+\frac{1}{3}\right)-1=\frac{1}{12}\)
\(x.\left(3+\frac{1}{3}\right)=\frac{1}{12}+1\)
\(x.\frac{10}{3}=\frac{13}{12}\)
\(x=\frac{13}{12}:\frac{10}{3}\)
\(x=\frac{13}{40}\)
28 tháng 7 2019
\(C=(\frac{2}{3}-\frac{1}{4}+\frac{5}{11}):(\frac{5}{12}+1-\frac{7}{11})\)
\(=\left(\frac{88}{132}-\frac{33}{132}+\frac{60}{132}\right):\left(\frac{55}{132}+\frac{132}{132}-\frac{84}{132}\right)=\left(\frac{115}{132}\right):\frac{103}{132}=\frac{115}{132}.\frac{132}{103}=\frac{115}{103}\)
\(D=1\frac{1}{3}+\frac{1}{8}:\left(0,75-\frac{1}{2}\right)-\frac{25}{100}.\frac{1}{2}=\frac{1}{3}+\frac{1}{8}:\frac{1}{4}-\frac{1}{8}=\frac{1}{3}+\frac{1}{2}-\frac{1}{8}=\frac{8+12-3}{24}=\frac{17}{24}\)
\(E=\left(-\frac{1}{2}\right)^2-\left(-2\right)^2-5^0=\frac{1}{4}-4-1=\frac{1-16-4}{4}=\frac{-19}{4}\)
\(1\frac{13}{15}.0,75-\left(\frac{11}{20}+\frac{25}{100}\right):\frac{3}{7}\)
= \(\frac{7}{5}-\frac{4}{5}:\frac{3}{7}\)
= \(\frac{7}{5}-\frac{28}{15}\)
= \(-\frac{7}{15}\)
\(1\frac{13}{15}\cdot0,75-\left(\frac{11}{20}+\frac{25}{100}\right):\frac{3}{7}\)
=>\(\frac{28}{15}\cdot\frac{3}{4}-\left(\frac{55}{100}+\frac{25}{100}\right)\cdot\frac{7}{3}\)
=>\(\frac{84}{60}-\frac{80}{100}\cdot\frac{7}{3}\)
=>\(\frac{7}{5}-\frac{4}{5}\cdot\frac{7}{3}=\frac{7}{5}-\frac{28}{15}\)
=>\(\frac{7}{5}-\frac{18}{14}=\frac{98}{60}-\frac{90}{60}=\frac{8}{60}=\frac{4}{30}\)