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\(\frac{a+10}{b}=\frac{3}{4}\)
\(\Rightarrow\frac{a}{b}+\frac{10}{b}=\frac{3}{4}\)
\(\Rightarrow\frac{10}{b}=\frac{3}{4}-\frac{a}{b}=\frac{3}{4}-\frac{7}{12}=\frac{1}{6}\)
\(\Rightarrow b=\frac{10\cdot6}{1}=60\)
\(\frac{a}{b}=\frac{7}{12}\Rightarrow a=\frac{7b}{12}=\frac{7\cdot60}{12}=35\)
\(\frac{-4}{7}+\left(\frac{-2}{3}+\frac{4}{7}\right)\)
\(=\frac{-4}{7}+\frac{-2}{3}+\frac{4}{7}\)
\(=\left(\frac{-4}{7}+\frac{4}{7}\right)+\frac{-2}{3}\)
\(=0+\frac{-2}{3}=\frac{-2}{3}\)
~ Hok tốt ~
\(\frac{-3}{10}.\frac{5}{12}+\frac{-3}{10}.\frac{7}{12}+5\frac{3}{10}\)
\(=\frac{-3}{10}.\frac{5}{12}+\frac{-3}{10}.\frac{7}{12}+\frac{53}{10}\)
\(=\frac{-3}{10}.\left(\frac{5}{12}+\frac{7}{12}\right)+\frac{53}{10}\)
\(=\frac{-3}{10}.1+\frac{53}{10}\)
\(=\frac{-3}{10}+\frac{53}{10}=\frac{50}{10}=5\)
~ Hok tốt ~
\(a,-\frac{4}{7}+\left(-\frac{2}{3}+\frac{3}{7}\right)=-\frac{4}{7}-\frac{2}{3}+\frac{3}{7}=-\frac{4}{7}+\frac{3}{7}-\frac{2}{3}=-\frac{1}{7}-\frac{2}{3}=-\frac{17}{21}\)
\(b,-\frac{3}{10}\cdot\frac{5}{12}+-\frac{3}{10}\cdot\frac{7}{12}+5\cdot\frac{3}{10}=\frac{3}{10}\cdot-\frac{5}{12}+\frac{3}{10}\cdot-\frac{7}{12}+5\cdot\frac{3}{10}=\frac{3}{10}\cdot\left(-\frac{5}{12}-\frac{7}{12}+5\right)\)
\(=\frac{3}{10}\cdot4=\frac{6}{5}\)
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\(\frac{a}{b}=\frac{7}{12}\Rightarrow a=\frac{7}{12}b\)
\(\Rightarrow\frac{a+10}{b}=\frac{\frac{7}{12}b+10}{b}=\frac{3}{4}\)
\(\Rightarrow4\left(\frac{7}{12}b+10\right)=3b\)
\(\frac{7}{3}b+40=3b\)
\(3b-\frac{7}{3}b=40\)
\(\frac{2}{3}b=40\)
\(b=40.\frac{3}{2}=60\)
\(a=\frac{7}{12}.60=35\)
\(a+b=60+35=95\)
\(\frac{a}{b}=\frac{7}{12}\Rightarrow\frac{a}{7}=\frac{b}{12}\)
Đặt \(\frac{a}{7}=\frac{b}{12}=k\)
\(\Rightarrow a=7k\)
\(b=12k\)
Ta có: \(\frac{a+10}{b}=\frac{3}{4}\)
\(\Rightarrow4\left(a+10\right)=3b\)
\(4a+40=3b\)
\(\Rightarrow3b-4a=40\)
\(3\cdot12k-4\cdot7k=40\)
\(36k-28k=40\)
\(\left(36-28\right)k=40\)
\(8k=40\)
\(k=40:8\)
\(k=5\)
\(\Rightarrow a=7k=7\cdot5=35\)
\(b=12k=12\cdot5=60\)
Vậy a+b= 36 + 60 = 96