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\(\frac{5\cdot8+10\cdot24+15\cdot32}{10\cdot16+20\cdot48+30\cdot64}\)
\(\frac{5\cdot8\left(1+2\cdot3+3\cdot4\right)}{10\cdot16\left(1+2\cdot3+3\cdot4\right)}\)
\(\frac{1}{4}\)
\(\frac{5.8+10.24+15.32}{10.16+20.48=30.64}\)
\(\frac{40+40.6+40.12}{160+160.6+160.12}\)
\(\frac{40.\left(1+6+12\right)}{160.\left(1+6+12\right)}\)
\(\frac{40}{140}=\frac{1}{4}\)
\(A=\frac{4116-14}{10290-35}=\frac{294.14-14}{35.294-35}=\frac{293.14}{293.35}=\frac{14}{35}=\frac{2}{5}\)
\(B=\frac{2929-101}{2.1919+404}=\frac{29.101-101}{38.101+4.101}=\frac{101.28}{101.42}=\frac{28}{42}=\frac{2}{3}\)
\(I=\frac{\frac{25}{17}-\frac{25}{27}-\frac{25}{37}-\frac{25}{47}}{\frac{45}{17}-\frac{45}{27}-\frac{45}{37}-\frac{45}{47}}\)
\(I=\frac{25.\left(\frac{1}{17}-\frac{1}{27}-\frac{1}{37}-\frac{1}{47}\right)}{45.\left(\frac{1}{17}-\frac{1}{27}-\frac{1}{37}-\frac{1}{47}\right)}\)
\(I=\frac{25}{45}=\frac{5}{9}\)