Tính:
\(\frac{\left(\frac{4}{7\cdot31}+\frac{6}{7\cdot41}+\frac{9}{10\cdot41}+\frac{10}{10\cdot57}\right)}{\left(\frac{7}{19\cdot31}+\frac{5}{19\cdot43}+\frac{3}{23\cdot43}+\frac{11}{23\cdot57}\right)}\)
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A=4/1.31+6/7.41+9/9.41+ 7/10.57
=20/35.31+30/35.41+45/45.41+35/50.57
=5(4/35.31+6/35.41+9/45.41+7/50.57)
=5(1/31-1/35+1/35-1/41+1/41-1/45+1/45-1/50+1/50-1/57)
=5(1/31-1/57)
B thì làm tương tự nhưng nhân với 2=> B=2(1/31-1/57)
=> A/B=5/2
\(B=\frac{4}{7\cdot31}+\frac{6}{7\cdot41}+\frac{9}{10\cdot41}+\frac{10}{10\cdot57}\)
B/5 = 4/5*7*31 + 6/5*7*41 + 9/5*10*410+7/5*10*57
B/5=4/31*35+6/35*41+9/41*50+7/50*57
B/5= 1/31-1/35+1/35-1/41+1/41-1/50+1/50-1/57
B/5=1/31-1/57
B/5=26/1767
B=130/1767
Vậy B = 130/1767
\(M=\frac{32}{323}\) \(N=\frac{86}{589}\) \(\frac{M}{N}=\frac{496}{731}\)
3) C thiếu đề
4) \(D=\frac{1}{9}-\left|\frac{-5}{23}\right|-\left(\frac{-5}{23}+\frac{1}{9}+\frac{25}{7}\right)+\frac{50}{4}-\frac{7}{30}\)
\(D=\frac{1}{9}-\frac{5}{23}+\frac{5}{23}-\frac{1}{9}-\frac{25}{7}+\frac{50}{4}-\frac{7}{30}\)
\(D=\frac{1}{9}-\frac{1}{9}-\frac{5}{23}+\frac{5}{23}+\frac{-25}{7}+\frac{50}{4}-\frac{7}{30}\)
\(D=0+0+\frac{125}{14}-\frac{7}{30}\)
\(D=\frac{913}{105}\)