Tính:
\(\frac{\left(\frac{4}{7\cdot31}+\frac{6}{7\cdot41}+\frac{9}{10\cdot41}+\frac{10}{10\cdot57}\right)}{\left(\frac{7}{19\cdot31}+\frac{5}{19\cdot43}+\frac{3}{23\cdot43}+\frac{11}{23\cdot57}\right)}\)
Tính tỉ số \(\frac{A}{B}\) biết : A = \(\frac{4}{7\cdot31}\)+\(\frac{6}{7\cdot41}\)+\(\frac{9}{10\cdot41}\)+\(\frac{7}{10\cdot57}\)và B = \(\frac{7}{19\cdot31}\)+ \(\frac{5}{19\cdot43}\) + \(\frac{3}{23\cdot43}\)+ \(\frac{11}{23\cdot57}\)
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
cho M = \(\frac{4}{31\cdot7}\)+\(\frac{6}{7\cdot41}\)+\(\frac{9}{10\cdot41}\)+\(\frac{7}{10\cdot57}\)và N = \(\frac{7}{19\cdot31}\)+\(\frac{5}{19\cdot43}\)+\(\frac{3}{23\cdot43}\)+\(\frac{11}{23\cdot57}\). Tính M/N
Bài 1 :Tính tỉ số A/B biết
a) \(A=\frac{4}{7\cdot31}+\frac{6}{7\cdot41}+\frac{9}{10\cdot41}+\frac{7}{10\cdot57}\)
\(B=\frac{7}{19\cdot31}+\frac{5}{19\cdot43}+\frac{3}{23\cdot43}+\frac{11}{23\cdot57}\)
b)\(A=\frac{40}{31\cdot39}+\frac{35}{39\cdot46}+\frac{30}{46\cdot52}+\frac{25}{52\cdot57}\)
\(B=\frac{91}{19\cdot31}+\frac{65}{19\cdot43}+\frac{39}{989}+\frac{143}{1311}\)
Bài 2 : Tính Avà B biết :
a) \(A=\frac{2006}{1\cdot2}+\frac{2006}{2\cdot3}+...+\frac{2006}{2006\cdot2007}\)
\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{2004\cdot2005\cdot2006}\)
Tính :
\(B=\left(\frac{7}{19\cdot31}+\frac{5}{19\cdot43}+\frac{3}{23\cdot43}+\frac{11}{23\cdot57}\right)\div\left(\frac{1}{31}-\frac{1}{57}\right)\)
\(B=\frac{4}{7\cdot31}+\frac{6}{7\cdot41}+\frac{9}{10\cdot41}+\frac{7}{10\cdot57}\)
\(B=\frac{4}{7\cdot31}+\frac{6}{7\cdot41}+\frac{9}{10\cdot41}+\frac{10}{10\cdot57}\)
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
Bạn Vũ Thành Huy giải hay wá :) Cảm ơn bạn.
\(B=\frac{4}{7.31}+\frac{6}{7.41}+\frac{9}{10.41}+\frac{7}{10.57}\)
\(B=\frac{5}{5}.\left(\frac{4}{7.31}+\frac{6}{7.41}+\frac{9}{10.41}+\frac{7}{10.57}\right)\)
\(B=5\left(\frac{4}{5.7.31}+\frac{6}{5.7.41}+\frac{9}{5.10.41}+\frac{7}{5.10.57}\right)\)
\(B=5\left(\frac{4}{35.31}+\frac{6}{35.41}+\frac{9}{50.41}+\frac{7}{50.57}\right)\)
\(B=5\left(\frac{4}{31.35}+\frac{6}{35.41}+\frac{9}{41.50}+\frac{7}{50.57}\right)\)
\(B=5\left(\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}\right)\)
\(B=5.\left(\frac{1}{31}-\frac{1}{57}\right)\)
\(B=5.\frac{26}{1767}\)
\(B=\frac{130}{1767}\)
B=\(\frac{4}{7\cdot31}+\frac{6}{7\cdot41}+\frac{9}{10\cdot41}+\frac{7}{10\cdot57}+\frac{13}{57\cdot14}\)
Tính \(\frac{M}{N}\)biết :
M = \(\frac{2}{7\cdot19}+\frac{6}{7\cdot27}+\frac{1}{4\cdot9}+\frac{7}{12\cdot43}+\frac{8}{17\cdot43}\)
N = \(\frac{9}{7\cdot19}+\frac{5}{7\cdot33}+\frac{7}{10\cdot33}+\frac{11}{10\cdot31}\)
\(M=\frac{32}{323}\) \(N=\frac{86}{589}\) \(\frac{M}{N}=\frac{496}{731}\)
Tính :
\(B=\left(\dfrac{7}{19\cdot31}+\dfrac{5}{19\cdot43}+\dfrac{3}{23\cdot43}+\dfrac{11}{23\cdot57}\right)\div\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)
Tính giá trị biểu thức:
3) C= \(\frac{5}{18}+\frac{8}{19}-\frac{7}{21}+\left(\frac{-10}{36}+\frac{11}{19}+\frac{1}{3}\right)-\frac{5}{8}\)\(\frac{5}{8}\)
4) D= \(\frac{1}{9}-\left|\frac{-5}{23}\right|-\left(\frac{-5}{23}+\frac{1}{9}+\frac{25}{7}\right)+\frac{50}{4}-\frac{7}{30}\)
3) C thiếu đề
4) \(D=\frac{1}{9}-\left|\frac{-5}{23}\right|-\left(\frac{-5}{23}+\frac{1}{9}+\frac{25}{7}\right)+\frac{50}{4}-\frac{7}{30}\)
\(D=\frac{1}{9}-\frac{5}{23}+\frac{5}{23}-\frac{1}{9}-\frac{25}{7}+\frac{50}{4}-\frac{7}{30}\)
\(D=\frac{1}{9}-\frac{1}{9}-\frac{5}{23}+\frac{5}{23}+\frac{-25}{7}+\frac{50}{4}-\frac{7}{30}\)
\(D=0+0+\frac{125}{14}-\frac{7}{30}\)
\(D=\frac{913}{105}\)
\(C=\frac{5}{18}+\frac{8}{19}-\frac{7}{21}+\left(\frac{-10}{36}+\frac{11}{19}+\frac{1}{3}\right)-\frac{5}{8}\)
\(C=\frac{5}{18}+\frac{8}{19}-\frac{1}{3}+\frac{-5}{18}+\frac{11}{19}+\frac{1}{3}-\frac{5}{8}\)
\(C=\left(\frac{5}{18}+\frac{-5}{18}\right)+\left(\frac{8}{19}+\frac{11}{19}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)-\frac{5}{8}\)
\(C=0+1+0-\frac{5}{8}\)
\(C=\frac{3}{8}\)