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\(\frac{4}{8.13}+\frac{4}{13.18}+\frac{4}{18.24}+...+\frac{4}{253.258}\)
\(=\frac{4}{5}\cdot\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+...+\frac{1}{253}-\frac{1}{258}\right)\)
\(=\frac{4}{5}\cdot\left(\frac{1}{8}-\frac{1}{258}\right)\)
\(=\frac{4}{5}\cdot\frac{125}{1032}\)
\(=\frac{25}{258}\)
\(\frac{4}{8.13}+\frac{4}{13.18}+\frac{4}{18.23}+...+\frac{4}{253.258}\)
\(=\frac{4}{5}\left(\frac{5}{8.13}+\frac{5}{13.18}+\frac{5}{18.23}+...+\frac{5}{253.258}\right)\)
\(=\frac{4}{5}\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+...+\frac{1}{253}-\frac{1}{258}\right)\)
\(=\frac{4}{5}\left(\frac{1}{8}-\frac{1}{258}\right)\)
\(=\frac{4}{5}.\frac{125}{1032}=\frac{25}{258}\)
\(A=\frac{1}{7\cdot12}+\frac{1}{12\cdot17}+\frac{1}{17\cdot22}+...+\frac{1}{52\cdot57}\)
\(A=\frac{1}{5}\left(\frac{5}{7\cdot12}+\frac{5}{12\cdot17}+\frac{5}{17\cdot22}+...+\frac{5}{52\cdot57}\right)\)
\(A=\frac{1}{5}\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{52}-\frac{1}{57}\right)\)
\(A=\frac{1}{5}\left(\frac{1}{7}-\frac{1}{57}\right)=\frac{1}{5}\cdot\frac{50}{399}=\frac{10}{399}\)
\(B=\frac{10}{8\cdot13}+\frac{10}{13\cdot18}+\frac{10}{18\cdot23}+...+\frac{10}{253\cdot258}\)
\(B=\frac{10}{5}\left(\frac{5}{8\cdot13}+\frac{5}{13\cdot18}+\frac{5}{18\cdot23}+...+\frac{5}{253\cdot258}\right)\)
\(B=2\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{253}-\frac{1}{258}\right)\)
\(B=2\left(\frac{1}{8}-\frac{1}{258}\right)=2\cdot\frac{125}{1032}=\frac{125}{516}\)
*Cái đây giải thích hơi bị " khó hiểu " :
Chỗ mẫu (12 - 7) = (17 - 12) = ... = (57 - 52) = 5
Tử là 1 , mẫu là 5 nên tử/mẫu = 1/5
Hay \(\frac{1}{5}\left(\frac{5}{7\cdot12}+\frac{5}{12\cdot17}+...+\frac{5}{52\cdot57}\right)\)
Còn bạn Trương Bùi Linh thì :
Mẫu = (13 - 8) = (18 - 13) = (23 - 18) = ... = 5
Tử là 10,mẫu là 5 => tử / mẫu = 10/5 = 2
\(A=\frac{34}{7.13}+\frac{51}{13.22}+\frac{85}{22.37}+\frac{68}{37.49}\)
\(=17.\left(\frac{2}{7.13}+\frac{3}{13.22}+\frac{5}{22.37}+\frac{4}{37.49}\right)\)
\(=17.\frac{1}{3}.\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)\)
\(=\frac{17}{3}.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+\frac{1}{22}-\frac{1}{37}+\frac{1}{37}-\frac{1}{49}\right)\)
\(=\frac{17}{3}.\left(\frac{1}{7}-\frac{1}{49}\right)\)
\(=\frac{17}{3}.\frac{6}{49}\)
\(=\frac{34}{49}\)
\(B=\frac{39}{7.16}+\frac{65}{16.31}+\frac{52}{31.43}+\frac{26}{43.49}\)
\(=13.\left(\frac{3}{7.16}+\frac{5}{16.31}+\frac{4}{31.43}+\frac{2}{43.49}\right)\)
\(=13.\frac{1}{3}.\left(\frac{9}{7.16}+\frac{15}{16.31}+\frac{12}{31.43}+\frac{6}{43.49}\right)\)
\(=\frac{13}{3}.\left(\frac{1}{7}-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}+\frac{1}{31}-\frac{1}{43}+\frac{1}{43}-\frac{1}{49}\right)\)
\(=\frac{13}{3}.\left(\frac{1}{7}-\frac{1}{49}\right)\)
\(=\frac{13}{3}.\frac{6}{49}\)
\(=\frac{26}{49}\)
Câu C sai đề rồi , phải là như thế này :
\(C=\frac{4}{8.13}+\frac{4}{13.18}+\frac{4}{18.23}+...+\frac{4}{253.258}\)
\(=\frac{4}{5}.\left(\frac{5}{8.13}+\frac{5}{13.18}+\frac{5}{18.23}+...+\frac{5}{253.258}\right)\)
\(=\frac{4}{5}.\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+...+\frac{1}{253}-\frac{1}{258}\right)\)
\(=\frac{4}{5}.\left(\frac{1}{8}-\frac{1}{258}\right)\)
\(=\frac{4}{5}.\frac{125}{1032}\)
\(=\frac{25}{258}\)
3.a)\(\dfrac{-1}{2}+\dfrac{5}{6}+\dfrac{1}{3}=\dfrac{-3}{6}+\dfrac{5}{6}+\dfrac{2}{6}=\dfrac{-3+5+2}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)
b)\(\dfrac{-3}{8}+\dfrac{7}{4}-\dfrac{1}{12}=\dfrac{-9}{24}+\dfrac{42}{24}-\dfrac{2}{24}=\dfrac{-9+42-2}{24}=\dfrac{31}{24}\)
c)\(\dfrac{3}{5}:\left(\dfrac{1}{4}.\dfrac{7}{5}\right)=\dfrac{3}{5}:\dfrac{7}{20}=\dfrac{3}{5}.\dfrac{20}{7}=\dfrac{12}{7}\)
d)\(\dfrac{10}{11}+\dfrac{4}{11}:4-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{4}{11}.\dfrac{1}{4}-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}=1-\dfrac{1}{8}=\dfrac{8}{8}-\dfrac{1}{8}=\dfrac{7}{8}\)
\(a,15\dfrac{3}{13}-\left(3\dfrac{4}{7}+8\dfrac{3}{13}\right)=15\dfrac{3}{13}-3\dfrac{4}{7}-8\dfrac{3}{13}=\left(15\dfrac{3}{13}-8\dfrac{3}{13}\right)-\dfrac{25}{7}=7-\dfrac{25}{7}=\dfrac{49}{7}-\dfrac{25}{7}=\dfrac{24}{7}\)
\(b,\left(7\dfrac{4}{9}+4\dfrac{7}{11}\right)-3\dfrac{4}{9}=\left(7\dfrac{4}{9}-3\dfrac{4}{9}\right)+4\dfrac{4}{9}=4+\dfrac{40}{9}=\dfrac{36}{9}+\dfrac{40}{9}=\dfrac{76}{9}\)
\(c,\dfrac{-7}{9}.\dfrac{4}{11}+\dfrac{-7}{9}.\dfrac{7}{11}+5\dfrac{7}{9}=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+\dfrac{52}{9}=\dfrac{-7}{9}.1+\dfrac{52}{9}=\dfrac{-7}{9}+\dfrac{52}{9}=\dfrac{45}{9}=5\)
\(d,50\%.1\dfrac{1}{3}.10.\dfrac{7}{35}.0,75=\dfrac{1}{2}.\dfrac{4}{3}.10.\dfrac{1}{5}.\dfrac{3}{4}=\left(\dfrac{1}{2}.\dfrac{1}{5}.10\right).\left(\dfrac{4}{3}.\dfrac{3}{4}\right)=1.1=1\)
\(e,\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{40.43}=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{40}-\dfrac{1}{43}=1-\dfrac{1}{43}=\dfrac{42}{43}\)
\(=\left[\dfrac{12}{7}\cdot\left(19+\dfrac{5}{8}-13-\dfrac{1}{4}\right)\right]\cdot\dfrac{4}{5}\)
\(=\dfrac{12}{7}\cdot\dfrac{51}{8}\cdot\dfrac{4}{5}=\dfrac{306}{35}\)
\(D=\dfrac{4}{8\cdot13}+\dfrac{4}{13\cdot18}+\dfrac{4}{18\cdot23}+...+\dfrac{4}{253\cdot258}\\ =\dfrac{4}{5}\cdot\dfrac{5}{8\cdot13}+\dfrac{4}{5}\cdot\dfrac{5}{13\cdot18}+\dfrac{4}{5}\cdot\dfrac{5}{18\cdot23}+...+\dfrac{4}{5}\cdot\dfrac{5}{253\cdot258}\\ =\dfrac{4}{5}\left(\dfrac{5}{8\cdot13}+\dfrac{5}{13\cdot18}+\dfrac{5}{18\cdot23}+...+\dfrac{5}{253\cdot258}\right)\\ =\dfrac{4}{5}\cdot\left(\dfrac{1}{8}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{23}+...+\dfrac{1}{253}-\dfrac{1}{258}\right)\\ =\dfrac{4}{5}\cdot\left(\dfrac{1}{8}-\dfrac{1}{258}\right)\\ =\dfrac{4}{5}\cdot\dfrac{125}{1032}\\ =\dfrac{25}{258}\)
ta có
Tính:
\(\dfrac{4}{8.13}+\dfrac{4}{13.18}+....+\dfrac{4}{253.258}\)
= 4 (\(\dfrac{1}{8.13}+\dfrac{1}{13.18}+.....+\dfrac{1}{253.258}\))
=\(\dfrac{4}{5}\left(\dfrac{1}{8}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{18}+...+\dfrac{1}{253}-\dfrac{1}{258}\right)\)
=\(\dfrac{4}{5}\left(\dfrac{1}{8}-\dfrac{1}{258}\right)\)
=\(\dfrac{25}{258}\)