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\(B=\frac{7}{3.13}+\frac{7}{13.23}+...+\frac{7}{53.63}\)
\(B=10.\left(\frac{1}{3.13}+\frac{1}{13.23}+....+\frac{1}{53.63}\right)\)
\(B=10.\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+...+\frac{1}{53}+\frac{1}{63}\right)\)
\(B=10.\left(\frac{1}{3}-\frac{1}{63}\right)\)
\(B=10.\frac{20}{63}\)
\(B=\frac{200}{63}\)
\(A=\frac{7}{3.13}+\frac{7}{13.23}+...+\frac{7}{53.63}\)
\(A=7.\left(\frac{1}{3.13}+\frac{1}{13.23}+...+\frac{1}{53.63}\right)\)
\(10A=7.\left(\frac{10}{3.13}+\frac{10}{13.23}+...+\frac{10}{53.63}\right)\)
\(10A=7.\left(\frac{1}{3}-\frac{1}{63}\right)\)
\(10A=7.\frac{20}{63}\)
\(10A=\frac{20}{9}\)
\(A=\frac{20}{9}:10\)
\(A=\frac{20}{9}.\frac{1}{10}\)
\(A=\frac{2}{9}\)
Vậy A=\(\frac{2}{9}\)
Chúc bạn học tốt~
A = \(\frac{7}{3.13}\)\(+\)\(\frac{7}{13.23}\)\(+\)\(\frac{7}{23.33}\)\(+\).......... \(+\)\(\frac{7}{53.63}\)
\(\Rightarrow\)A = 7 . (\(\frac{1}{3.13}\)\(+\)\(\frac{1}{13.23}\)\(+\)\(\frac{1}{23.33}\)\(+\)...... \(\frac{1}{53.63}\))
\(\Rightarrow\)A = \(\frac{1}{10}\). 7 . ( \(\frac{1}{3}\)\(-\)\(\frac{1}{13}\)\(+\)\(\frac{1}{13}\)\(-\)\(\frac{1}{23}\)\(+\)\(\frac{1}{23}\)\(-\)\(\frac{1}{33}\)\(+\)....... \(+\)\(\frac{1}{53}\)\(-\)\(\frac{1}{63}\))
\(\Rightarrow\)A = \(\frac{7}{10}\). ( \(\frac{1}{3}\)\(-\)\(\frac{1}{63}\))
\(\Rightarrow\)A = \(\frac{7}{10}\). \(\frac{20}{63}\)
\(\Rightarrow\)A = \(\frac{2}{9}\)
Chúc các anh em học tốt !!!
\(\frac{7}{3.13}+\frac{7}{13.23}+\frac{7}{23.33}+\frac{7}{33.43}\)
\(=\frac{7}{10}\left(\frac{10}{3.13}+\frac{10}{13.23}+\frac{10}{23.33}+\frac{10}{33.43}\right)\)
\(=\frac{7}{10}\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+\frac{1}{33}-\frac{1}{43}\right)\)
\(=\frac{7}{10}\left(\frac{1}{3}-\frac{1}{43}\right)\)
\(=\frac{7}{10}\left(\frac{43}{129}-\frac{3}{129}\right)\)
\(=\frac{7}{10}.\frac{40}{129}\)
\(=\frac{28}{129}\)
mk làm đúng rồi nha, ko tin bấm thử máy tính
7/3.13 + 7/13.23 + 7/23.33 + 7/33.43
= 7/10.(1/3-1/13+1/13-1/23+1/23-1/33+1/33-1/43)
= 7/10.(1/3-1/43)
= 7/10 . 14/43
= 49/215
phần a dễ bạn tự làm đi tử thì bạn tính như bình thường còn mẫu thì:7.(\(\frac{1}{3.13}\)+\(\frac{1}{13.23}\)+\(\frac{1}{23.33}\))
\(\frac{7}{10}\).(\(\frac{1}{3}\)-\(\frac{1}{33}\))=\(\frac{7}{33}\)
b)(1+1/3+1/5+..+1/199)-(1/2+1/4+...+1/200)
(1+1/2+1/3+...+1/199+1/200)-(1/2+1/2+1/4+1/4+...+1/200+1/200)
=1+1/2+1/3+...+1/199+1/200-(1+1/2+1/3+...+1/100)
=1/101+1/102+...+1/200
\(S=\frac{5}{3.13}+\frac{5}{13.23}+.....+\frac{5}{83.93}\)
\(2S=\frac{2.5}{3.13}+\frac{2.5}{13.23}+....+\frac{2.5}{83.93}\)
\(2S=\frac{10}{3.13}+\frac{10}{13.23}+.....+\frac{10}{83.93}\)
\(2S=\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+....+\frac{1}{83}-\frac{1}{93}\)
\(2S=\frac{1}{3}-\frac{1}{93}=\frac{30}{93}\)
\(S=\frac{30}{93}.\frac{1}{2}=\frac{15}{93}\)
Sửa đề:
\(S=\frac{5}{3.13}+\frac{5}{13.23}+.....+\frac{5}{83.93}\)
\(S=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+....+\frac{1}{83}-\frac{1}{93}\right)\)
\(S=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{93}\right)\)
\(S=\frac{1}{2}.\left(\frac{31}{93}-\frac{1}{93}\right)\)
\(S=\frac{1}{2}.\frac{10}{31}\)
\(S=\frac{5}{31}\)
a)\(=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}\)
\(=\left(\frac{-3}{7}+\frac{3}{7}\right)-\left(\frac{15}{26}+\frac{2}{13}\right)\)
\(=0-\frac{19}{26}\)
\(=-\frac{19}{26}\)
c)\(=\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)
\(=\frac{-11}{23}.2-\frac{1}{23}\)
\(=\frac{-22}{23}-\frac{1}{23}\)
\(=-1\)
A=\(\frac{7}{10}\)*(\(\frac{10}{3\cdot13}\)+\(\frac{10}{13\cdot23}\)+\(\frac{10}{23\cdot33}\)+\(\frac{10}{43\cdot53}\)+\(\frac{10}{53\cdot63}\))
A=\(\frac{7}{10}\)*(\(\frac{1}{3}\)-\(\frac{1}{13}\)+\(\frac{1}{13}\)-\(\frac{1}{23}\)+\(\frac{1}{23}\)-\(\frac{1}{33}\)+\(\frac{1}{43}\)-\(\frac{1}{53}\)+\(\frac{1}{53}\)-\(\frac{1}{63}\))
A=\(\frac{7}{10}\)*(\(\frac{1}{3}\)-\(\frac{1}{33}\)+\(\frac{1}{43}\)-\(\frac{1}{63}\))
A=\(\frac{7}{10}\)*(\(\frac{10}{33}\)+\(\frac{20}{2709}\))
A=\(\frac{7}{10}\)*\(\frac{9250}{29799}\)
A=\(\frac{925}{4257}\)
trình bày sai
mà cũng giống giống
Sai chỗ nào
Đề sai thì có lý
mệt ghê
sao mệt
qua nt riêng nha bạn
bn sai rôi đó
CM công thức :
\(\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{n+a}{n\left(n+a\right)}-\dfrac{n}{n\left(n+a\right)}=\dfrac{a}{n\left(n+a\right)}\)
\(A=\dfrac{7}{3.13}+\dfrac{7}{13.23}+\dfrac{7}{23.33}+...+\dfrac{7}{53.63}\)
\(A=\dfrac{7}{10}\left(\dfrac{7}{3.13}+\dfrac{7}{13.23}+\dfrac{7}{23.33}+...+\dfrac{7}{53.63}\right)\)
\(A=\dfrac{7}{10}\left(\dfrac{1}{3}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{33}+...+\dfrac{1}{53}-\dfrac{1}{63}\right)\)
\(A=\dfrac{7}{10}\left(\dfrac{1}{3}-\dfrac{1}{63}\right)\)
\(A=\dfrac{7}{10}.\dfrac{20}{63}=\dfrac{2}{9}\)
vậy A=\(\dfrac{2}{9}\)
chết quên !!!
cho mk sửa : ở dòng thứ 3 cho mk sửa tử trong ngoặc là 10 nha!!!!
XIN TRÂN THÀNH CẢM ƠN!!!
Ế !!!hình như đề sai!!! nếu đề đúng thì .....mk làm sai rồi nha!!!XIN LỖI !!!
