Tìm n :
A = \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+.....+\frac{1}{n}\)
Biết A = \(\frac{39}{40}\)
Cho A = \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{n}\)
Biết A = \(\frac{39}{40}\). Tìm n
A = 1/1x2 +1/2x3 + 1/3x4 + 1/4x5 + 1/5x6 + .... + 1/y x n = 39/40
A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + ... + 1/y - 1/n = 39/40
A = 1 - 1/n = 39/40
A = 1 - 39/40 = 1/n
A = 1/40 = 1/n
=> n = 40
Ta có : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{n}\) = \(\frac{39}{40}\)
= \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{ax\left(a+1\right)}=\frac{39}{40}\) ( có : n = a x ( a+1) )
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{a}-\frac{1}{a+1}=\frac{39}{40}\)
=\(\frac{1}{1}-\frac{1}{a+1}=\frac{39}{40}\)
( ta triệt tiêu tất cả các phân số ở giữa ) VD: trừ 1/2 rồi lại cộng 1/2 thì còn lại 0
\(\frac{1}{a+1}=\frac{1}{1}-\frac{39}{40}\)
\(\frac{1}{a+1}=\frac{1}{40}\)
a+1 = 40
a = 40 - 1
a = 39
vì a x (a+1) = n
nên 39 x 40 = n
n = 1560
\(ĐS:1560\)
CHÚC BẠN HỌC GIỎI
Cho \(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{n}\)
Biết \(A=\frac{39}{40}\)Tìm n
khong the co n vi 40 khong bang so tu nhien lien tiep nao ca
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{n}\)
biết A=\(\frac{39}{40}\)
tìm n
n = 1/1560 nha Nguyễn Minh Khánh
Cho M = \(\frac{1}{2}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+ \(\frac{1}{20}\)+ \(\frac{1}{30}\)+ ... + \(\frac{1}{n}\)
Tìm n biết M =\(\frac{39}{40}\)
Ta có: 1/2=1/1*2 m*m+1=n
Nên 1/1*2 + 1/2*3 + 1/3*4 +... +1/m*m+1
(1-1/2) + (1/2-1/3) + ... (1/m-1/m+1)
Triệt tiêu đi còn1- 1/m+1 =39/40
Suy ra 1/m+1 = 1/40
Vậy m=39
n = 39* (39+1) = 39*40= 1560
a) \(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
b)Cho \(A=\frac{n+2}{n-5}\left(n\inℤ;n\ne5\right)\). Tìm n để \(A\inℤ\)
c) Tìm \(x\inℕ\), biết:
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x.\left(x+1\right)}=\frac{6}{7}\)
a/ \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}\)
=> \(A=\frac{9}{10}\)
b/ \(A=\frac{n+2}{n-5}=\frac{n-5+7}{n-5}=\frac{n-5}{n-5}+\frac{7}{n-5}\)
=> \(A=1+\frac{7}{n-5}\)
Để A nguyên => 7 chia hết cho n-5 => n-5=(-7; -1; 1; 7)
=> n=(-2; 4, 6, 8)
Bài 1 :Thực hiện phép tính :
a) M =(\(\frac{-6}{13}+\frac{15}{26}-\frac{47}{39}-\frac{1}{78}\)) : (\(99\frac{17}{65}-100\frac{5}{52}+\frac{1}{130}\))
b) N = \(\frac{(\frac{3}{5}-0,435+\frac{1}{200}):\left(-0,04\right)}{30,75+\frac{1}{12}+3\frac{1}{6}}\)
c) P = (\(\frac{-5}{6}:\frac{-10}{11}\))+\(\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{\frac{-2}{12}-\frac{10}{24}+\frac{14}{39}}\)
Bài 2 : Thực hiện phép tính :V
a) P =\(\frac{\frac{1}{5}-\frac{1}{9}+\frac{1}{13}}{\frac{9}{5}-1+\frac{9}{13}}+\frac{\frac{10}{7}-\frac{10}{11}-\frac{10}{17}}{\frac{12}{7}-\frac{12}{11}-\frac{12}{17}}\)
b) Q = \(\frac{\frac{1}{14}-\frac{1}{30}-\frac{1}{46}}{\frac{2}{35}-\frac{2}{75}-\frac{2}{115}}:\frac{\frac{3}{8}-\frac{15}{17}+\frac{30}{31}}{\frac{1}{6}-\frac{20}{51}+\frac{40}{93}}\)
có rất nhiều câu dễ ở trong đề sao bạn Ko thử làm đi rồi câu nào khó lại hỏi
Tìm A biết : \(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(\Rightarrow A=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(\Rightarrow A=1-\frac{1}{7}\)
\(\Rightarrow A=\frac{6}{7}\)
nhớ tick nha!!!!!!!!!!!!!
A = 1/1*2 +1/2*3 +1/3*4 +1/4*5+ 1/5*6 +1/6*7
A = 1/1 - 1/2 +1/2 -1/3 +1/3 -1/4 +1/4 -1/5 +1/5 -1/6 +1/6 -1/7
A = 1 - 1/7
A= 6/7
A=6/7. Ta lấy tất cả cộng vào là xong.
Ở vòng 12 lớp 5 bài 2
1) Tính:\(A=3-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}\)
2) Tìm tất cả các số nguyên tố x,y sao cho x2 - 6y2 - 1 = 0
3) Cho \(n\in N\)biết n-10; n+4. n+60 đều là số nguyên tố. CMR: n+90 là số nguyên tố
4) Tính nhanh
\(A=\left(\frac{7}{9}+1\right)\left(\frac{7}{20}+1\right)\left(\frac{7}{33}+1\right).....\left(\frac{7}{10800}+1\right)\)
Các bn giúp mk nhanh lên nhé
\(A=3-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}\)
\(A=3-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)\)
\(A=3-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\right)\)
\(A=3-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)
\(A=3-\left(1-\frac{1}{8}\right)\)
\(A=3-\frac{5}{8}\)
\(A=\frac{19}{8}\)
e,\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{42}\right)\)
\(\Rightarrow A=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}=4-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(\Rightarrow A=4-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)=4-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(\Rightarrow A=4-\left(\frac{1}{1}-\frac{1}{7}\right)=4-\frac{6}{7}=3\frac{1}{7}\)
BN mún hỏi j vậy, đây k phải câu hỏi, mà có thì phải là toán lớp 6