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Tính nhanh :
\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)
\(A=2\left(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+\frac{1}{8\cdot10}+\frac{1}{10\cdot12}+\frac{1}{12\cdot14}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{14}\right)\)
\(A=2\cdot\frac{3}{7}\)
\(A=\frac{6}{7}\)
\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)
\(A=\frac{2}{8}+\frac{2}{24}+\frac{2}{48}+\frac{2}{80}+\frac{2}{120}+\frac{2}{168}\)
\(A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}+\frac{2}{12.14}\)
\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\)
\(A=\frac{1}{2}-\frac{1}{14}\)
\(A=\frac{3}{7}\)
_Chúc bạn học tốt_
\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)
\(=\frac{1}{2}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{7}\right)\)
\(=\frac{1}{2}.\frac{6}{7}=\frac{3}{7}\)
Đặt \(C=\frac{1}{2}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{84}\)
\(\Rightarrow\frac{C}{2}=1+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}\)
\(\Rightarrow C.\frac{1}{2}=1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}\)
\(\Rightarrow C.\frac{1}{2}=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\)
\(\Rightarrow C.\frac{1}{2}=1+\frac{1}{2}-\frac{1}{7}\)
\(\Rightarrow C=\left(1+\frac{1}{2}-\frac{1}{7}\right).2\)
Ax1/2=1/2+1/6+1/12+1/20+1/30+1/42
Ax1/2=1/1.2+1/2.3+1/3.4+...+1/6.7
Ax1/2=1-1/2+1/2-1/3+1/3-1/4+....+1/6-1/7
Ax1/2=1-1/7=6/7
=>A=6/7x2=12/7
A= \(\frac{1}{4}\)+ \(\frac{1}{12}\)+ \(\frac{1}{24}\)+ \(\frac{1}{40}\)+ \(\frac{1}{60}\)+ \(\frac{1}{84}\)
2A=\(\frac{1}{2}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+\(\frac{1}{20}\)+ \(\frac{1}{30}\)+ \(\frac{1}{42}\)
2A=\(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ \(\frac{1}{4.5}\)+ \(\frac{1}{5.6}\)+ \(\frac{1}{6.7}\)
2A=\(1-\frac{1}{2}\)\(+\frac{1}{2}\)\(-\frac{1}{3}\)\(+...+\)\(\frac{1}{6}\)\(-\frac{1}{7}\)
2A=\(1-\frac{1}{7}\)
2A=\(\frac{6}{7}\)
A=\(\frac{6}{7}:2\)
\(\Rightarrow A=\frac{3}{7}\)
Vậy \(A=\frac{3}{7}\)
Chúc bạn học tốt nhé!
\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+...+\frac{1}{4900}\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\frac{49}{50}=\frac{49}{100}\)
\(B=\)\(\frac{3+33+333+3333+33333}{4+44+444+4444+44444}\)
\(B=\frac{3.1+3.11+3.111+3.1111+3.11111}{4.1+4.11+4.111+4.1111+4.11111}\)
\(B=\frac{3.\left(1+11+111+1111+11111\right)}{4.\left(1+11+111+1111+11111\right)}\)
\(B=\frac{3}{4}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)
\(A.2=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\right).2\)
\(A.2=\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)
=>\(A.2-A=\left(\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\right)-\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\right)\)
\(A=\frac{2}{3}-\frac{1}{192}\)
\(A=\frac{127}{192}\)
\(\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)
Đặt \(C=\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)
\(C=\frac{1995.1990.1997.1993.997}{1997.1993.1994.1995.995}\)
\(C=\frac{1990.997}{1994.995}\)
\(C=\frac{995.2+997}{997.2+995}=1\)
\(B=\frac{3+33+333+3333+ 33333}{4+44+444+4444+44444}\)
\(\Rightarrow B=\frac{3\left(1+11+111+1111+11111\right)}{4\left(1+11+111+1111+11111\right)}=\frac{3}{4}\)
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
a) 5/30+15/90+25/150+35/210+45/270
=1/6+1/6+1/6+1/6+1/6
=1/6 x 5
=5/6
b) 1/2+1/6+1/12+1/20+....+1/56
=1/1x2+1/2x3+1/3x4+1/4x5+.....1/7x8
=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.......-1/7+1/7-1/8
=1/1-1/8
=7/8
c) mình chịu
A=\(\frac{3}{7}\)
CÒN CÁCH LÀM ĐANG CHƯA BIẾT
Trả lời:
\(A=\frac{3}{7}\)
Hmm chứ ko phải là cứ cộng hết vào là đc ạ hay phải tính nhanh?
:p