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Ta có: \(A=\frac{2008+\frac{2007}{2}+\frac{2006}{3}+....+\frac{2}{2007}+\frac{1}{2008}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2008}+\frac{1}{2009}}\)
Xét tử : \(2008+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)
\(=\left(1+1+...+1\right)+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)( có 2008 số hạng 1 )
\(=\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+...+\left(1+\frac{2}{2007}\right)+\left(1+\frac{1}{2008}\right)+1\)
\(=\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2007}+\frac{2009}{2008}+\frac{2009}{2009}\)
\(=2009\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)\)
Ghép tử và mẫu....
Vậy A = 2009
Xét ct trước :D
\(\frac{2}{\left[\left(n-1\right)n\left(n+1\right)\right]}=\frac{1}{\left[\left(n-1\right)n\right]}-\frac{1}{\left[n\left(n+1\right)\right]}\)
Sau khi xét ct rồi thì /Bùm/ Ta được:
\(2M=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{10.11.12}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{10.11}+\frac{1}{11.12}\)
\(=\frac{1}{1.2}-\frac{1}{11.12}\)
\(=\frac{65}{132}\)
\(\Rightarrow M=\frac{65}{264}\)
Ok rồi nhé :)
\(C=\frac{2^{19}.27^3-15.4^9.9^4}{6^9.2^{10}-12^{10}}\)
\(C=\frac{2^{19}.\left(3^3\right)^3-3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}-\left(3.2^2\right)^{10}}\)
\(C=\frac{2^{19}.3^9-3.5.2^{18}.3^8}{2^9.3^9.2^{10}-3^{10}.2^{20}}\)
\(C=\frac{2^{19}.3^9-3^9.2^{18}.5}{2^{19}.3^9-3^{10}.2^{20}}\)
\(C=\frac{2^{18}.3^9\left(2-5\right)}{2^{18}.3^9\left(2-3.2^2\right)}\)
\(C=\frac{-3}{-10}=\frac{3}{10}\)
a/ (-3,2).\(\frac{-15}{64}\)+(0,8-2\(\frac{4}{5}\)):1\(\frac{23}{24}\)
=(\(\frac{-16}{5}\)).\(\frac{-15}{64}\)+(\(\frac{4}{5}\)-\(\frac{14}{5}\)):\(\frac{47}{24}\)
=(\(\frac{-16}{5}\)).\(\frac{-15}{64}\)+(-2):\(\frac{47}{24}\)
= \(\frac{3}{4}\)+\(\frac{-48}{47}\)
=\(\frac{-51}{188}\)
b/ 1\(\frac{13}{15}\).3.(0,5)\(^2\).3+(\(\frac{8}{15}\)-1\(\frac{19}{60}\)):1\(\frac{23}{24}\)
= \(\frac{28}{15}\).3.\(\frac{1}{4}\).3+(\(\frac{8}{15}\)-\(\frac{79}{60}\)):\(\frac{47}{24}\)
= \(\frac{28}{15}\).3.\(\frac{1}{4}\).3+(\(\frac{-47}{60}\)):\(\frac{47}{24}\)
= \(\frac{28}{5}\).\(\frac{1}{4}\).3+(\(\frac{-47}{60}\)):\(\frac{47}{24}\)
= \(\frac{7}{5}\).3+(\(\frac{-47}{60}\)):\(\frac{47}{24}\)
= \(\frac{21}{5}\)+(\(\frac{-47}{60}\)):\(\frac{47}{24}\)
= \(\frac{21}{5}\)+(\(\frac{-2}{5}\))
= \(\frac{19}{5}\)
mk làm hơi dài dòng chút
CHÚC BẠN HỌC TỐT
A em tự tính nhé
b) B = 1+ 3 + 32+...+399
3B = 3+ 32+33+...+3100
do đó 3B-B= (3+32+33+...+3100) - ( 1+3+32+...+399)
2B= 3100-1
B= (3100-1) : 2
c) \(C=1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x.\left(x+1\right)}\)
\(C=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}\)
\(C=1+\frac{1}{2}.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)\)
\(C=1+\frac{1}{2}.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}\right)\)
\(C=1+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)\)
\(C=1+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{x+1}\right)\)
Phần c thế này thôi vì ko có giá trị x cụ thể .
d) \(D=\frac{9}{8}.\frac{16}{15}.\frac{25}{24}.....\frac{8100}{8099}\)
\(D=\frac{9.16.25....8100}{8.15.24....8099}\)
\(D=\frac{3.3.4.4.5.5....90.90}{2.4.3.5.4.6.....89.91}\)
\(D=\frac{\left(3.4.5...90\right).\left(3.4.5...90\right)}{\left(2.3.5...89\right).\left(4.5.6...91\right)}\)
\(D=\frac{3.4.5...90}{2.3.4...89}.\frac{3.4.5...90}{4.5.6...91}\)
\(D=\frac{90}{2}.\frac{3}{91}\)
\(D=45.\frac{3}{91}=\frac{135}{91}\)
Các bạn giỏi toán và thầy cô giúp em với ạ
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{10}}\)
\(3A=3+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{10}}+\frac{1}{3^{11}}\)
\(3A-A=\left(3+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{11}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{10}}\right)\)
\(3A-A=3+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{10}}+\frac{1}{3^{11}}-\frac{1}{3}-\frac{1}{3^2}-...-\frac{1}{3^{10}}\)
\(2A=3-\frac{1}{3^{10}}\)
\(A=\frac{3-\frac{1}{3^{10}}}{2}\)
LỚP 6 MÌNH NGHĨ BẠN NÊN TÌM HIỂU THÊM PHẦN NÀY VỚI DÃY SỐ THEO QUY LUẬT NHÉ. CÓ BÀI NÀO KHÓ THÌ NÓI MÌNH GIẢI CHO. NHÉ