\(G=\frac{1}{2x9}+\frac{1}{9x7}+\frac{1}{7x19}+.....+\frac{1}{252x509}\)
Tính G
Những câu hỏi liên quan
Qfrac{1}{2x9}+frac{1}{9x7}+frac{1}{7x19}+.......+frac{1}{252x509}Rfrac{1}{10x9}+frac{1}{18x13}+frac{1}{26x17}+.....+frac{1}{802x405}Sfrac{2}{4x7}-frac{3}{5x9}+frac{2}{7x10}-frac{3}{9x13}+.........+frac{2}{301x304}+frac{3}{401x403}
Đọc tiếp
Q=\(\frac{1}{2x9}\)+\(\frac{1}{9x7}\)+\(\frac{1}{7x19}\)+.......+\(\frac{1}{252x509}\)
R=\(\frac{1}{10x9}\)+\(\frac{1}{18x13}\)+\(\frac{1}{26x17}\)+.....+\(\frac{1}{802x405}\)
S=\(\frac{2}{4x7}\)\(-\)\(\frac{3}{5x9}\)+\(\frac{2}{7x10}\)\(-\)\(\frac{3}{9x13}\)+.........+\(\frac{2}{301x304}\)+\(\frac{3}{401x403}\)
Q=\(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{7}+\frac{1}{7}-\frac{1}{19}+...+\frac{1}{252}-\frac{1}{509}\)
=\(\frac{1}{2}-\left(\frac{1}{9}+\frac{1}{9}\right)-\left(\frac{1}{7}+\frac{1}{7}\right)-...-\left(\frac{1}{252}+\frac{1}{252}\right)-\frac{1}{509}\)
=\(\frac{1}{2}-0+0+0+...+0-\frac{1}{509}\)
=\(\frac{1}{2}-\frac{1}{509}\)
=\(\frac{507}{1018}\)
MẤY CÂU KHÁC THÌ TƯƠNG TỰ, CHÚC BẠN MAY MẮN!!!:))
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a/ 1/2x9 + 1/9x7+1/7x19+...........+1/252x509
b/1/10x9+1/18x13+1/26x17+..................+1/802x405
Ai BIẾT CÂU NÀY GIÚP MIK VỚI
a) \(\frac{1}{2.9}+\frac{1}{9.7}+\frac{1}{7.19}+...+\frac{1}{202.509}=\frac{2}{4.9}+\frac{2}{9.14}+\frac{2}{14.19}+...+\frac{2}{504.509}\)
\(=\frac{2}{5}\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{504.509}\right)\)
\(=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{504}-\frac{1}{509}\right)=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{509}\right)\)
\(=\frac{2}{5}.\frac{505}{2036}=\frac{101}{1018}\)
b) \(\frac{1}{10.9}+\frac{1}{18.13}+...+\frac{1}{802.405}=\frac{2}{10.18}+\frac{2}{18.26}+...+\frac{2}{802.810}\)
\(=\frac{2}{8}\left(\frac{8}{10.18}+\frac{8}{18.26}+...+\frac{8}{802.810}\right)=\frac{1}{4}\left(\frac{1}{10}-\frac{1}{18}+\frac{1}{18}-\frac{1}{26}+...+\frac{1}{802}-\frac{1}{810}\right)\)
\(=\frac{1}{4}\left(\frac{1}{10}-\frac{1}{810}\right)=\frac{1}{4}.\frac{40}{405}=\frac{10}{405}\)
Bạn vào câu hỏi tương tự tham khảo !
Tính a,b,c,d,e,g,h,i,k
\(\frac{654}{12254}=1+\frac{a}{1+\frac{b}{1+\frac{c}{1+\frac{d}{1+\frac{e}{1+\frac{g}{1+\frac{h}{1+\frac{i}{1+k}}}}}}}}\)
\(\frac{654}{12254}=\frac{12254-11600}{12254}=1+\frac{-11600}{12254}=1+\frac{1}{\frac{12254}{-11600}}=1+\frac{1}{1+\frac{23854}{-11600}}=1+\frac{1}{1+\frac{1}{-\frac{11600}{23854}}}=\)sức gõ công thức có hạn, cứ theo đó mà làm tiếp, đảm bảo sẽ ra ngay kết quả
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\(G=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)
\(H=\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)
tính tỉ số G và H
\(G=\left(\frac{1}{3}-1\right)\left(\frac{1}{6}-1\right)\left(\frac{1}{15}-1\right)\left(\frac{1}{21}-1\right)\left(\frac{1}{36}-1\right)\)
Tính G
\(G=\left(\frac{1}{3}-1\right)\left(\frac{1}{6}-1\right)\left(\frac{1}{15}-1\right)\left(\frac{1}{21}-1\right)\left(\frac{1}{36}-1\right)\)
\(\Leftrightarrow G=-\frac{2}{3}.\frac{-5}{6}.\frac{-14}{15}.\frac{-20}{21}.\frac{-35}{36}\)
\(\Leftrightarrow G=\frac{-2.}{3}.\frac{5}{6}.\frac{14}{15}.\frac{20}{21}.\frac{35}{36}\)
\(\Leftrightarrow G=\frac{-2.5.2.7.2.2.5.5.7}{3.2.3.3.5.3.7.2.3.2.3}\)
\(\Leftrightarrow G=\frac{-2^4.5^3.7^2}{2^3.3^6.5.7}\)
\(\Leftrightarrow G=\frac{-2.5^2.7}{3^6}\)
\(\Leftrightarrow G=\frac{-350}{729}\)
P/s : Xin lỗi vì cách giải cùi bắp của mình :((
\(B=\frac{1}{3x10}+\frac{1}{10x17}+...+\frac{1}{73x80}-\frac{1}{2x9}-\frac{1}{9x16}-...-\frac{1}{23x30}\)
Tìm B giúp mìh nha!!!!
Ai nhanh và giống cách giải của cấp 1 mình tik cho!!!
1.Cho G= \(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{2^{1024}}\right)\)và H=\(\frac{1}{2^{2047}}\)Tính G+H
cho \(G=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{2^{1024}}\right)\)và \(H=\frac{1}{2^{2047}}\). Tính G+H
Cho \(G=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{2^{1024}}\right)\)và \(H=\frac{1}{2^{2047}}\). Tính G+H