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Ta có
\(\frac{1}{3^{400}}=\frac{1}{\left(3^4\right)^{100}};\frac{1}{4^{300}}=\frac{1}{\left(4^3\right)^{100}}\)
\(\Rightarrow\frac{1}{3^4}< \frac{1}{4^3}\left(3^4>4^3\right)\\
\Rightarrow\frac{1}{3^{400}}< \frac{1}{4^{300}}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{x\left(x+1\right)}=\frac{215}{216}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{215}{216}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{215}{216}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{215}{216}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{216}\)
\(\Leftrightarrow x=216-1=215\)
\(C=\frac{3}{4}x\frac{8}{9}x\frac{15}{16}x...x\frac{9999}{10000}\)
\(C=\frac{3}{4}x\frac{4x2}{3x3}x\frac{3x5}{2x8}x...x\frac{99x101}{100x100}\)
\(C=...\) ( Tự làm tiếp )
\(E=1\frac{1}{3}x1\frac{1}{8}x1\frac{1}{15}x1\frac{1}{24}x...x1\frac{1}{99}\)
\(E=\frac{4}{3}x\frac{9}{8}x\frac{16}{15}x\frac{25}{24}x...x\frac{100}{99}\)
\(E=....\)( tương tự câu C )
\(\Rightarrow4x-\left(\frac{4}{5.7.9}+\frac{4}{7.9.11}+...+\frac{4}{99.101.103}\right)=\frac{2}{83224}=\frac{1}{41612}\)
\(4x-\left(\frac{9-5}{5.7.9}+\frac{11-7}{7.9.11}+...+\frac{103-99}{99.101.103}\right)=\frac{1}{41612}\)
\(4x-\left(\frac{1}{5.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.11}+...+\frac{1}{99.101}-\frac{1}{101.103}\right)=\frac{1}{41612}\)
\(4x-\left(\frac{1}{5.7}-\frac{1}{101.103}\right)=\frac{1}{41612}\)
Từ đó tìm ra x
nhân cả 2 vế của đẳng thức với 1/2 ta được
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{x\left(x+1\right)}=\frac{2014}{2015}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}=\frac{2014}{2015}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-......+\frac{1}{x}-\frac{1}{x+1}=\frac{2014}{2015}\)
\(=\frac{1}{2}-\frac{1}{x+1}=\frac{2014}{2015}\)
\(=>\frac{1}{x+1}=\frac{1}{2}-\frac{2014}{2015}\)
\(\frac{1}{x+1}=-\frac{2013}{4030}\)
hay \(1:\left(x+1\right)=-\frac{2013}{4030}\)
\(x+1=-\frac{4030}{2013}\)
\(=>x=-\frac{6043}{2013}\)
1. Tìm x
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}=x\)
\(\Rightarrow\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}=x\)
\(\Rightarrow1-\frac{1}{100}=x\)
\(\Rightarrow x=\frac{99}{100}\)
\(2.Tính\)
\(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
học vui!!
Xin lỗi nha. Bài 1 mk làm sai. Lại nè:
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}=x\)
\(\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)=x\)
\(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)=x\)
\(\frac{1}{3}.\left(1-\frac{1}{100}\right)=x\)
\(\frac{1}{3}\cdot\frac{99}{100}=x\)
\(\frac{33}{100}=x\)
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)....\left(1+\frac{1}{99}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.......\frac{100}{99}\)
\(=\frac{3.4.5....100}{2.3.4.....99}\)
\(=\frac{100}{2}=50\)
cảm ơn