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\(\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}+\frac{2}{11\times13}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)
\(=\frac{1}{3}-\frac{1}{13}\)
\(=\frac{13}{39}-\frac{3}{39}=\frac{13-3}{39}=\frac{10}{39}\)
\(A=11\frac{2}{1.3}+11\frac{2}{3.5}+11\frac{2}{5.7}+11\frac{2}{7.9}+11\frac{2}{9.11}\)
\(A=11+\frac{2}{1.3}+11+\frac{2}{3.5}+11+\frac{2}{5.7}+11+\frac{2}{7.9}+11+\frac{2}{9.11}\)
\(A=55+\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(A=55+\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(A=55+\frac{1}{1}-\frac{1}{11}\)
\(A=56-\frac{1}{11}\)
\(A=\frac{616}{11}-\frac{1}{11}=\frac{615}{11}\)
Vậy A = 615 / 11
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{x\cdot(x+2)}=\frac{100}{101}\)
\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{100}{101}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{100}{101}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{101}\)
\(\Leftrightarrow x+2=101\Leftrightarrow x=99\)
Vậy x = 99
=2/2 - 2/3 - 2/4 - .........- 2/38 - 2/39 - 2/40
= 2/2 - 2/40
=1 - 2/40
=38/40
= 19/20
a) (2/5 + 7/8)+3/5 b) 19/11 +( 5/13 + 3/11)
=2/5 + 7/8 + 3/5 = 19/11 + 5/13 + 3/11
= ( 2/5 +3/5) +7/8 = ( 19/11 + 3/11) + 5/13
= 1 + 7/8 = 21/11 + 5/13
=8/8 + 7/8 =..................
vậy ..................
=15/8
a,Đặt \(A=\frac{1}{1\times4}+\frac{1}{4\times7}+...+\frac{1}{97\times100}\)
\(\Rightarrow3A=\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{3}{97\times100}\)
\(\Rightarrow3A=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)
\(\Rightarrow3A=1-\frac{1}{100}=\frac{99}{100}\)
\(\Rightarrow A=\frac{99}{300}\)
b, \(\frac{1}{2}\times\frac{2}{3}\times...\times\frac{99}{100}=\frac{1\times2\times...\times99}{2\times3\times...\times1000}=\frac{1}{100}\)
c, \(\frac{3}{4}\times\frac{8}{9}\times...\times\frac{99}{100}=\frac{1.3}{2.2}\times\frac{2.4}{3.3}\times...\times\frac{9.11}{10.10}=\frac{1.2.....9}{2.3.....10}\times\frac{3.4.....11}{2.3.....10}=\frac{1}{10}\times\frac{11}{2}=\frac{11}{20}\) (dấu . là dấu nhân)
b
Q=\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{9900}\)
Rồi giải tương tự như câu a là được
M=\(5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=5\left(1-\frac{1}{100}\right)=5.\frac{99}{100}=\frac{99}{20}\)
\(B=\frac{2}{8}+\frac{2}{24}+\frac{2}{48}+...+\frac{2}{18\cdot20}\)
\(B=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{18\cdot20}\)
\(B=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{18}-\frac{1}{20}\)
\(B=\frac{1}{2}-\frac{1}{20}\)
\(B=\frac{9}{20}\)
=))
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A=\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\left(\frac{1}{9}-\frac{1}{9}\right)-\frac{1}{10}\)
\(A=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{101}-\frac{1}{103}\)
\(A=\frac{1}{3}-\frac{1}{103}\)
\(A=\frac{100}{309}\)
\(A=\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{99\times101}+\frac{2}{101\times103}\)
\(A=1\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}+\frac{1}{101}-\frac{1}{103}\right)\)
\(A=1\times\left(\frac{1}{3}-\frac{1}{103}\right)\)
\(A=1\times\frac{100}{309}\)
\(A=\frac{100}{309}\)