\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
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\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(=\frac{1}{2}\times\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\times\frac{12}{13}\)
\(=\frac{6}{13}\)
1/3 + 1/3×5 + 1/5×7 + 1/7×9 + 1/9×11 + 1/11×13
=> 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 +1/9 -1/11 + 1/11 + -1/13
=> 2/3 - 1/13 = 23/39
Đúng 100% nha. k mk nha
\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{99.101}\)
\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(A=\frac{1}{2}.\frac{98}{303}\)
\(A=\frac{49}{303}\)
A= \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
2A=\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
2A=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
2A=\(\frac{1}{3}-\frac{1}{101}\)
2A=\(\frac{98}{303}\)
A=\(\frac{98}{303}.\frac{1}{2}\)
A=\(\frac{49}{303}\)
Chúc bạn học tốt!
Đặt \(A=1\frac{7}{15}-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}-\frac{1}{143}-\frac{1}{195}\)
\(\Rightarrow A=\frac{22}{15}-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\right)\)
Đặt \(B=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(\Rightarrow B=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\)
\(\Rightarrow2B=2\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\right)\)
\(\Rightarrow2B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}+\frac{2}{13\cdot15}\)
\(\Rightarrow2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(\Rightarrow2B=1-\frac{1}{15}\)
\(\Rightarrow2B=\frac{14}{15}\)
\(\Rightarrow B=\frac{14}{15}:2\Rightarrow B=\frac{7}{15}\)
\(\Rightarrow A=\frac{22}{15}-\frac{7}{15}\Rightarrow A=\frac{15}{15}=1\)
<=> \(\left(\frac{1}{3\cdot5}+\frac{1}{5.7}+...+\frac{1}{13\cdot15}\right)+x=\frac{17}{15}\)
<=> \(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{15}\right)+x=\frac{17}{15}\)
<=>\(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{15}\right)+x=\frac{17}{15}\)
<=> \(\frac{2}{15}+x=\frac{17}{15}\)
=> x = 1
(1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15)+x=17/15
[2.(1/3-1/5+1/5-1/7+...+1/13-1/15)]+x=17/15
[2.(1/3-1/15)]+x=17/15
(2.4/15)+x=17/15
6/15+x=17/15
x=17/15-6/15
x=11/15
=> 2(1/15+1/35+1/63+1/99)x=2
=>(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11)x=2
=>8/33x=2
=>x=2:8/33
=>x=8,25
\(\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\cdot x=1\)
\(\left(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}\right)\cdot x=1\)
\(\left[\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\right]\cdot x=1\)
\(\left[\frac{1}{2}\left(\frac{1}{3}-\frac{1}{11}\right)\right]\cdot x=1\)
\(\left[\frac{1}{2}\cdot\frac{8}{33}\right]\cdot x=1\)
\(\frac{4}{33}\cdot x=1\)
\(\Rightarrow x=\frac{1}{\frac{4}{33}}=\frac{33}{4}\)
\(\Rightarrow A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{99.101}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{88}{303}\)
\(\Rightarrow A=\frac{44}{303}\)
\(A=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{99\times101}\)
\(\Rightarrow2A=\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{99\times101}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)
=> A = 98/203 : 2 = 49/303
\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
= \(\frac{2}{21}+\frac{1}{63}+\frac{1}{99}\)
= \(\frac{1}{9}+\frac{1}{99}\)
= \(\frac{11}{99}+\frac{1}{99}\)
= \(\frac{12}{99}\)
= \(\frac{2}{9}\)
1/3.5+1/5.7+1/7.9+1/9.11 (. là nhân bn nha)
2(1/3.5+1/5.7+1/7.9+1/9.11)
2/3.5+2/5.7+2/7.9+2/9.11
1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11
1/3-1/11
11/33-3/33
8/33
tk nha bn