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Đặt A = 8.10 + 10.12 + 12.14 + ....... + 98.100
=> 6A = 8.10.12 - 8.10.12 + 10.12.14 - 10.12.14 + ...... + 98.100.102
=> 6A = 98.100.102
=> A = 98.100.102/6
=> A = 166600
c.1.2.3+2.3.4+4.5.6+5.6.7=6+24+120+210
=30+120+210
=150+210
=360
=>2A=2(1/2x4+1/4.6+1/6.8+1/8.10+1/10.12+1/12.14)
=> 2A=2/2.4 + 2/4.6 + 2/6.8 + 2/8.10 + 2/10.12 + 2/12.14
=> 2a =1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7
=> 2A =1-1/7
=>2A=16/17
=> A= 8/17
Mình chắc chắn . Chúc bạn học tốt
\(A=\frac{1}{2.4}\)\(+\frac{1}{4.6}\)\(+\frac{1}{6.8}\)\(+\frac{1}{8.10}\)\(+\frac{1}{10.12}\)\(+\frac{1}{12.14}\)
\(\Rightarrow2A=2.\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}+\frac{1}{12.14}\right)\)
\(\Rightarrow2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}+\frac{2}{12.14}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{14}=\frac{7}{14}-\frac{1}{14}=\frac{6}{14}\)
\(\Rightarrow2A=\frac{6}{14}\)
\(\Rightarrow A=\frac{3}{14}\)
Gọi biều thức trên là A, ta có:
A=(1/2.4+1/4.6+1/6.8+1/8.10+1/10.12)x=2
2A=(2/2.4+2/4.6+2/6.8+2/8.10+2/10.12)x=2
2A=(1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10+1/10-1/12)x=2
2A=(1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10+1/10-1/12)x=2
2A=(1/2-1/12)x=2
2A=5/12x=2
=>A=5/24x=1
=>x=1:5/24=24/5
=>1/2.(5/12).x=1
5/24.x=1
x=1:5/24
x=24/5
lưu ý, 1/2.5/12 là tính xong phần 1/2.4 +...+1/10.12 rùi nhé
Ta có \(A=\dfrac{2}{1.3}-\dfrac{2}{2.4}+\dfrac{2}{3.5}-\dfrac{2}{4.6}+\dfrac{2}{5.7}-\dfrac{2}{6.8}+\dfrac{2}{7.9}-\dfrac{2}{8.10}+\dfrac{2}{9.11}-\dfrac{2}{10.12}\)
\(\Rightarrow A=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\right)-\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+\dfrac{2}{8.10}+\dfrac{2}{10.12}\right)\) \(\Rightarrow A=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)-\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{12}\right)\) \(\Rightarrow A=\left(1-\dfrac{1}{11}\right)-\left(\dfrac{1}{2}-\dfrac{1}{12}\right)\)
\(\Rightarrow A=1-\dfrac{1}{11}-\dfrac{1}{2}+\dfrac{1}{12}\)
\(\Rightarrow A=\dfrac{9}{22}+\dfrac{1}{12}\)
\(\Rightarrow A=\dfrac{65}{132}\)
Mà \(\dfrac{65}{132}< 1\) \(\Rightarrow A< 1\)
Vậy \(A< 1\)
\(\frac{3}{4.6}+\frac{3}{6.8}+\frac{3}{8.10}+\frac{3}{10.12}+\frac{3}{12.14}\)
=\(3.\left(\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}+\frac{1}{12.14}\right)\)
=\(\frac{3}{2}.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}+\frac{2}{12.14}\right)\)
=\(\frac{3}{2}.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\right)\)
=\(\frac{3}{2}.\left(\frac{1}{4}-\frac{1}{14}\right)\)
=\(\frac{3}{2}.\left(\frac{7}{28}-\frac{2}{28}\right)\)
=\(\frac{3}{2}.\frac{5}{28}=\frac{15}{56}\)
\(\sqrt[]{\frac{ }{ }\frac{ }{ }\hept{\begin{cases}\\\end{cases}}\hept{\begin{cases}\\\\\end{cases}}\orbr{\begin{cases}\\\end{cases}}^2}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=\dfrac{1}{2}\cdot\dfrac{4}{10}=\dfrac{2}{10}=\dfrac{1}{5}\)
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(S=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}=\frac{1}{8}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}.\left(\frac{58}{45}\right)\)
\(S=\frac{29}{45}\)
\(S=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)
\(S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\)
\(S=\frac{1}{2}-\frac{1}{10}\)
\(S=\frac{2}{5}\)