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+ \(\frac{1}{n\times\left(n+2\right)}=\frac{\left(n+2\right)-n}{n\times\left(n+2\right)}\)
\(=\frac{n+2}{n\times\left(n+2\right)}-\frac{n}{n\times\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+2}\)
+ \(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}\)
\(=1-\frac{1}{3}+1-\frac{1}{15}+1-\frac{1}{35}+1-\frac{1}{63}+1-\frac{1}{99}\)
\(=5-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)
\(=5-\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}\right)\)
\(=5-\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)\)
\(=5-\frac{1}{2}\times\left(1-\frac{1}{11}\right)\)
\(=5-\frac{1}{2}+\frac{1}{22}=\frac{50}{11}\)
\(\frac{2}{3}\cdot y-\frac{12}{3}:\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)=\frac{1}{3}\)\(\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}\right)=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+\frac{9-7}{7\cdot9}+\frac{11-9}{9\cdot11}+\frac{13-11}{11\cdot13}\right)=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(1+\frac{1}{3}-\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{7}+\frac{1}{9}-\frac{1}{9}+\frac{1}{11}-\frac{1}{11}+\frac{1}{13}\right)\)\(=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(\frac{1}{1}+\frac{1}{3}\right)=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\frac{4}{3}\)\(=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4\cdot\frac{3}{4}=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-3=\frac{1}{3}\)
\(\frac{2}{3}\cdot y=\frac{1}{3}+3\)
\(\frac{2}{3}\cdot y=\frac{10}{3}\)
\(y=\frac{10}{3}:\frac{2}{3}\)
y=5
\(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{25}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
\(=\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)
\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...+\dfrac{1}{11}-\dfrac{1}{13}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)
\(=\left(1-\dfrac{1}{3}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)
\(=\dfrac{2}{3}\cdot\dfrac{1}{2}+\dfrac{1}{25}\)
\(=\dfrac{1}{3}+\dfrac{1}{25}\)
\(=\dfrac{28}{75}\)
3/8 x 5/4 =15/32
4/7 x 2/13=8/91
cách làm là lấy tử nhân tử mẫu nhân mẫu
cái này học từ lớp 4 rồi còn gì
Đặt \(B=\frac{3}{15}+\frac{3}{35}+\frac{3}{63}+\frac{3}{99}+\frac{3}{143}\)
\(\Leftrightarrow B=\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+\frac{3}{11.13}\)
\(\Leftrightarrow2B=3\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(\Leftrightarrow2B=3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(\Leftrightarrow2B=3\left(\frac{1}{3}-\frac{1}{13}\right)=1-\frac{3}{13}=\frac{10}{13}\)
\(\Leftrightarrow A=1+\frac{3}{15}+\frac{3}{35}+\frac{3}{63}+\frac{3}{99}+\frac{3}{143}=1+\frac{10}{13}=\frac{23}{13}\)
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\)
\(=1-\frac{1}{9}=\frac{8}{9}\)
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\)( chỗ này tối giản nha )
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\)
~ Hok tốt ~
B=2/1.3 + 2/3.5 + 2/5.7 +...+ 2/299.301
B=1-1/3+1/3-1/5+1/5-1/7+...+1/299-1/301=1-1/301=300/301
\(Ta có: \frac{2}{3}=\frac{1}{1}-\frac{1}{3}\);
\(\frac{2}{15}=\frac{1}{3}-\frac{1}{5}\);
\(\frac{2}{35}=\frac{1}{5}-\frac{1}{7}\) ; ... ; \(\frac{2}{89999}=\frac{1}{299}-\frac{1}{301}\).
=> B= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{299}-\frac{1}{301}\)
=> B=\(\frac{1}{1}-\frac{1}{301}\)
=> B=\(\frac{300}{301}\)