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(2/1+2/3) + (2/3+2/5) + (2/5+2/7) + ...+ (2/77+2/79) 2/1 - 2/79 156/79
a)\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}\cdot\frac{402}{2015}\)
\(=\frac{603}{2015}\)
b)\(=\frac{4}{5}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{98}\right)\)
\(=\frac{4}{5}\left(\frac{1}{3}-\frac{1}{98}\right)\)
\(=\frac{4}{5}\cdot\frac{95}{294}\)
\(=\frac{38}{147}\)
a) Gọi tổng trên là A
A = \(\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+...+\frac{3}{2013.2015}\)
A == \(\frac{3}{5}-\frac{3}{7}+\frac{3}{7}-\frac{3}{9}+\frac{3}{9}-\frac{3}{11}+...+\frac{3}{2013}-\frac{3}{2015}\)
Vì một số trừ cho a rồi cộng cho a sẽ bằng chính số đó nên:
A = \(\frac{3}{5}-\frac{3}{2015}\)
A = \(\frac{1209}{2015}-\frac{3}{2015}\)
A = \(\frac{1206}{2015}\)
b) Gọi tổng trên là B
B = \(\frac{4}{3.8}+\frac{4}{8.13}+\frac{4}{13.15}+...+\frac{4}{93.98}\)
B = \(\frac{4}{3}-\frac{4}{8}+\frac{4}{8}-\frac{4}{13}+\frac{4}{13}-\frac{4}{15}+...+\frac{4}{93}-\frac{4}{98}\)
Vì một số trừ cho a rồi cộng cho a sẽ bằng chính số đó nên:
B = \(\frac{4}{3}-\frac{4}{98}\)
B = \(\frac{686}{294}-\frac{12}{294}\)
B = \(\frac{674}{294}=\frac{337}{147}\)
\(P=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)
a) Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=1-\frac{1}{101}=\frac{100}{101}\)
\(A=\frac{100}{101}\div2=\frac{50}{101}\)
b) Đặt \(B=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+\frac{1}{12.15}\)
\(3B=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+\frac{3}{12.15}\)
\(3B=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\)
\(3B=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)
\(B=\frac{4}{15}\div3=\frac{4}{45}\)
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=1-\frac{1}{101}=\frac{100}{101}\)
\(A=\frac{100}{101}\div2=\frac{50}{101}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{9}\right)\)
\(=\frac{1}{2}.\frac{8}{9}\)
\(=\frac{4}{9}\)
Đặt: A=1/1.3+1/3.5+1/5.7+1/7.9
2A=2/1.3+2/3.5+2/5.7+2/7.9
2A=1-1/3+1/3-1/5+1/5-1/7+1/7-1/9
2A=1-1/9
2A=8/9
A=4/9
2/3.5 + 2/5.7 + 2/7.9 + ... + 2/41.43
= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/41 - 1/43
= 1/3 - 1/43
= 40/129
ỦNG HỘ NHA
\(\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+\frac{2}{17.19}+\frac{2}{19.21}\right)\) ) . 462 - x = 19
\(\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}+\frac{1}{19}-\frac{1}{21}\right)\) . 462 - x = 19
\(\left(\frac{1}{11}-\frac{1}{21}\right)\) . 462 - x = 19
... Chúc bạn học tốt !
\(\left(\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+\frac{2}{15\cdot17}+\frac{2}{17\cdot19}+\frac{2}{19\cdot21}\right)\)\(\cdot462-x=19\)
\(\left(\frac{1}{11\cdot13}+\frac{1}{13\cdot15}+\frac{1}{15\cdot17}+\frac{1}{17\cdot19}+\frac{1}{19\cdot21}\right)\)\(\cdot462-x=19\)
\(\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}+\frac{1}{19}-\frac{1}{21}\right)\)\(\cdot462-x=19\)
\(\left(\frac{1}{11}-\frac{1}{21}\right)\cdot462-x=19\)
\(\frac{10}{231}\cdot462-x=19\)
\(20-x=19\)
\(x=20-19\)
\(x=1\)
\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+\frac{2}{15.17}\)
\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{17}\)
\(A=1-\frac{1}{17}\)
\(A=\frac{16}{17}\)
\(B=\frac{4}{1.3}+\frac{4}{3.5}+...+\frac{4}{9.11}+\frac{4}{11.13}\)
\(B=\frac{4}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(B=\frac{4}{2}\left(1-\frac{1}{13}\right)\)
\(B=\frac{4}{2}\cdot\frac{12}{13}\)
\(B=\frac{24}{13}\)
=> A= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}\)
=> A= \(\frac{1}{1}-\frac{1}{17}\)
=> A= \(\frac{16}{17}\)
\(\Rightarrow B=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(\Rightarrow B=2.\left(\frac{1}{1}-\frac{1}{13}\right)\)
\(\Rightarrow B=2.\frac{12}{13}\)
\(\Rightarrow B=\frac{24}{13}\)