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\(A=\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{2001\cdot2005}\)
\(A=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{2001}-\dfrac{1}{2005}\)
\(A=1-\dfrac{1}{2005}=\dfrac{2004}{2005}\)
\(B=\dfrac{3}{10\cdot12}+\dfrac{3}{12\cdot14}+...+\dfrac{3}{998\cdot1000}\)
\(\dfrac{2}{3}B=\dfrac{2}{10\cdot12}+...+\dfrac{2}{998\cdot1000}\)
\(\dfrac{2}{3}B=\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-...+\dfrac{1}{998}-\dfrac{1}{1000}\)
\(\dfrac{2}{3}B=\dfrac{1}{10}-\dfrac{1}{1000}=\dfrac{99}{1000}\)
\(B=\dfrac{99}{1000}:\dfrac{2}{3}=\dfrac{297}{2000}\)
\(A=\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{2001.2005}\)
\(\Rightarrow A=4\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{2001.2005}\right)\)
\(\Rightarrow A=4.\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{2001}-\dfrac{1}{2005}\right)\)
\(\Rightarrow A=1-\dfrac{1}{2005}\)
\(\Rightarrow A=\dfrac{2004}{2005}\)
2.
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{93}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+\frac{1}{2}.\left(\frac{1}{7}-\frac{1}{9}\right)+...+\frac{1}{2}.\left(\frac{1}{2x+1}-\frac{1}{2x+3}\right)=\frac{15}{93}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}\right)=\frac{15}{93}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2x+3}\right)=\frac{15}{93}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{2x+3}=\frac{15}{93}:\frac{1}{2}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Rightarrow\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)
\(\Rightarrow\frac{1}{2x+3}=\frac{1}{93}\)
\(\Rightarrow\)2x + 3 = 93
\(\Rightarrow\)2x = 93 - 3
\(\Rightarrow\)2x = 90
\(\Rightarrow\)x = 90 : 2 = 45
\(H=\frac{3}{1.5}+\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{33.37}\)
= \(\frac{3}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{33}-\frac{1}{37}\right)\)
= \(\frac{3}{4}\left(1-\frac{1}{37}\right)\)
= \(\frac{3}{4}.\frac{36}{37}=\frac{27}{37}\)
\(a,\dfrac{3}{5}+\dfrac{3}{5\cdot9}+\dfrac{3}{9\cdot13}+....+\dfrac{3}{97\cdot101}\)
\(=\dfrac{3}{4}\cdot\left(\dfrac{4}{5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+....+\dfrac{4}{97\cdot101}\right)\)
\(=\dfrac{3}{4}\cdot\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+....+\dfrac{1}{97}-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{4}\cdot\left(1-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{4}\cdot\dfrac{100}{101}\)
\(=\dfrac{75}{101}\)
\(b,\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{4}\right)\cdot....\cdot\left(1+\dfrac{1}{99}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot....\cdot\dfrac{100}{99}\)
\(=\dfrac{100}{2}=50\)
Tính nhanh:
a) \(\dfrac{3}{5}+\dfrac{3}{5.9}+\dfrac{3}{9.13}+...+\dfrac{3}{97.101}\)
= \(\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)
= \(\dfrac{3}{4}\left(1-\dfrac{1}{101}\right)\)
= \(\dfrac{3}{4}\times\dfrac{100}{101}\)
= \(\dfrac{75}{101}\)
b) \(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)...\left(\dfrac{1}{98}+1\right)\left(\dfrac{1}{99}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{99}{98}.\dfrac{100}{99}\)
\(=\dfrac{3.4.5...99.100}{2.3.4...98.99}\)
\(=\dfrac{100}{2}\)
\(=50\)
\(B=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)\(=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+3^8.2^8.2^2.5}\)\(\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+3^8.2^{10}.5}=\frac{2^{10}.\left(3^8-3^9\right)}{2^{10}\left(3^8+3^8.5\right)}\)
\(=\frac{3^8-3^9}{3^8+3^8.5}=\frac{3^8-3^8.3}{3^8\left(1+5\right)}=\frac{3^8\left(1-3\right)}{3^8\left(1+5\right)}=\frac{1-3}{1+5}=\frac{-2}{6}=\frac{-1}{3}\)
Tử số : \(4^5.9^4-2.6^9\)
\(=\left(2^2\right)^5.\left(3^2\right)^4-2.2^9.3^9\)
\(=2^{10}.3^8-2^{10}.3^9\)
Mẫu số : \(2^{10}.3^8+6^8.20\)
\(=2^{10}.3^8+2^8.3^8.2^2.5\)
\(=2^{10}.3^8+2^{10}.3^8.5\)
\(=2^{10}.3^8.6\)
3A=32-33+34-35+............+32010-32011
3A+A=(32-33+34-35+.........+32010-32011)+(3-32+33-34+.................+32009-32010)
4A=3-32011
A=\(\frac{3-3^{2011}}{4}\)
