3/3.5+3/5.7+...+3/61.63
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Ta đặt
\(A=1\times3+3\times5+...+61\times63\)
\(6A=1\times3\times6+3\times5\times6+....+61\times63\times6\)
\(6A=1\times3\times6+3\times5\times\left(7-1\right)+...+61\times63\times\left(65-59\right)\)
\(6A=1\times3\times6+3\times5\times7-1\times3\times5+...+61\times63\times65-59\times61\times63\)
\(6A=1\times3\times6-1\times3\times5+61\times63\times65\)
\(6A=3+61\times63\times65\)
\(6A=3\times\left(1+61\times21\times65\right)\)
\(2A=83266\)
\(A=83266\div2=41633\)
\(1-\dfrac{2}{3.5}-\dfrac{2}{5.7}-...-\dfrac{2}{61.63}-\dfrac{2}{63.65}\)
\(=1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{63}-\dfrac{1}{65}\right)\)
\(=1-\left(\dfrac{1}{3}-\dfrac{1}{65}\right)\)
\(=1-\dfrac{62}{195}\)
\(=\dfrac{133}{195}\)
Ta có: \(1-\dfrac{2}{3\cdot5}-\dfrac{2}{5\cdot7}-\dfrac{2}{7\cdot9}-...-\dfrac{2}{61\cdot63}-\dfrac{2}{63\cdot65}\)
\(=1-\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{61\cdot63}+\dfrac{2}{63\cdot65}\right)\)
\(=1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{61}-\dfrac{1}{63}+\dfrac{1}{63}-\dfrac{1}{65}\right)\)
\(=1-\left(\dfrac{1}{3}-\dfrac{1}{65}\right)\)
\(=1-\dfrac{62}{195}\)
\(=\dfrac{133}{195}\)
Ta có: \(B=1-\dfrac{2}{3.5}-\dfrac{2}{5.7}-\dfrac{2}{7.9}-...-\dfrac{2}{61.63}-\dfrac{2}{63.65}\)
\(=1-\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{61.63}+\dfrac{2}{63.65}\right)\)
\(=1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{63}-\dfrac{1}{65}\right)\)
\(=1-\left(\dfrac{1}{3}-\dfrac{1}{65}\right)\)
\(=1-\dfrac{62}{195}=\dfrac{133}{195}.\)
Vậy \(B=\dfrac{133}{195}.\)
=3.2/1.3.2+3.2/3.5.2+...+3.2/49.51
=3/2.(2/1.3+2/3.5+2/5.7+...+2/49.51)
=3/2.(1-1/3+1/3-1/5+...+1/49-1/51)
=3/2.(1-1/51)
=3/2.50/51
=25/17
CHÚC BẠN HỌC GIỎI
K MÌNH NHÉ
A=3/2(2/3.5+2/5.7+...+2/61.63)
=3/2(1/3-1/5+1/5-1/7+...+1/61-1/63)= 3/2(1/3-1/63)=3/2 x 20/63=10/21
Đs: 10/21