tính 2^2/3.5 +3^2/5.7 +...+49^2/97.99
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a.
\(M=1.\left[\frac{1}{3}-\frac{1}{5}+.....\frac{1}{97}-\frac{1}{99}\right]\)
\(M=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
b.
\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right]\)
\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{199}\right]=\frac{291}{995}\)
mk đầu tiên nha bạn
\(\frac{x-2}{3}+\frac{x-2}{3.5}+\frac{x-2}{5.7}+...+\frac{x-2}{97.99}=\frac{-49}{99}\)
<=>\(\left(x-2\right)\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)=-\frac{49}{99}\)
<=>\(\left(x-2\right)\cdot\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)=-\frac{49}{99}\)
<=>\(\left(x-2\right)\cdot\frac{1}{2}\cdot\left(1-\frac{1}{99}\right)=-\frac{49}{99}\)
<=>\(\left(x-2\right)\cdot\frac{49}{99}=-\frac{49}{99}\)
<=>x-2=-1
<=>x=1
2/1.3 + 2/3.5 + 2/5.7 +...+ 2/97.99
=(1/1-1/3)+(1/3-1/5)+(1/5-1/7)+...+(1/97-1/99)
=1-1/99=98/99
=3.(3/1.3+3/3.5+3/5.7+...+3/95.97+3/97.99)
=3(1-1/3+1/3-1/5+1/5-1/7+...+1/95-1/97+1/97-1/99)
=3[(1-1/99)+(1/5-1/5)+(1/7-1/7)+...+(1/97-1/97)]
=3(1-1/99)=3(99/99-1/99)=3.98/99=1.98/33=98/33
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
~ Hok tốt ~
\(\)
2/3.5+ 2 /5.7+ 2/7.9+...+ 2/97.99
=1/3-1/5+1/5-1/7+1/7-1/9+....+1/97-1/99
=1/3-1/99
=32/99
2/3.5 + 2/5.7 +...+ 2/97.99
=1/3 - 1/5 + 1/5 - 1/7 + ... + 1/97 - 1/99
=1/3 - 1/99
=32/99
= 2/ 3.5 = 1/3 - 1/5 ; 2/ 5.7 = 1/5 - 1/7 ;........; 2/ 97.99 = 1/97 - 1/99
= 1/3 - 1/99 = 32/99