2/3.5 + 2/5.7 +...+ 2/97.99
các bạn cho mk hỏi câu này
2/3.5+2/5.7+2/7.9+...+2/97.99
thì mk sẽ viết thành
1/3.5+1/5.7+1/7.9+...+1/97.99
hay
2.(1/3.5+1/5.7+1/7.9+...+1/97.99)
giúp mk với
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
~ Hok tốt ~
\(\)
Viết thành 2 . (1/3.5 + 1/5.7 + 1/7.9 + ...+ 1/97.99
2/3.5+2/5.7+...2/97.99 =?
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
1.Tính hợp lí
a/ 2/3.5 + 2/5.7 + 2/7.9 +...+2/97.99
b/ 1/3.5 + 1/5.7 + 1/7.9 +...+1/97.99
c/1/18 + 1/54 + 1/108 +...+1/990
2.Chứng minh rằng: 1/14 + 1/42 + 1/43 +...+1/79 + 1/80 > 7.12
2/3.5 + 2/5.7 + 2/7.9 + ... + 2/97.99 = ?
vipboyss5: \(\frac{32}{99}\)chứ ko phải 33
1/3.5+1/5.7+1/7.9+...+1/97.99
=1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99
=1/3-1/99
=33/99-1/99
=32/99
M=2/3.5 + 2/5.7+2/7.9 +....+ 2/97.99 =?
M=(1/3-1/5)+(1/5+1/7)+...+(1/97+1/99)
M=1/3+(1/5-1/5)+...+(1/97-1/97)-1/99
M=1/3-1/99
M=32/99
Tính: 2/1.3+ 2/3.5+ 2/5.7+ ......... + 2/97.99
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
g=1.3^2+3.5^2+5.7^2+...+97.99^2
k) (2 ^ 2)/3.5 + (2 ^ 2)/5.7 + (2 ^ 2)/7.9 +...+ 2^ 2 97.99
\(S=\dfrac{2^2}{3x5}+\dfrac{2^2}{5x7}+\dfrac{2^2}{7x9}+...+\dfrac{2^2}{97x99}\)
\(\dfrac{S}{2}=\dfrac{2}{3x5}+\dfrac{2}{5x7}+\dfrac{2}{7x9}+...+\dfrac{2}{97x99}\)
\(\dfrac{S}{2}=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}...+\dfrac{1}{97}-\dfrac{1}{99}=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)
S=\(\dfrac{64}{99}\)
Chứng tỏ rằng : B = 2/3.5+2/5.7+2/7.9+...+2/97.99
A = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/95 - 1/97 + 1/97 - 1/99
A = 1/3 - 1/99
A = 32/99
BẠN TICK CHO MÌNH NHA !
\(B=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
\(=2.(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99})\)
\(=2.(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99})\)
\(=2.(\dfrac{1}{3}-\dfrac{1}{99})\)
\(=2.\dfrac{1}{297}\)
=\(\dfrac{2}{297}\)