Chứng tỏ rằng : B = 2/3.5+2/5.7+2/7.9+...+2/97.99
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
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
Cho biểu thức:
S=2/3+2/3.5+2/5.7+2/7.9+....+2/97.99
Hãy chứng tỏ rằng:S>1
Ta có S=2/3+2/3.5+2/5.7+2/7.9+...+2/97.99
=2/3+1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99
=2/3+1/3+(1/5-1/5)+(1/7-1/7)+...+(1/97-1/97)+1/99
=1+0+0+0+...+0+1/99
=1+1/99
=100/99
Mà 100/99>1.Suy ra S>1
Vậy S>1
S=1-1/3 + 1/3 - 1/5 + ... + 1/97 - 1/99
=1 - 1/99 => S<1
Cho A = 2/1.3 + 2/3.5 + 2/5.7 + 2/7.9 +.....+ 2/97.99 và B = 1^2 / 1.2 x 2^2/2.3 x 3^2 / 3.4 x 4^2 /4.5 x .... x 98^2 / 98.99. Chứng tỏ A = 98B
Ta có :\(B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}.....\frac{98^2}{98.99}=\frac{\left(1.2.3.4...98\right).\left(1.2.3.4...98\right)}{\left(1.2.3.4...98\right).\left(2.3.4.5...99\right)}=\frac{1}{99}\)
Lại có A = \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}\)
Lại có \(A:B=\frac{98}{99}:\frac{1}{99}=98\)
=> A = 98B
các bạn có về sweet home
Chứng minh rằng: 2/3.5 + 2/5.7 + 2/7.9 + ... + 2/97.99 > 8/25
Giúp mk với!!!!!
PLEASE!!!!!🙏🙏🙏👏
\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\)\(=\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+...+\frac{99-97}{97\cdot99}\)\(=\frac{5}{3\cdot5}-\frac{3}{3\cdot5}+\frac{7}{5\cdot7}-\frac{5}{5\cdot7}+...+\frac{99}{97\cdot99}-\frac{97}{97\cdot99}\)\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)\(=\frac{1}{3}-\frac{1}{99}\)\(=\frac{32}{99}>\frac{8}{25}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
Nhận thấy : \(\frac{32}{99}>\frac{8}{25}\left(32>8;99>25\right)\)
Tính:
a) M=2/3.5+2/5.7+2/7.9+...+2/97.99
b) N=3/5.7+3/7.9+3/9.11+...+3/197.199
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
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
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=\(\dfrac{64}{99}\)