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
Tính các tổng sau :
A=2/3.5+2/5.7+2/7.9+....+2/97.99
B=1/6+1/12+1/20+1/30+1/42+1/56
1/3.5 + 1/5.7 + 1/7.9 + ... + 1/97.99
= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)
= \(\dfrac{\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}}{2}\)
= \(\dfrac{\dfrac{1}{3}-\dfrac{1}{99}}{2}=\dfrac{16}{99}\)
Bài kia mk nhầm sorry
Tính giá trị biểu thức sau: A= 1/1.3+1/3.5+1/5.7+1/7.9+...+1/97.99
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{98}{99}\)
\(=\dfrac{49}{99}\)
1/3.5+1/5.7+1/7.9+...+1/97.99 = ?
\(\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{97\times99}\)
\(=\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{9}\right)+...+\left(\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{99-3}{297}\)
\(=\frac{96}{297}=\frac{32}{99}\)
A=1/1.2+1/2.3+1/3.4+1/4.5+...+1/98.99+1/99.100
B=2/1.3+2/3.5+2/5.7+2/7.9+...+2/97.99+2/99.101
C=1/2+1/4+1/8+1/16+...+1/1024+1/2048
D=1/2+1/6+1/18+1/54+1/4374+1/13122 ( . (DẤU CHẤM LÀ ''NHÂN'')
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{!}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
\(C=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....+\frac{1}{1024}+\frac{1}{2048}\)
\(\Rightarrow\)\(2C=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{512}+\frac{1}{1024}\)
\(\Rightarrow\)\(2C-C=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\right)\)
\(\Leftrightarrow\)\(C=1-\frac{1}{2048}=\frac{2047}{2048}\)
Câu A bạn quên 1/4.5 kìa , với câu D đâu >>>
tính giá trị biểu thức sau:A=1/1.3+1/3.5+1/5.7+1/7.9+...+1/97.99
A= \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+\(\dfrac{1}{7.9}\)+...+\(\dfrac{1}{97.99}\)
2A= 1 - \(\dfrac{1}{3}\)+\(\dfrac{1}{3}\) - \(\dfrac{1}{5}\)+\(\dfrac{1}{5}\) - \(\dfrac{1}{7}\)+\(\dfrac{1}{7}\) - \(\dfrac{1}{9}\)+...+\(\dfrac{1}{97}\)-\(\dfrac{1}{99}\)
2A= 1-\(\dfrac{1}{99}\)
2A= \(\dfrac{98}{99}\)
A= \(\dfrac{98}{99}\) : 2
A=\(\dfrac{49}{99}\)
a ) tính nhanh:M=2/3.5+2/5.7+2/7.9+.....+2/97.99
b)cho A =1/22+1/32+1/42+.....+1/20142+1/20152+1/20162
chứng minh rằng A không phải là số tự nhiên
a/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=1/3-1/99
M=32/99
b)ta có 1/2.3+1/3.4+1/4.5+..+1/2015.2016+1/2016.2017<A
=>1/2-1/3+1/3-1/4+1/4-1/5+..+1/2015-1/2016+1/2016-1/2017<a
1/2-1/2017<A
2/15/4034<A (1)
Ta có
1/1.2+1/2.3+1/3.4+1/4.5+..+1/2015.2016>A
=>1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+..+1/2015-1/2016>A
1-1/2016
2015/2016>A (2)
Từ (1) và (2)=>A không phải là số tự nhiên(đpcm)
a) \(M=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{97.99}\)
\(M=\frac{2}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\right)\)
\(M=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(M=\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)
b) (+)Hiển nhiên A > 0 (1)
(+) Tổng quát: \(\frac{1}{n^2}<\frac{1}{\left(n-1\right).n}\)
Ta có: \(A=\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{2015^2}+\frac{1}{2016^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2014.2015}+\frac{1}{2015.2016}\)
\(\Rightarrow A<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2016}=1-\frac{1}{2016}=\frac{2015}{2016}<1\)
=>A < 1 (2)
Từ (1);(2)=>0 < A <1
=>A ko là số tự nhiên (đpcm)
M=2/3.5 + 2/5.7+2/7.9 +....+ 2/97.99
giải thích cho mik đoạn : 1/3-1/5+1/5-1/7