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\(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
\(M=2.(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99})\)
\(M=2.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
\(M=2.\dfrac{32}{99}\)
\(M=\dfrac{64}{99}\)
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\(\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right)-x=-\dfrac{100}{99}\)
\(\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)-x=-\dfrac{100}{99}\)
\(\left(1-\dfrac{1}{99}\right)-x=-\dfrac{100}{99}\)
\(\dfrac{98}{99}-x=-\dfrac{100}{99}\)
\(x=\dfrac{98}{99}-\left(-\dfrac{100}{99}\right)\)
\(x=\dfrac{198}{99}\)
Vậy \(x=\dfrac{198}{99}\)
= 2 . ( \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+ ..... + \(\frac{1}{97}\)- \(\frac{1}{99}\)
= 2 . ( \(\frac{1}{3}\)- \(\frac{1}{99}\))
= 2 . \(\frac{2}{3}\)
= \(\frac{4}{3}\)
32% = \(\frac{32}{100}\)= \(\frac{8}{25}\)
\(\frac{4}{3}\)> \(\frac{8}{25}\)=> \(\frac{2}{3.5}\)+ \(\frac{2}{5.7}\)+ \(\frac{2}{7.9}\)+ ..... + \(\frac{2}{97.99}\)> 32%
\(A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(A=\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}=\frac{800}{2475}\)
\(32\%=\frac{8}{25}=\frac{792}{2475}\)
\(\frac{800}{2475}>\frac{792}{2475}\Rightarrow\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}>32\%\)
Đặt : \(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)
Do \(\frac{32}{99}>32\%\)nên \(A>32\%\left(đpcm\right)\)
a)
\(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{24.25}\)
\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
\(=\dfrac{1}{5}-\dfrac{1}{25}\)
\(=\dfrac{4}{25}\)
b)
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=1-\dfrac{1}{101}\)
\(=\dfrac{100}{101}\)
a) \(\dfrac{1}{5.6}=\dfrac{1}{5}-\dfrac{1}{6}\)
⇒ \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{24.25}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}=\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)b) \(\dfrac{2}{1.3}=1-\dfrac{1}{3}\)
tương tự
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}\)
a, \(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(=\dfrac{1}{3}-\dfrac{1}{99}\)
\(=\dfrac{32}{99}\)
Vậy \(M=\dfrac{32}{99}\)
b, Ta có: \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2012^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2011.2012}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\)
\(=1-\dfrac{1}{2012}< 1\) (1)
Do mỗi phân số đều lớn hơn 0 nên \(A>0\) (2)
Từ (1), (2) \(\Rightarrow0< A< 1\)
\(\Rightarrow A\notin N\left(đpcm\right)\)
Vậy...
a, \(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{2}{97}-\dfrac{2}{99}\\ =\dfrac{1}{3}-\dfrac{2}{99}=\dfrac{31}{99}\)
2)
a) M = \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
M = \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
M = \(\dfrac{1}{3}-\dfrac{1}{99}\)
M = \(\dfrac{32}{99}\)
1) Quãng đường AB là :
\(36\dfrac{1}{4}\)km/h . 3.2h = 116 ( km )
Thời gian người ấy đi từ A đến B lúc về là :
116 : 40 = 2.9 ( giờ )
Đ/S : 2.9 giờ
Trả lời
a)\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...\dfrac{2}{99.101}\)
=\(2.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\right)\)
=\(2.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
= \(2.\left(\dfrac{1}{1}-\dfrac{1}{101}\right)\)
=\(2.\dfrac{100}{101}\)
=\(\dfrac{200}{101}\)
a, \(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{37.39}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\)
\(=\dfrac{1}{3}-\dfrac{1}{39}\)
\(=\dfrac{12}{39}\)
Vậy \(A=\dfrac{12}{39}\)
b,\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{73.76}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{73}-\dfrac{1}{76}\)
\(=1-\dfrac{1}{76}\)
\(=\dfrac{75}{76}\)
Vậy \(B=\dfrac{75}{76}\)
a) Ta có :
\(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+....................+\dfrac{2}{37.39}\)
\(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...................+\dfrac{1}{37}-\dfrac{1}{39}\)
\(A=\dfrac{1}{3}-\dfrac{1}{39}=\dfrac{4}{13}\)
b) Ta có :
\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+..................+\dfrac{3}{73.76}\)
\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+..................+\dfrac{1}{73}-\dfrac{1}{76}\)
\(B=1-\dfrac{1}{76}=\dfrac{75}{76}\)
~ Học tốt ~
Giải:
Biến đổi vế trái BĐT:
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(=\dfrac{1}{3}-\dfrac{1}{99}\)
\(=\dfrac{32}{99}\)
Vì \(\dfrac{32}{99}>\dfrac{32}{100}\)
\(\Leftrightarrow\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}>\dfrac{32}{100}\)
\(\Leftrightarrow\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}>32\%\)
Vậy ...
Thanks