Tính
a)\(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)
b)\(\frac{18}{2.5}+\frac{18}{5.8}+...+\frac{18}{203.206}\)
Bài 1:Tính nhanh. \(\frac{4}{3.5}+\frac{4}{5.7}+....+\frac{4}{97.99}\)
Bài 2: Tìm x. a) \(2.4^x-18=110\)
b) \(\left(\frac{3}{2}.x-1\right)^5=1\)
Bài 1:
\(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)
\(=2\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=2.\frac{32}{99}=\frac{64}{99}\)
Bài 2:
a) \(2.4^x-18=110\)
\(\Leftrightarrow2.4^x=128\)
\(\Leftrightarrow4^x=64\)
\(\Leftrightarrow4^x=4^3\Leftrightarrow x=3\)
Vậy x = 3
b) \(\left(\frac{3}{2}x-1\right)^5=1\)
\(\Leftrightarrow\frac{3}{2}x-1=1\)
\(\Leftrightarrow\frac{3}{2}x=2\)
\(\Leftrightarrow x=\frac{4}{3}\)
Vậy \(x=\frac{4}{3}\)
a) 4/3.5 + 3/5.7 + .... + 4/97.99
= 4( 1/3.5 +1/5.7 + ... + 1/97.99 )
= 4 . 1/2 . 2 ( 1/3.5 +1/5.7 + ... + 1/97.99 )
= 4/2 ( 2/3.5 + 2/5.7 + .... + 2/97.99 )
= 2 ( 5-3/3.5 + 7-5/5.7 + ..... + 99-97/97.99 )
= 2 (5/3.5 - 3/3.5 + 7/5.7 - 5/5.7 + .... + 99/97.99 - 97/97.99 )
= 2 ( 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/97 - 1/99 )
= 2 ( 1/3 -1/99 )
= 2 (33/99 - 1/99 )
= 2 . 32/99
= 32.2/99
=64/99
b) 2. 4x - 18 = 110
2. 4x = 110 + 18
2. 4x = 128
4x = 128 : 2
4x = 64
4x = 43
=> x = 3
\(\frac{4}{3.5}+\frac{4}{5.7}+....+\frac{4}{97.99}\)
tính tổng
Ta có :
\(\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{97.99}\)
\(=\)\(2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\right)\)
\(=\)\(2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\)\(2\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=\)\(2.\frac{32}{99}\)
\(=\)\(\frac{64}{99}\)
Chúc bạn học tốt ~
Tìm giá trị của biểu thức \(P=\frac{2}{1.3}-\frac{4}{3.5}+\frac{6}{5.7}+\frac{8}{7.9}+...-\frac{96}{95.97}+\frac{98}{97.99}\)
đề 4:
b) M = \(\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{97.99}\)
ghi rõ cách làm bài
de y \(\frac{1}{3}\)-\(\frac{1}{5}\)=\(\frac{2}{3.5}\)
tuong tu suy ra
M=\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+......+\(\frac{1}{97}\)-\(\frac{1}{99}\)
M=\(\frac{1}{3}\)-\(\frac{1}{99}\)
M=\(\frac{32}{99}\)
a) A=\(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)
b) Chứng tỏ M không thuộc N
M=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\)
Cảm ơn mn nhìu ạk
a. \(A=\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)
\(\Rightarrow A=2\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(\Rightarrow A=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Rightarrow A=2\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(\Rightarrow A=2.\frac{32}{99}\)
\(\Rightarrow A=\frac{64}{99}\)
b. \(M=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\notin N\)
Ta cần chứng minh M < 1 và M khác 0
Dễ thấy M khác 0 ( 1 )
Ta có : \(M=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(M< 1-\frac{1}{2020}=\frac{2019}{2020}\)
=> M < 1 ( 2 )
Từ ( 1 ) và ( 2 ) => Đpcm
Bài 1.So sánh A và B biết: A=\(\frac{10^{17}+1}{10^{18}+1}\) B=\(\frac{10^{18}+1}{10^{19}+1}\)
Bài 2.So sánh S=\(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)với 4
Bài 3.Cho A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)Chứng minh rằng A<\(\frac{3}{4}\)
Bài 4.
a)Tính nhanh tổng sau:A=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)
b)Tìm x biết:\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{1008}{2017}\)
mn giúp mk nha mk đang cần gấp
ai nhanh mk sẽ tick cho
tk mn
Bài 1 :
Ta có :
\(A=\frac{10^{17}+1}{10^{18}+1}=\frac{\left(10^{17}+1\right).10}{\left(10^{18}+1\right).10}=\frac{10^{18}+10}{10^{19}+10}\)
Mà : \(\frac{10^{18}+10}{10^{19}+10}>\frac{10^{18}+1}{10^{19}+1}\)
Mà \(A=\frac{10^{18}+10}{10^{19}+10}\)nên \(A>B\)
Vậy \(A>B\)
Bài 2 :
Ta có :
\(S=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)
\(\Rightarrow S=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2013+3}{2013}\)
\(\Rightarrow S=1-\frac{1}{2014}+1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{3}{2013}\)
\(\Rightarrow S=4+\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)
Vì \(\frac{1}{2013}>\frac{1}{2014}>\frac{1}{2015}>\frac{1}{2016}\)nên \(\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)
Nên : \(M>4\)
Vậy \(M>4\)
Bài 3 :
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{100^2}\)
Suy ra : \(A< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+....+\frac{1}{99.101}\)
\(\Rightarrow A< \frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{99.101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-......-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+......+\frac{1}{101}\right)\right]\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}\right)\)
\(\Rightarrow A< \frac{3}{4}\)
Vậy \(A< \frac{3}{4}\)
Bài 4 :
\(a)A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{1}{2015.2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{2016}{2017}\)
\(\Rightarrow A=\frac{1008}{2017}\)
Vậy \(A=\frac{1008}{2017}\)
\(b)\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{x\left(x+2\right)}=\frac{1008}{2017}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{x.\left(x+2\right)}=\frac{2016}{2017}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2016}{2017}\)
\(1-\frac{1}{x+2}=\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{2017}\)
\(\Rightarrow x+2=2017\)
\(\Rightarrow x=2017-2=2015\)
Vậy \(x=2015\)
Thực hiện phép tính :
C = \(\frac{4}{3.5}+\frac{4}{5.7}+......+\frac{4}{97.99}\)
mk làm đc,vs điều kiện bạn phải tick cho mk khi mk giải xog, ko đc tick cho ai
tich minh cho minh len thu 8 tren bang sep hang cai
Chứng tỏ:
\(\frac{4}{3.5}\)+ \(\frac{4}{5.7}\)+ \(\frac{4}{7.9}\)+.................+ \(\frac{4}{97.99}\)> 65%
ta co : 65%=0,65
goi A= 4.(1/3.5+1/5.7+1/7.9+............+1/97.99)
2A=4.( 2/3.5+2/5.7+2/7.9+...............+2/97.99)
2A=4.(1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99)
2A=4.(1/3-1/99)
2A=4.(33/=99+1/99)
2A=4.34/99
2A=136/99
A=136/99:2
A=68/99=0,69=0,68
Vi A=0,68 > 0,65
=> A > 65%
\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(\frac{3}{41}-\frac{12}{47}+\frac{27}{53}\)
------------------
\(\frac{4}{41}-\frac{16}{47}+\frac{36}{53}\)
a) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(=5.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
\(=5.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right):2\)
\(=5.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right):2\)
\(=5.\left(1-\frac{1}{101}\right):2=5.\frac{100}{101}:2=\frac{500}{101}.\frac{1}{2}\)\(=\frac{250}{101}\)
b) \(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(=3\left(\frac{1}{3.6}+\frac{1}{6.9}+...+\frac{1}{30.33}\right)\)\(.\frac{1}{3}\)
\(=(\frac{3}{3.6}+\frac{3}{6.9}+...+\frac{3}{30.33}).\frac{1}{3}\)
\(=(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}).\frac{1}{3}\)
\(=(\frac{1}{3}-\frac{1}{33}).\frac{1}{3}=\frac{10}{33}.\frac{1}{3}=\frac{10}{99}\)
câu c bạn có thể viết rõ được ko
Làm nốt câu c hộ các bạn:
\(\frac{\frac{3}{41}-\frac{12}{47}+\frac{27}{53}}{\frac{4}{41}-\frac{16}{47}+\frac{36}{53}}=\frac{3\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}{4\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}=\frac{3}{4}\)