bài 1:tìm X
phần a, 1/1x2 + 1/ 2x 3 + 1/ 3 x4 + ....+ 1/ X x( X +1)= 499 / 500
phần b,1/1x3 + 1/3x5 + 1/ 5 x 7+ ....+ 1/X x ( X + 2 )= 20 /41
tìm x biết
a, (1/1x2+1/2x3+1/5x4+...+1/99x100) X=1/1x2+2x3+3x4+...+98x99
b, X/1x3+X/3x5+X/5x7+...+X/2013x2015=4/2015
c, X+1/2015+X+2/2016=X+3/2017+X+4/2018
b) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)
\(=\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{2015-2013}{2013.2015}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{2015}\right)=\frac{1007}{2015}\)
Phương trình tương đương với:
\(\frac{1007X}{2015}=\frac{4}{2015}\Leftrightarrow X=\frac{4}{1007}\)
c) \(\frac{x+1}{2015}+\frac{x+2}{2016}=\frac{x+3}{2017}+\frac{x+4}{2018}\)
\(\Leftrightarrow\frac{x+1}{2015}-1+\frac{x+2}{2016}-1=\frac{x+3}{2017}-1+\frac{x+4}{2018}-1\)
\(\Leftrightarrow\frac{x-2014}{2015}+\frac{x-2014}{2016}=\frac{x-2014}{2017}+\frac{x-2014}{2018}\)
\(\Leftrightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)
Mọi người ơi. Vy cần gấp mấy câu này. Bạn nào giải giúp Vy với.
1. Tính
M=5+5^3+5^5+...+5^47+5^49
2. Tìm số tự nhiên x, biết:
1+3+3^2+3^3+...+3^x=265720
1x2+2x3+3x4+...+ x ( x+1) (x+2) = 353430
1x3+3x5+5x7+...+ (2x+1) (2x+3)
M = 5 + 53 + 55 + ... + 547 + 549
52M = 52(5 + 53 + 55 + ... + 547 + 549)
25M = 53 + 55 + 57 + ... + 549 + 551
25M - M = ( 53 + 55 + 57 + ... + 549 + 551) - (5 + 53 + 55 + ... + 547 + 549)
24M = 551 - 5
M = \(\frac{5^{51}-5}{24}\)
Bài 3 : Tính nhanh :
a ) A = 2 + 4 + 6 + 8 + ....+ 1000
b ) 1/ 20 + 1/30 + 1/42 + 1/50 + 1/72 + 1/90 + 1/110 + 1/132
c ) 2 / 1x3 + 2 / 3x5 + 2/5 x 7 + .....+ 2 / 97 x 99 + 2/99 x 101
a) A = 2 + 4 + 6 + 8 + ... + 1000
Ta có : A = 2 + 4 + 6 + 8 + ... + 1000 ( có 500 số )
= (1000 + 2) . 500 : 2 = 250500
c) \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
[ 1/3 + 1/5 ] + [ 1/6 - 1/5 ]
3/16 x 7/5 + 3/5 x 9/16
1/1x2 + 1/2x3 + 1/ 3x4 + ...... + 1/2020x2021
1/1x3 + 1/3x5 + 1/5x7 + ..... + 1/2021 x 2023
3/2 x 1/7 x 5/4 + 15/2 x 6/7 x 1/4
NHỜCÁC BN LM GIÚP MIK GẤP CHIỀU NAY MIK CẦN ÒI !!!!!~~~
TÍNH BẰNG CÁCH NHANH NHẤT NHA CÁC BN
a) \(\left(\frac{1}{3}+\frac{1}{5}\right)+\left(\frac{1}{6}-\frac{1}{5}\right)=\left(\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)=\frac{1}{2}\)
b) \(\frac{3}{16}\times\frac{7}{5}+\frac{3}{5}\times\frac{9}{16}=\frac{21}{80}+\frac{27}{80}=\frac{48}{80}=\frac{3}{5}\)
c) \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2020\times2021}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2020}-\frac{1}{2021}\)
\(=1-\frac{1}{2021}=\frac{2020}{2021}\)
d) \(\frac{1}{1\times3}+\frac{1}{3\times5}+...+\frac{1}{2021\times2023}=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+...+\frac{2}{2021\times2023}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2021}-\frac{1}{2023}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{2023}\right)=\frac{1}{2}\times\frac{2022}{2023}=\frac{1011}{2023}\)
e) \(\frac{3}{2}\times\frac{1}{7}\times\frac{5}{4}+\frac{15}{2}\times\frac{6}{7}\times\frac{1}{4}==\frac{15}{56}+\frac{80}{56}=\frac{95}{56}\)
a,\((\) 1\(-\) \(\dfrac{1}{3}\)\()\)x\((\)1\(-\)\(\dfrac{2}{5}\)\()\)x\((\)1\(-\)\(\dfrac{2}{7}\)\()\)x\((\)1\(-\)\(\dfrac{2}{9}\)\()\)
b,\(\dfrac{1}{1x3}\) + \(\dfrac{1}{3x5}\) + \(\dfrac{1}{5x7}\) + \(\dfrac{1}{7x9}\)
a) \(\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{2}{5}\right)\times\left(1-\dfrac{2}{7}\right)\times\left(1-\dfrac{2}{9}\right)\)
\(=\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\times\left(\dfrac{5}{5}-\dfrac{2}{5}\right)\times\left(\dfrac{7}{7}-\dfrac{2}{7}\right)\times\left(\dfrac{9}{9}-\dfrac{2}{9}\right)\)
\(=\dfrac{2}{3}\times\dfrac{3}{5}\times\dfrac{5}{7}\times\dfrac{7}{9}\)
\(=\dfrac{2\times3\times5\times7}{3\times5\times7\times9}\)
\(=\dfrac{2}{9}\)
b) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}\)
\(=1-\dfrac{1}{9}\)
\(=\dfrac{9}{9}-\dfrac{1}{9}\)
\(=\dfrac{8}{9}\)
a)1/1x3+1/3x5+1/5x7+...+1/Xx(x+3)=99/200
b)1/1x3+1/3x5+1/5x7+...+1/Xx(x+2)
a) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{x\times\left(x+3\right)}=\dfrac{99}{200}\)
Ta có: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=1-\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=\dfrac{1}{100}\)
\(\Rightarrow x+1=100\)
\(x=100-1\)
\(x=99\)
Công thức: \(\dfrac{1}{a\times b}=\) 1/ khoảng cách giữa a và b \(\times\left(\dfrac{1}{a}-\dfrac{1}{b}\right)\)
* Bạn làm theo công thức và vẫn dụng câu b nhé.
Tìm x, biết:1/1 x 3+1/3x5+1/5x7+...+1/x(x+2)=20/41
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+3\right)}=\frac{20}{41}\)
\(2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}\right)=2.\frac{20}{41}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{40}{41}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{40}{41}\)
\(1-\frac{1}{x+2}=\frac{40}{41}\)
\(\frac{1}{x+2}=1-\frac{40}{41}\)
\(\frac{1}{x+2}=\frac{1}{41}\)
=> x + 2 = 41
=> x = 41 - 2
=> x = 39
Vẫy x = 39
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)
=> \(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)
=> \(1-\frac{1}{x+2}=\frac{40}{41}\)
=> \(\frac{1}{x+2}=\frac{1}{41}\)
=> x + 2 = 41
=> x = 39
a.(x*2,4-4,2):x=1
b.1/1x3+1/3x5+1/5x7+...+1/x(x+2)
c.7/3+7/4-7/11 +5/7
5/3+5/9-5/11
\(\left(x\cdot2,4-4,2\right)\div x=1\)
\(\Rightarrow x\cdot2,4-4,2=x\)
\(\Rightarrow x\cdot2,4=x+4,2\)
\(\Rightarrow\frac{12x}{5}=\frac{5x+21}{5}\)
\(\Rightarrow12x=5x+21\)
\(\Rightarrow12x-5x=21\)
\(\Rightarrow7x=21\Rightarrow x=\frac{21}{7}=3\)
Vậy x = 3
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{x\left(x+2\right)}\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{x+2}\right)\)
\(=\frac{1}{2}\cdot\frac{x+1}{x+2}\)
\(=\frac{x+1}{2x+2}\)
Đề sai bn ơi : Sửa : \(\frac{\frac{7}{3}+\frac{7}{4}-\frac{7}{11}}{\frac{5}{3}+\frac{5}{4}-\frac{5}{11}}+\frac{5}{7}\)
\(=\frac{7\cdot\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{11}\right)}{5\cdot\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{11}\right)}+\frac{5}{7}\)
\(=\frac{7}{5}+\frac{5}{7}=\frac{49}{35}+\frac{25}{35}=\frac{74}{35}\)
1 tim x , biet
a) x - 7va 5/8 = 1 va 1/4
b) x + 7 va 5/8 = 9 va 1/4
c) ( x - 7 va 5/8 ) : 1/2 = 3
d) x / 1x3 + x/3x5+.....+x/97x99=99
giup mik nhanh nha . mik can gap lam
\(a,x-7\frac{5}{8}=1\frac{1}{4}\)
=> \(x-\frac{61}{8}=\frac{5}{4}\)
=> \(x=\frac{5}{4}+\frac{61}{8}\)
=> \(x=\frac{10}{8}+\frac{61}{8}=\frac{71}{8}=8\frac{7}{8}\)
\(b,x+7\frac{5}{8}=9\frac{1}{4}\)
=> \(x+\frac{43}{5}=\frac{37}{4}\)
=> \(x=\frac{37}{4}-\frac{43}{5}=\frac{13}{20}\)
\(c,\left[x-7\frac{5}{8}\right]:\frac{1}{2}=3\)
=> \(\left[x-\frac{61}{8}\right]=3\cdot\frac{1}{2}\)
=> \(\left[x-\frac{61}{8}\right]=\frac{3}{2}\)
=> \(x-\frac{61}{8}=\frac{3}{2}\)
=> \(x=\frac{3}{2}+\frac{61}{8}=\frac{12}{8}+\frac{61}{8}=\frac{73}{8}=9\frac{1}{8}\)
d, \(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+\frac{x}{5\cdot7}+...+\frac{x}{97\cdot99}=99\)
=> \(\frac{x}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right]=99\)
=> \(\frac{x}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=99\)
=> \(\frac{x}{2}\left[1-\frac{1}{99}\right]=99\)
=> \(\frac{x}{2}\cdot\frac{98}{99}=99\)
=> \(\frac{98x}{198}=99\)
=> 98x = 99 . 198
=> 98x = 19602
=> x = 19602 : 98 = 9801/49
a) \(x-7\frac{5}{8}=1\frac{1}{4}\)
=> \(x=\frac{5}{4}+\frac{61}{8}\)
=> \(x=\frac{71}{8}\)
b) \(x+7\frac{5}{8}=9\frac{1}{4}\)
=> \(x=\frac{37}{4}-\frac{61}{8}\)
=> \(x=\frac{13}{8}\)
c) \(\left(x-7\frac{5}{8}\right):\frac{1}{2}=3\)
=> \(x-\frac{61}{8}=3.\frac{1}{2}\)
=> \(x-\frac{61}{8}=\frac{3}{2}\)
=> \(x=\frac{3}{2}+\frac{61}{8}\)
=> \(x=\frac{73}{8}\)
d) \(\frac{x}{1.3}+\frac{x}{3.5}+...+\frac{x}{97.99}=99\)
=> \(x.\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)=99\)
=> \(\frac{1}{2}x\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)=99\)
=> \(x\left(1-\frac{1}{99}\right)=99:\frac{1}{2}\)
=> \(x.\frac{98}{99}=198\)
=> \(x=198:\frac{98}{99}=\frac{9801}{49}\)
\(b,\text{ }x+7\frac{5}{8}=9\frac{1}{4}\)
\(x=9\frac{1}{4}-\left(7+\frac{5}{8}\right)\)
\(x=9+\frac{1}{4}-7-\frac{5}{8}\)
\(x=\left(9-7\right)+\left(\frac{1}{4}-\frac{5}{8}\right)\)
\(x=2+-\frac{3}{8}\)
\(x=2\frac{-3}{8}\)