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10 tháng 4 2019
\(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{x.\left(x+2\right)}=\frac{24}{35}\)
\(\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{x.\left(x+2\right)}\right)=\frac{24}{35}\)
\(\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{x+2}\right)=\frac{24}{35}\)
\(\frac{3}{10}-\frac{3}{2x+4}=\frac{24}{35}\)
\(\frac{3}{2x+4}=\frac{-27}{70}\)
tự làm nốt
DS
26 tháng 6 2016
2.[1/42+1/56+1/72+...+1/x.(x+1)]=2/9
1/6.7+1/7.8+1/8.9+....+1/x.(x+1)=1/9
1/6-1/7+1/7-1/8+1/8-1/9+.....+1/x-1/x+1=1/9
1/6-1/x+1=1/9
1:(x+1)=1/6-1/9
x+1=1:(1/18)
x+1=18
x=18-1
x=17
Vậy x=17
Chúc em học tốt
Ủng hộ anh nha^^
26 tháng 6 2016
2/42 + 2/56 + 2/72 + ... + 2/x.(x+1) = 2/9
2.[1/42 + 1/56 + 1/72 + ... + 1/x.(x+1)] = 2/9
1/6.7 + 1/7.8 + 1/8.9 + ... + 1/x.(x+1) = 2/9 : 2
1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + ... + 1/x - 1/x+1 = 2/9 . 1/2
1/6 - 1/x+1 = 1/9
1/x+1 = 1/6 - 1/9
1/x+1 = 6/36 - 4/36
1/x+1 = 2/36 = 1/18
=> x+1=18
=> x=18-1
=> x=17
Vậy x=17
26 tháng 6 2016
ta xét VT=\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=2\left(\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{x\left(x+1\right)}\right)\)
=\(2\left(\frac{7-6}{6\cdot7}+\frac{8-7}{7\cdot8}+...+\frac{\left(x+1\right)-x}{x\left(x+1\right)}\right)=2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}\right)\)
=\(2\left(\frac{1}{6}-\frac{1}{x+1}\right)\)= 2*1/9
=> \(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
<=> \(\frac{1}{x+1}=\frac{1}{18}\)
<=> x+1=18
=> x=17
KT
23 tháng 7 2018
\(\left(1-\frac{1}{35}\right)\left(1-\frac{1}{36}\right)\left(1-\frac{1}{37}\right)...\left(1-\frac{1}{2010}\right)\left(1-\frac{1}{2011}\right)\)
\(=\frac{34}{35}.\frac{35}{36}.\frac{36}{37}.....\frac{2009}{2010}.\frac{2010}{2011}\)
\(=\frac{34}{2011}\)
\(\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}+\frac{109}{110}+\frac{131}{132}+\frac{155}{156}\)
\(=1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}+1-\frac{1}{110}+1-\frac{1}{132}+1-\frac{1}{156}\)
\(=7-\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}\right)\)
\(=7-\left(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}\right)\)
\(=7-\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{12}-\frac{1}{13}\right)\)
\(7-\left(\frac{1}{6}-\frac{1}{13}\right)=6\frac{71}{78}\)
3 tháng 4 2018
\(\frac{x-2}{12}+\frac{x-2}{20}+\frac{x-2}{30}+\frac{x-2}{42}+\frac{x-2}{56}+\frac{x-2}{72}=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{3}{9}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\frac{2}{9}=\frac{16}{9}\)
\(x-2=\frac{16}{9}:\frac{2}{9}\)
\(x-2=\frac{16}{9}\cdot\frac{9}{2}\)
\(x-2=8\)
\(x=8+2\)
\(x=10\)
Vậy \(x=10\)
17 tháng 4 2020
\(\left(x-2\right)\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=\)\(=\frac{16}{9}\)
\(\left(x-2\right)\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\left(\frac{2}{9}\right)=\frac{16}{9}\)
2(x-2)=16
x-2=8
x=10
24 tháng 2 2015
ta có:$\frac{x-1}{12}+\frac{x-1}{20}+\frac{x-1}{30}+\frac{x-1}{42}+\frac{x-1}{56}+\frac{x-1}{72}=\frac{16}{9}$
=> x+1(1/12+1/20+1/30+1/42+1/56+1/72)=16/9
=> x+1.2/9=16/9
=> x+1 = (16/9):(2/9)
=> x+1 = 8
=> x = 9
5 tháng 4 2017
thông cảm mình ko đánh được dấu ngoặc tròn
[x-1].[1/12+1/20+1/30+1/42+1/56+1/72] =16/9
[x-1].[1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9]=16/9
[x-1].[1/3-1/9]=16/9
[x-1].2/9=16/9
x-1=16/9:2/9
x-1=8
x=7
Vậy x=7
28 tháng 5 2016
=> x-2.(1/12+1/20+1/30+1/42+1/56+1/72)=16/9
Đặt : Sáng = 1/12+1/20+1/30+1/42+1/56+1/72
=> Sáng = 1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9
=> Sáng = 1.(1/3-1/4+1/4-1/5+...+1/8-1/9
=> Sáng = 91.(1/3-1/9)
=> Sáng = 2/9
Thay Sáng vô biểu thức 1/12+1/20+1/30+1/42+1/56+1/72
Ta được :
x-2.2/9=16/9
giờ thì tự làm nha
28 tháng 5 2016
=> x-2.(1/12+1/20+1/30+1/42+1/56+1/72)=16/9
Đặt : Sáng = 1/12+1/20+1/30+1/42+1/56+1/72
=> Sáng = 1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9
=> Sáng = 1.(1/3-1/4+1/4-1/5+...+1/8-1/9
=> Sáng = 91.(1/3-1/9)
=> Sáng = 2/9
Thay Sáng vô biểu thức 1/12+1/20+1/30+1/42+1/56+1/72
Ta được :
x-2.2/9=16/9
giờ thì tự làm nha
Ai k mk mk k lại
HK
3
1 tháng 4 2018
Ta có :
\(\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}+35=2^5\)
\(\Leftrightarrow\)\(\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}=2^5-35\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{2004}+1\right)+\left(\frac{x+2}{2003}+1\right)+\left(\frac{x+3}{2002}+1\right)=32-35+3\)
\(\Leftrightarrow\)\(\frac{x+2005}{2004}+\frac{x+2005}{2003}+\frac{x+2005}{2002}=-3+3\)
\(\Leftrightarrow\)\(\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\right)=0\)
Vì \(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\ne0\)
Nên \(x+2005=0\)
\(\Rightarrow\)\(x=-2005\)
Vậy \(x=-2005\)
Chúc bạn học tốt ~
1 tháng 4 2018
Ta có: \(\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}+35=2^5\)
\(\Rightarrow\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}=2^5-35\)
\(\Rightarrow\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}=-3\)
\(\Rightarrow\frac{x+1}{2004}+1+\frac{x+2}{2003}+1+\frac{x+3}{2002}+1=-3+3\)
\(\Rightarrow\frac{x+1+2004}{2004}+\frac{x+2+2003}{2003}+\frac{x+3+2002}{2002}=0\)
\(\Rightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}+\frac{x+2005}{2002}=0\)
\(\Rightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\right)=0\)
Vì \(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\ne0\)
Nên x + 2005 = 0
=> x = -2005
Vậy x = -2005
Ta có: \(\frac{x}{35}-\frac{x}{42}=\frac{1}{2}\)
⇔\(\frac{x}{35}-\frac{x}{42}-\frac{1}{2}=0\)
⇔\(\frac{6x}{210}-\frac{5x}{210}-\frac{105}{210}=0\)
\(\Leftrightarrow x-105=0\)
\(\Leftrightarrow x=105\)
Vậy: x=105