Tìm x, biết:
a) \(\frac{29}{30}\)-\((\frac{13}{23}\)+x\()\)=\(\frac{7}{69}\)
b) \(\frac{1}{2}\)+\(\frac{1}{6}\)+\(\frac{1}{12}\)+...+\(\frac{1}{x\left(x+1\right)}\)=\(\frac{2011}{2012}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
30 tháng 3 2018
a) \(x+\frac{7}{12}=\frac{17}{18}-\frac{1}{9}\)
\(\Rightarrow x+\frac{7}{12}=\frac{5}{6}\)
\(\Rightarrow x=\frac{5}{6}-\frac{7}{12}\)
\(\Rightarrow x=\frac{1}{4}\)
b) \(\frac{29}{30}-\left(\frac{13}{23}+x\right)=\frac{7}{69}\)
\(\Rightarrow\frac{13}{23}+x=\frac{29}{30}-\frac{7}{69}\)
\(\Rightarrow\frac{13}{23}+x=\frac{199}{230}\)
\(\Rightarrow x=\frac{199}{230}-\frac{13}{23}\)
\(\Rightarrow x=\frac{3}{10}\)
30 tháng 3 2018
a)\(x+\frac{7}{12}=\frac{17}{18}-\frac{1}{9}\)
\(x+\frac{7}{12}=\frac{5}{6}\)
\(x=\frac{5}{6}-\frac{7}{12}\)
\(x=\frac{1}{4}\)
b)\(\frac{29}{30}-\left(\frac{13}{23}+x\right)=\frac{7}{69}\)
\(\left(\frac{13}{23}+x\right)=\frac{29}{30}-\frac{7}{69}\)
\(\left(\frac{13}{23}+x\right)=\frac{199}{230}\)\(x=\frac{199}{230}-\frac{13}{23}\)
\(x=\frac{3}{10}\)
8 tháng 1 2017
a)
\(2^x\left(1+2+2^2+2^3\right)=480\)
\(2^x.15=480\Rightarrow2^x=\frac{480}{15}=32=2^5\Rightarrow x=5\)
24 tháng 3 2019
Bài 1: Tìm x biết:
1) x +\(\frac{7}{12}\)= \(\frac{17}{18}\)- \(\frac{1}{9}\) 2) \(\frac{29}{30}\)- (\(\frac{13}{23}\)+ x) = \(\frac{7}{69}\)
x +\(\frac{7}{12}\)= \(\frac{15}{18}\) \(\frac{13}{23}\)+ x = \(\frac{29}{30}\)- \(\frac{7}{69}\)
x = \(\frac{15}{18}\)- \(\frac{7}{12}\) \(\frac{13}{23}\)+ x = \(\frac{199}{230}\)
x = \(\frac{1}{4}\) x = \(\frac{3}{10}\)
2 tháng 3 2020
\(a,⇔\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
\(⇔(x-23)(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27})=0\)
\(⇔x-23=0\) (vì \(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}>0\))
\(⇔x=23\)
\(b,⇔\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}+\frac{x+100}{95}=0\)
\(⇔(x+100)(\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95})=0\)
\(⇔x+100=0\) (vì \(\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95}>0\))
\(⇔x=-100\)
\(c,⇔(\frac{x+1}{2012}+1)+(\frac{x+2}{2011}+1)=(\frac{x+3}{2010}+1)+(\frac{x+4}{2009}+1)\)
\(⇔\frac{x+2013}{2012}+\frac{x+2013}{2011}-\frac{x+2013}{2010}-\frac{x+2013}{2009}=0\)
\(⇔(x+2013)(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009})=0\)
\(⇔x+2013=0\) (vì \(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}<0\))
\(⇔x=-2013\)
KS
2 tháng 3 2020
\(\frac{201-x}{99}+\frac{203}{97}=\frac{205}{95}+3\)
\(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
\(\frac{2-x}{2010}-1=\frac{1-x}{2011}-\frac{x}{2012}\)
Giúp mk với ạ
a) \(\frac{29}{30}\)- (\(\frac{13}{23}\)+X)=\(\frac{7}{69}\)
\(\frac{13}{23}\)+X=\(\frac{29}{30}\)-\(\frac{7}{69}\)
\(\frac{13}{23}\)+X=\(\frac{199}{230}\)
X=\(\frac{199}{230}\)-\(\frac{13}{23}\)
X=\(\frac{3}{10}\)
b)1/2+1/6+1/12+...+1/x(x+1)=2011/2012
=>1/1.2+1/2.3+1/3.4+...+1/x(x+1)=2011/2012
=>1-1/2+1/2-1/3+1/3+1/4+...+1/x+1/x+1=2011/2012
=>1-1/x+1=2011/2012
=>1/x+1=1-2011-2012
=>1/x+1=2012/2012-2011/2012
1/x+1=1/2012
=>x+1=2012
=>x=2011
a) 3/10