T
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
MH
1 tháng 10 2015
a.\(\Rightarrow\left(\frac{3}{5}+x\right):\frac{2}{7}=\frac{3}{35}-\frac{2}{7}\)
\(\Rightarrow\left(\frac{3}{5}+x\right):\frac{2}{7}=-\frac{1}{5}\)
\(\Rightarrow\frac{3}{5}+x=-\frac{1}{5}.\frac{2}{7}\)
\(\Rightarrow\frac{3}{5}+x=-\frac{2}{35}\)
\(\Rightarrow x=-\frac{2}{35}-\frac{3}{5}\)
Vậy \(x=-\frac{23}{35}\).
b. => 5x-1=0 hoặc 2x-1/3=0
=> 5x=1 hoặc 2x=1/3
=> x=1/5 hoặc x=1/6
c. \(\Rightarrow\frac{1}{7}:x=\frac{3}{14}-\frac{3}{7}\)
\(\Rightarrow\frac{1}{7}:x=-\frac{3}{14}\)
\(\Rightarrow x=\frac{1}{7}:\left(-\frac{3}{14}\right)\)
Vậy \(x=\frac{-2}{3}\).
10 tháng 5 2016
c)\(\frac{1}{2}x+\frac{1}{8}x=\frac{3}{4}\)
\(\Rightarrow x.\left(\frac{1}{2}-\frac{1}{8}\right)=\frac{3}{4}\)
\(\Rightarrow x.\frac{3}{8}=\frac{3}{4}\)
=>x\(=\frac{3}{4}:\frac{3}{8}\)
=>x=\(2\)
10 tháng 5 2016
a)\(x+\frac{1}{6}=\frac{-3}{8}\)
=>\(x=\frac{-3}{8}-\frac{1}{6}\)
=>\(x=\frac{-9}{24}-\frac{4}{24}\)
=>\(x=\frac{-13}{24}\)
b)\(2-\left|\frac{3}{4}-x\right|=\frac{7}{12}\)
=>\(\left|\frac{3}{4}-x\right|=2-\frac{7}{12}\)
=>\(\left|\frac{3}{4}-x\right|=\frac{24}{12}-\frac{7}{12}\)
\(\Rightarrow\left|\frac{3}{4}-x\right|=\frac{17}{12}\)
TH1: \(\frac{3}{4}-x=\frac{17}{12}\)
=>x=\(\frac{3}{4}-\frac{17}{12}\)
=>x=\(x=-\frac{2}{3}\)
TH2:\(\frac{3}{4}-x=-\frac{17}{12}\)
=>\(x=\frac{3}{4}-\left(-\frac{17}{12}\right)\)
=>x=\(x=\frac{13}{6}\)
Dzồi nhìu phết
DN
1
ML
12 tháng 8 2015
\(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)
=>\(\left(x-\frac{1}{3}\right)^2=\frac{1}{4}=\left(\frac{1}{2}\right)^2=\left(-\frac{1}{2}\right)^2\)
=>\(x-\frac{1}{3}=\frac{1}{2}\)hoặc \(x-\frac{1}{3}=-\frac{1}{2}\)
=>x=1/2+1/3hoawcj x=-1/2+1/3
=>x=5/6 hoặc x=-1/6
12 tháng 5 2016
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{n\left(n+1\right)}=\frac{1999}{2001}\) <=>\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.......+\frac{2}{n\left(n+1\right)}=\frac{1999}{2001}\)
<=>\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{n\left(n+1\right)}\right)=\frac{1999}{2001}\)
<=>\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}+.....\frac{1}{n}-\frac{1}{n-1}\right)=\frac{1999}{2001}\)
<=>\(2\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{1999}{2001}\)
<=>\(\frac{1}{2}-\frac{1}{n+1}=\frac{1999}{4002}\)
<=>\(\frac{1}{n+1}=\frac{1}{2001}\)
<=>n+1 =2001
<=>n = 2000
12 tháng 5 2016
ta có:
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{n\left(n+1\right)}=\frac{1999}{2001}\)
\(\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n\left(n+1\right)}\right)=\frac{1}{2}.\frac{1999}{2001}\)
\(\frac{1}{2.3}+\frac{1}{2.6}+\frac{1}{2.10}+...+\frac{1}{n\left(n+1\right)}=\frac{1999}{4002}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}=\frac{1999}{4002}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n\left(n+1\right)}=\frac{1999}{4002}\)
\(\frac{1}{2}-\frac{1}{n+1}=\frac{1999}{4002}\)
\(\frac{1}{n+1}=\frac{1}{2}-\frac{1999}{4002}\)
\(\frac{1}{n+1}=\frac{1}{2001}\)
=>\(n+1=2001\)
=>\(n=2000\)