Tìm x biết \(\frac{x+5}{x+3}
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30 tháng 1 2020
b) \(\left|5x-3\right|-x=7\)
\(\Rightarrow\left|5x-3\right|=7+x\)
\(\Rightarrow\orbr{\begin{cases}5x-3=7+x\\5x-3=-\left(7+x\right)\end{cases}\Rightarrow\orbr{\begin{cases}5x-3=7+x\\5x-3=-7-x\end{cases}\Rightarrow}\orbr{\begin{cases}5x-x=7+3\\5x+x=-7+3\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}4x=10\\6x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{2}{3}\end{cases}}}\)
Vậy ....................
30 tháng 1 2020
Bạn ơi !!! ý A tham khảo tại link này nè :
https://h.vn/hoi-dap/question/394208.html
~ Học tốt ~
NN
1
21 tháng 2 2020
\(\frac{x+7}{3}+\frac{x+5}{4}=\frac{x+3}{5}+\frac{x+1}{6}\)
\(\Leftrightarrow\left(\frac{x+7}{3}+2\right)+\left(\frac{x+5}{4}+2\right)=\left(\frac{x+3}{5}+2\right)+\left(\frac{x+1}{6}+2\right)\)
\(\Leftrightarrow\frac{x+13}{3}+\frac{x+13}{4}=\frac{x+13}{5}+\frac{x+13}{6}\)
\(\Leftrightarrow\frac{x+13}{3}+\frac{x+13}{4}-\frac{x+13}{5}-\frac{x+13}{6}=0\)
\(\Leftrightarrow\left(x+13\right)\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)
\(\Rightarrow x+13=0\)( vì \(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\ne0\))
\(\Leftrightarrow x=-13\)
Vậy nghiệm của phương trình x=-13
1 tháng 7 2018
3 + \(\frac{3}{20}\)+ \(\frac{3}{13}\) + \(\frac{3}{2013}\)
X x = \(\frac{5}{3}\)
5 + \(\frac{5}{20}\) + \(\frac{5}{13}\) + \(\frac{5}{2013}\)
3 x ( 1 + \(\frac{1}{20}\) + \(\frac{1}{13}\) + \(\frac{1}{2013}\) )
X x = \(\frac{5}{3}\)
5 x ( 1 + \(\frac{1}{20}\) +\(\frac{1}{13}\) + \(\frac{1}{2013}\) )
X x \(\frac{3}{5}\) = \(\frac{5}{3}\) => X = \(\frac{25}{9}\) vậy X = \(\frac{25}{9}\)
1 tháng 7 2018
Ta có : \(X.\frac{3+\frac{3}{20}+\frac{3}{13}+\frac{3}{2013}}{5+\frac{5}{20}+\frac{5}{13}+\frac{5}{2013}}=\frac{5}{3}\)
\(\Leftrightarrow X.\frac{3\left(1+\frac{1}{20}+\frac{1}{13}+\frac{1}{2013}\right)}{5\left(1+\frac{1}{20}+\frac{1}{13}+\frac{1}{2013}\right)}=\frac{5}{3}\)
\(\Leftrightarrow X.\frac{3}{5}=\frac{5}{3}\Rightarrow X=\frac{5}{3}:\frac{3}{5}=\frac{5}{3}.\frac{5}{3}=\frac{25}{9}\)
M
1
29 tháng 4 2017
=> \(\left(3x-\frac{3}{2}\right)-\left(5x-3\right)=-x+\frac{1}{5}
\)
=> \(3x-\frac{3}{2}-5x+3=-x+\frac{1}{5}\)
=> \(3x-5x-\frac{3}{2}+3=-x+\frac{1}{5}\)
=> \(-2x-\left(\frac{3}{2}-3\right)=-x+\frac{1}{5}\)
=> \(-2x-\frac{-3}{2}=-x+\frac{1}{5}\)
=> \(-2x+\frac{3}{2}=-x+\frac{1}{5}\)
=> \(-2x+x=\frac{1}{5}-\frac{3}{2}\)(Chuyển vế)
=> \(x.\left(-2+1\right)=\frac{-13}{10}\)
=> \(x.\left(-1\right)=\frac{-13}{10}\)
=> \(x=\frac{-13}{10}:\left(-1\right)\)
=> \(x=\frac{13}{10}\)
NT
Tìm x biết: \(\frac{x+1}{9}+\frac{x+4}{6}+\frac{x+5}{5}=\frac{x+2}{8}+\frac{x+3}{7}+\frac{x+6}{4}.\)
4
13 tháng 8 2017
\(\frac{x+1}{9}+\frac{x+4}{6}+\frac{x+5}{5}=\frac{x+2}{8}+\frac{x+3}{7}+\frac{x+6}{4}\)
\(\Rightarrow\frac{x+1}{9}+\frac{x+4}{6}+\frac{x+5}{5}+3=\frac{x+2}{8}+\frac{x+3}{7}+\frac{x+6}{4}+3\)
\(\Rightarrow\left(\frac{x+1}{9}+1\right)+\left(\frac{x+4}{6}+1\right)+\left(\frac{x+5}{5}+1\right)=\left(\frac{x+2}{8}+1\right)\)\(+\left(\frac{x+3}{7}+1\right)+\left(\frac{x+6}{4}\right)\)
\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}=\frac{x+10}{8}+\frac{x+10}{7}+\frac{x+10}{4}\)
\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}\right)=\left(x+10\right)\left(\frac{1}{8}+\frac{1}{7}+\frac{1}{4}\right)\)
\(\Rightarrow\left(x+10\right)\frac{43}{90}=\left(x+10\right)\frac{29}{56}\)
\(\Rightarrow x+10=0\)
\(\Rightarrow x=-10\)
S
13 tháng 8 2017
cộng 3 vào cả hai vế nên phương trình vẫn bằng nhau
Ta có \(\frac{x+1}{9}+1+\frac{x+4}{6}+1+\frac{x+5}{5}+1=\frac{x+2}{8}+1+\frac{x+3}{7}+1+\frac{x+6}{4}+1\)
\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}=\frac{x+10}{8}+\frac{x+10}{7}+\frac{x+10}{4}\)
\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}-\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{4}=0\)
\(\Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}-\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
mà \(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}-\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)
\(\Rightarrow x+10=0\)
\(\Leftrightarrow x=-10\)
NP
9
8 tháng 1 2017
Ta co
\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+349}{5}-4\)=0
\(\Rightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Rightarrow\left(x+329\right).\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Vì \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}>0\)
\(\Rightarrow x+329=0\Rightarrow x=-329\)
Vay \(x=-329\)
3 tháng 8 2015
ta có \(1+\frac{x+5}{1995}+1+\frac{x+4}{1996}+1+\frac{x+3}{1997}=1+\frac{x+1995}{5}+1+\frac{x+1996}{4}+1+\frac{x+1997}{3}\)
\(=\frac{x+2000}{1995}+\frac{x+2000}{1996}+\frac{x+2000}{1997}=\frac{x+2000}{5}+\frac{x+2000}{4}+\frac{x+2000}{3}\)
\(=\left(x+2000\right)\left(\frac{1}{1995}+\frac{1}{1996}+\frac{1}{1997}\right)=\left(x+2000\right)\left(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}\right)\) (1)
Xét \(\frac{1}{1995}+\frac{1}{1996}+\frac{1}{1997}\ne\frac{1}{5}+\frac{1}{4}+\frac{1}{3}vàx+2000=x+2000\) (2)
từ \(\left(1\right)\Leftrightarrow x+2000=0\) ( để (1) là đúng )
\(\Rightarrow x=2000\)
T
1
24 tháng 12 2016
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+329+20}{5}=0\)
\(\Leftrightarrow\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+329}{5}+4=0\)
\(\Leftrightarrow\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+329}{5}=0\)
\(\Leftrightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
\(\Leftrightarrow x+329=0\). Do \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\ne0\)
\(\Leftrightarrow x=-329\)
LA
1
AN
29 tháng 12 2016
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\frac{x+2}{327+1}+\frac{x+3}{326+1}+\frac{x+4}{325+1}+\frac{x+5}{324+1}+\frac{x+349}{5-4}=0\)
\(\Leftrightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\left(x+329\right)\cdot\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Do \(\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)
Nên \(x+329=0\Rightarrow x=-329\)
CG
2
2 tháng 9 2018
(x + 2) / 327 + (x + 3) / 326 + (x + 4) / 325 + (x + 5) / 324 + (x + 349) / 5 = 0
<=> (x + 2) / 327 +1+ (x + 3) / 326 +1+ (x + 4) / 325 +1+ (x + 5) / 324 +1+ (x + 349) / 5 -4 = 0
<=> (x+ 329)/327 + (x+ 329)/326 + (x+ 329)/325 + (x+ 329)//324 + (x+ 329)/5 =0
<=> (x+ 329).(1/327 + 1/ 326 + 1/325 + 1/324 +1/5) =0
Do (1/327 + 1/ 326 + 1/325 + 1/324 +1/5) >0 nên x+ 329 =0 => x= -329
NN
24 tháng 6 2020
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)
\(\Leftrightarrow\frac{x+2+327}{327}+\frac{x+3+326}{326}+\frac{x+4+325}{325}+\frac{x+5+324}{324}+\frac{x+349-20}{5}=0\)
\(\Leftrightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Vì \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}>0\)
\(\Rightarrow x+329=0\)\(\Leftrightarrow x=-329\)
Vậy \(x=-329\)
\(\frac{x+5}{x+3}
x+5
x+3