tìm x
\(\frac{9+x}{13-x}\)= \(\frac{5}{6}\)
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
28 tháng 3 2018
\(\frac{9+x}{13-x}=\frac{5}{6}\)
=> ( 9 + x ) .6 = ( 13 - x ) .5
=> 9.6 + 6x = 13.5 - 5x
=> 54 + 6x = 65 - 5x
=> 6x + 5x = 65 - 54
=> 11x = 11
=> x = 1
NL
21 tháng 4 2017
\(\Leftrightarrow\)\(x-\left(\frac{13x}{18}-\frac{4}{18}\right)=\frac{4}{9}\)
\(\Leftrightarrow\)\(\frac{18x}{18}-\frac{13x}{18}+\frac{4}{18}=\frac{4}{9}\)
\(\Leftrightarrow\)\(\frac{5x}{18}=\frac{4}{9}-\frac{4}{18}\)
\(\Leftrightarrow\)\(\frac{5x}{18}=\frac{2}{9}\)
\(\Leftrightarrow\)\(5x=\frac{18.2}{9}\)
\(\Leftrightarrow\)\(5x=4\)
\(\Leftrightarrow\)\(x=\frac{4}{5}\)
VN
3
11 tháng 8 2018
\(\frac{9+x}{13-x}=\frac{5}{6}\)
<=> (9+x).6 = (13 - x).5
<=> 54 + 6x = 65 - x.5
<=> 11x = 65 - 54
<=> x = 11 : 11
<=> x = 1
28 tháng 9 2017
a) Ta có : \(\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}=\frac{x+5}{11}+\frac{x+5}{13}\)
\(\Rightarrow\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}-\left(\frac{x+5}{11}+\frac{x+5}{13}\right)=0\)
\(\Rightarrow\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}-\frac{x+5}{11}-\frac{x+5}{13}=0\)
\(\Rightarrow\left(x+5\right)\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}-\frac{1}{11}-\frac{1}{13}\right)=0\)
Do \(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}-\frac{1}{11}-\frac{1}{13}\ne0\)
\(\Rightarrow x+5=0\Rightarrow x=-5\)
Vậy x = -5
b) Ta có : \(\frac{x+2}{100}+\frac{x+3}{99}+\frac{x+4}{98}=\frac{x+5}{97}+\frac{x+6}{96}+\frac{x+7}{95}\)
\(\Rightarrow\frac{x+2}{100}+\frac{x+3}{99}+\frac{x+4}{98}+3=\frac{x+5}{97}+\frac{x+6}{96}+\frac{x+7}{95}+3\)
\(\Rightarrow\frac{x+2}{100}+1+\frac{x+3}{99}+1+\frac{x+4}{98}+1=\frac{x+5}{97}+1+\frac{x+6}{96}+1+\frac{x+7}{95}+1\)
\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}=\frac{x+102}{97}+\frac{x+102}{96}+\frac{x+102}{95}\)
\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}-\left(\frac{x+102}{97}+\frac{x+102}{96}+\frac{x+102}{95}\right)=0\)
\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}-\frac{x+102}{97}-\frac{x+102}{96}-\frac{x+102}{95}\)
\(\Rightarrow\left(x+102\right)\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
Do \(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\)
\(\Rightarrow x+102=0\Rightarrow x=-102\)
Vậy x = -102
c) Ta có : (x + 2) - (x + 3) = x + 2 - x - 3
= x - x + 2 - 3
= -1
mà (x + 2) - (x + 3) > 0 => không tồn tại x sao cho (x + 2) - (x + 3) > 0
d) Ta có : \(\left(x-5\right)\left(x+\frac{7}{3}\right)\ge0\)
\(\Rightarrow\orbr{\begin{cases}x\ge5\\x\ge\frac{-7}{3}\end{cases}}\)
\(\Rightarrow x\ge\frac{-7}{3}\)
Vậy \(x\ge\frac{-7}{3}\)
tổng không đổi,tổng lúc đầu là:13+9=22
9+x là:22:(5+6)x5=10
vậy x=1
ta có
\(\frac{9+X}{13-X}\)=\(\frac{5}{6}\)\(\Leftrightarrow\)6(9+X)=5(13-x)\(\Leftrightarrow\)54+6x=65-5x\(\Leftrightarrow\)11x=11\(\Leftrightarrow\)x=1