\(x=1\)

\(\frac{9+x}{13-x}=\frac{5}{6}\)

<=> (9+x).6 = (13 - x).5

<=> 54 + 6x =  65 - x.5

<=> 11x        = 65 - 54

<=> x            = 11 : 11

<=> x            = 1

\(\frac{9+x}{13-x}=\frac{5}{6}\)

=> 6(9 + x) = 5(13 - x)

=> 54 + 6x = 65 - 5x

=> 54 - 65 = - 5x + 6x

=> - 9 = - 1x

=> x = 9

vậy_

bằng 2

đáp án đúng = 2

ms làm hôm nay xong

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 \(x\ge\frac{-7}{3}\)

\(\frac{x+9}{13-x}=\frac{6}{5}\)

\(\Rightarrow\orbr{\begin{cases}x+9=6\\13-x=5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-3\\x=8\end{cases}}\)

\(\frac{120-x}{30}=3\frac{1}{2}\)

\(\Rightarrow\frac{120-x}{30}=\frac{7}{2}\)

\(\Rightarrow\frac{120-x}{30}=\frac{105}{30}\)

\(\Rightarrow120-x=105\)

\(\Rightarrow x=120-105\)

\(\Rightarrow x=15\)

(x+2)/17+(x+4)/15+(x+6)/13=(x+8)/11+(x+10)/9+(x+12)/7

=>(x+2+17)/17+(x+4+15)/15+(x+6+13)/13=(x+8+11)/11+(x+10+9)/9+(x+12+7)/7

=>(x+19)/17+(x+19)/15+(x+19)/13=(x+19)/11+(x+19)/9+(x+19)/7

=>(x+19)/17+(x+19)/15+(x+19)/13-(x+19)/11-(x+19)/9-(x+19)/7=0

=>(x+19)*(1/17+1/15+1/13-1/11-1/9-1/7)=0

=>x+19=0

=>x=19

áp dụng tc tỉ lệ thức ta có :

\(\Leftrightarrow\frac{671x+2804}{3315}=\frac{239x+2462}{693}\Rightarrow\left(671x+2804\right)693=3315\left(239x+2462\right)\)

=>(671x+2804)693=693(671x+2804) (VT)

<=>693(671x+2804)=3315(239x+2462)

=>465003x+1943172=792285x+8161530

=>-327282x=621835

=>x=621835:(-327282)

=>x=-19

\(\frac{9+x}{13-x}=\frac{5}{6}\)

=> ( 9 + x ) .6 = ( 13 - x ) .5

=> 9.6 + 6x = 13.5 - 5x

=> 54 + 6x = 65 - 5x

=> 6x + 5x = 65 - 54

=> 11x = 11

=> x = 1

54 + 6x=65 -5x

<=> 11=11x

<=> x=1

x=11 nha

vế trái có bị sai ko em. chị nghĩ câu hỏi nên là: x+2/17 + x+4/15 + x+6/13=x+8/11 + x+10/9 + x+12/7