a) 13-[7-(x+8)]=11         

    [7-(x+8)]=13-11

    [7-(x+8)]=2

    (x+8)=7-2

    (x+8)=5

     x=5-8

     x=-3

a: =>2x-1=4 hoặc 2x-1=-4

=>2x=5 hoặc 2x=-3

=>x=5/2 hoặc x=-3/2

d: =>x=|2|=2

e: \(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\x-y=0\end{matrix}\right.\Rightarrow x=y=1\)

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 \(x\ge\frac{-7}{3}\)

e, ta có \(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x^2}{9}=\frac{y^2}{4}\)

AĐTCTSBN ta có \(\frac{x^2}{9}=\frac{y^2}{4}=\frac{x^2+y^2}{9+4}=\frac{52}{13}=4\)

\(\Rightarrow\hept{\begin{cases}x=2\cdot3=6\\y=2\cdot2=4\end{cases}}\)

a) Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{x}{7}=\frac{y}{4}=\frac{x-y}{7-4}=\frac{30}{3}=10\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{7}=10\Leftrightarrow x=70\\\frac{y}{4}=10\Leftrightarrow y=40\end{cases}}\)

b,\(\frac{x^2}{9}=\frac{y}{4}\)

=\(\frac{x^2+y}{9+4}\)=\(\frac{100}{13}\)

a) x + 9 = -7

x = -7 – 9

x = -16

b) x – 4 = -8

x = -8 + 4

x = -4

c) | x | - 2 = 5

| x | = 5 + 2

| x | = 7

x = ±7

d) | x – 2 | = 5

x – 2 = 5 hoặc x – 2 = -5

x = 5 + 2 hoặc x = -5 + 2

x = 7 hoặc x = -3

a) ( x - 266 ): 5 = 7 

    ( x - 266 )     = 7 x 5

    ( x - 266 )     =  35     

      x                 = 35 + 266

      x                 = 301

b) ( x - 129 ) x 4 = 36

    ( x - 129 )       = 36 : 4

    ( x - 129 )       = 9

      x                   = 9 + 129

      x                   = 138

c) 786 - x = 5 x 4 : 2

    786 - x = 20 : 2

    786 - x = 10

             x = 786 - 10

             x = 776

d) x + 24 : 4 x 3 = 285

    x + 6 x 3        = 285

    x + 18            = 285

    x                    = 285 - 18 

    x                    = 267