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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 \(x\ge\frac{-7}{3}\)

\(\frac{2}{5}+\frac{3}{5}:x=\frac{9}{11}=>\frac{3}{5}:x=\frac{9}{11}-\frac{2}{5}=\frac{23}{55}=>x=\frac{3}{5}:\frac{23}{55}=\frac{3}{5}.\frac{55}{23}=\frac{33}{23}\)

a) | x+5| - 6 = 9
|x+5| = 15
TH1: x + 5 = 15 => x = 10
TH2: x + 5 = -15 => x = -20
KL:...
phần b bn làm tương tự nha
a) |x + 5| - 6 = 9
|x + 5| = 9 + 6
|x + 5| = 15
=> x + 5 = \(\pm\) 15
TH1: x + 5 = 15
x = 15 - 5
x = 10
TH2: x + 5 = -15
x = -15 - 5
x = -20
Vậy x \(\in\){10; -20}
b) |x - \(\frac{2}{5}\)| + \(\frac{3}{4}\)= \(\frac{11}{4}\)
|x - \(\frac{2}{5}\)| = \(\frac{11}{4}\)- \(\frac{3}{4}\)
|x - \(\frac{2}{5}\)| = 2
=> x - \(\frac{2}{5}\)= \(\pm\)2
TH1: x - \(\frac{2}{5}\)= 2
x = 2 + \(\frac{2}{5}\)
x = \(\frac{12}{5}\)
TH2: x - \(\frac{2}{5}\)= -2
x = -2 + \(\frac{2}{5}\)
x = \(\frac{-8}{5}\)
Vậy x \(\in\){\(\frac{12}{5}\); \(\frac{-8}{5}\)}

\(3x.\left(x-\frac{2}{3}\right)=0\)
\(\Leftrightarrow3x=0\)hoặc \(x-\frac{2}{3}=0\)
\(3x=0\Rightarrow x=0\)
\(x-\frac{2}{3}=0\Rightarrow x=0+\frac{2}{3}=\frac{2}{3}\)
Vậy..
\(\frac{33}{23}\)