a, \(\frac{23+x}{201-x}=\frac{3}{5}\)
\(\Rightarrow\left(23+x\right)5=3\left(201-x\right)\)
\(\Rightarrow115+5x=603-3x\)
\(\Rightarrow5x+3x=603-115\)
\(\Rightarrow8x=448\Rightarrow x=61\)
Vậy x = 81
\(x+20=\frac{5}{7}\left(3x-20\right)\)
\(\Leftrightarrow7x+140=15x-100\)
\(\Leftrightarrow15x-7x=140+100\)
\(\Leftrightarrow8x=240\Rightarrow x=30\)
Vậy x = 30
\(a,\frac{23+x}{201-x}=\frac{3}{5}\)
\(\frac{5.\left(23+x\right)}{5.\left(201-x\right)}=\frac{3.\left(201-x\right)}{5.\left(201-x\right)}\)
\(5.\left(23+x\right)=3.\left(201-x\right)\)
\(115+5x=603-3x\)
\(5x+3x=603-115\)
\(8x=488\)
\(x=61\)
\(b,x+20=\frac{5}{7}\left(3x-20\right)\)
\(\frac{7\left(x+20\right)}{7}=\frac{5\left(3x-20\right)}{7}\)
\(7x-140=15x-100\)
7x-15x=-100+140
-8x=40
x=-5
a, \(23.5+5x=603-3x\)
\(5x+3x=603-115\)
\(8x=498\)
\(x=\frac{498}{8}\)
