a/
→\(3.9=\left(x-5\right).\left(x-5\right)^2\)
→\(27=\left(x-5\right)^3\)
→\(3^3=\left(x-5\right)^3\)
→\(3=x-5\)
→x=5+3
→x=8
b/
→\(3.\left(-24\right)=\left(x+1\right).12\)
→\(-72=12x+12\)
→\(-72-12=12x\)
→\(-84=12x\)
→x=-7
c/
→\(\left(-8\right).\left(-2\right)=\left(2x+1\right).\left(2x+1\right)\)
→\(16=\left(2x+1\right)^2\)
→\(4^2=\left(2x+1\right)^2\)
→\(4=2x+1\)
→2x=3
→x=\(\dfrac{3}{2}\)
d/
→\(2.18=\left(4-x\right).\left(4-x\right)\)
→\(36=\left(4-x\right)^2\)
→\(6^2=\left(4-x\right)^2\)
→\(6=4-x\)
→x=4-6
→x=-2
\(a,\dfrac{3}{x-5}=\dfrac{\left(x-5\right)^2}{9}\\ \Rightarrow\left(x-5\right)\left(x-5\right)^2=3.9\\ \Leftrightarrow\left(x-5\right)^3=27\\ \Leftrightarrow\left(x-5\right)^3=3^3\\ \Leftrightarrow x-5=3\\ \Leftrightarrow x=8\\ b,\dfrac{x+1}{3}=\dfrac{-24}{12}\\ \Leftrightarrow\dfrac{x+1}{3}=\dfrac{-6}{3}\\ \Leftrightarrow x+1=-6\\ \Leftrightarrow x=-7\\ c,\dfrac{2x+1}{-2}=\dfrac{-8}{2x+1}\\ \Leftrightarrow\left(2x+1\right)\left(2x+1\right)=\left(-2\right)\left(-8\right)\\ \Leftrightarrow\left(2x+1\right)^2=16\\ \Leftrightarrow\left[{}\begin{matrix}2x+1=4\\2x+1=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\\ d,\dfrac{2}{4-x}=\dfrac{4-x}{18}\\ \Leftrightarrow\left(4-x\right)\left(4-x\right)=2.18\\ \Leftrightarrow\left(4-x\right)^2=36\\ \Leftrightarrow\left[{}\begin{matrix}4-x=6\\4-x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=10\end{matrix}\right.\)
