a) \(x^2+6x+9=144\)
\(\Leftrightarrow\left(x+3\right)^2=12^2\)
\(\Leftrightarrow x+3=12\)
\(\Leftrightarrow x=9\)
\(\text{a) }x^2+6x+9=144\\ \Leftrightarrow\left(x^2+6x+9\right)-144=0\\ \Leftrightarrow\left(x+3\right)^2-12^2=0\\ \Leftrightarrow\left(x+3+12\right)\left(x+3-12\right)=0\\ \Leftrightarrow\left(x+15\right)\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+15=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-15\\x=9\end{matrix}\right.\)
Vậy tập nghiệm phương trình là \(S=\left\{9;-15\right\}\)
\(\dfrac{x-19}{1999}+\dfrac{x-23}{1995}+\dfrac{x+82}{700}=5\\ \Leftrightarrow\left(\dfrac{x-19}{1999}-1\right)+\left(\dfrac{x-23}{1995}-1\right)+\left(\dfrac{x+82}{700}-3\right)=0\\ \Leftrightarrow\dfrac{x-2018}{1999}+\dfrac{x-2018}{1995}+\dfrac{x-2018}{700}=0\\ \Leftrightarrow\left(x-2018\right)\left(\dfrac{1}{1999}+\dfrac{1}{1995}+\dfrac{1}{700}\right)=0\\ \Leftrightarrow x-2018=0\left(\text{Vì }\dfrac{1}{1999}+\dfrac{1}{1995}+\dfrac{1}{700}\ne0\right)\\ \Leftrightarrow x=2018\)
Vậy nghiệm của phương trình là \(x=2018\)
\(\text{c) }x^3-3x^2+4=0\\ \Leftrightarrow x^3-2x^2-x^2+4=0\\ \Leftrightarrow\left(x^3-2x^2\right)-\left(x^2-4\right)=0\\ \Leftrightarrow x^2\left(x-2\right)-\left(x+2\right)\left(x-2\right)=0\\ \Leftrightarrow\left(x^2-x-2\right)\left(x-2\right)=0\\ \Leftrightarrow\left(x^2-2x+x-2\right)\left(x-2\right)=0\\ \Leftrightarrow\left[\left(x^2-2x\right)+\left(x-2\right)\right]\left(x-2\right)=0\\ \Leftrightarrow\left[x\left(x-2\right)+\left(x-2\right)\right]\left(x-2\right)=0\\\Leftrightarrow \left(x+2\right)\left(x-2\right)^2=0\\\Leftrightarrow\left[{}\begin{matrix}x+2=0\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right. \)
Vậy tập nghiệm phương trình là \(S=\left\{-2;2\right\}\)
