`~rai~`
\(e)\sqrt{x^2-4x+5}+\sqrt{9y^2-6y+1}=1\quad(1)\\\text{Ta có:}\\+)x^2-4x+5\\=(x^2-4x+4)+1\\=(x-2)^2+1\ge 1\quad\forall x\\\Rightarrow \sqrt{(x-2)^2+1}\ge\sqrt{1}\quad\forall x\\\Rightarrow \sqrt{x^2-4x+5}\ge 1\quad\forall x\\+)9y^2-6y+1\\=(3y-1)^2\ge 0\quad\forall y\\\Rightarrow \sqrt{(3y-1)^2}\ge 0\quad\forall y\\\Rightarrow \sqrt{9y^2-6y+1}\ge 0\quad\forall y\\\Rightarrow \sqrt{x^2-4x+5}+\sqrt{9y^2-6y+1}\ge 1+0=1\\\text{mà theo (1),ta lại có:}\\\sqrt{x^2-4x+5}+\sqrt{9y^2-6y+1}=1\\\Rightarrow \begin{cases}(x-2)^2=0\\(3y-1)^2=0\end{cases}\\\Leftrightarrow \begin{cases}x-2=0\\3y-1=0\end{cases}\\\Leftrightarrow \begin{cases}x=2\\y=\dfrac{1}{3}.\end{cases}\\\text{Vậy (x;y)=}\left(2;\dfrac{1}{3}\right).\\f)\sqrt{x^2+2x+1}-\sqrt{4x^2-4x+1}=0\\\Leftrightarrow \sqrt{x^2+2x+1}=\sqrt{4x^2-4x+1}\\\Leftrightarrow (\sqrt{x^2+2x+1})^2=(\sqrt{4x^2-4x+1})^2\\\Leftrightarrow x^2+2x+1=4x^2-4x+1\\\Leftrightarrow 4x^2-4x+1-x^2-2x-1=0\\\Leftrightarrow 3x^2-6x=0\\\Leftrightarrow 3x(x-2)=0\\\Leftrightarrow \left[\begin{array}{I}3x=0\\x-2=0\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=0\\x=2.\end{array}\right.\\\text{Vậy phương trình có tập nghiệm là S=}\{0;2\}.\)
