\(\begin{array}{l} {\left( {{x^2} + x} \right)^2} + 4\left( {{x^2} + x} \right) = 12\\ \Leftrightarrow {\left( {{x^2} + x} \right)^2} + 2\left( {{x^2} + x} \right).2 + {2^2} = 12 + 4\\ \Leftrightarrow {\left( {{x^2} + x + 2} \right)^2} = 16\\ \Leftrightarrow \left[ \begin{array}{l} {x^2} + x + 2 = 4\\ {x^2} + x + 2 = - 4 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} {x^2} + x - 2 = 0 \Leftrightarrow \left[ \begin{array}{l} x = 1\\ x = - 2 \end{array} \right.\\ {x^2} + x + 6 = 0\left( {VN} \right) \end{array} \right. \end{array}\)
b) \(x-\sqrt{2}+3.\left(x^2-2\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)+3.\left[x^2-\left(\sqrt{2}\right)^2\right]=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)+3.\left(x-\sqrt{2}\right).\left(x+\sqrt{2}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right).\left(1+3+x+\sqrt{2}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right).\left(4+x+\sqrt{2}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right).\left(x+4+\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{2}=0\\x+4+\sqrt{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0+\sqrt{2}\\x=0-4-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-4-\sqrt{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{\sqrt{2};-4-\sqrt{2}\right\}.\)
Chúc bạn học tốt!
\(\begin{array}{l} x - \sqrt 2 + 3\left( {{x^2} - 2} \right) = 0\\ \Leftrightarrow \left( {x - \sqrt 2 } \right) + 3\left( {x - \sqrt 2 } \right)\left( {x + \sqrt 2 } \right) = 0\\ \Leftrightarrow \left( {x - \sqrt 2 } \right)\left[ {1 + 3\left( {x + \sqrt 2 } \right)} \right] = 0\\ \Leftrightarrow \left[ \begin{array}{l} x - \sqrt 2 = 0\\ 1 + 3\left( {x + \sqrt 2 } \right) = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \sqrt 2 \\ x + \sqrt 2 = - \dfrac{1}{3} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \sqrt 2 \\ x = - \dfrac{{1 - 3\sqrt 2 }}{3} \end{array} \right. \end{array}\)
\(\begin{array}{l} x - \sqrt 2 + 3\left( {{x^2} - 2} \right) = 0\\ \Leftrightarrow \left( {x - \sqrt 2 } \right) + 3\left( {x - \sqrt 2 } \right)\left( {x + \sqrt 2 } \right) = 0\\ \Leftrightarrow \left( {x - \sqrt 2 } \right)\left[ {1 + 3\left( {x + \sqrt 2 } \right)} \right] = 0\\ \Leftrightarrow \left[ \begin{array}{l} x - \sqrt 2 = 0\\ 1 + 3\left( {x + \sqrt 2 } \right) = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \sqrt 2 \\ x + \sqrt 2 = - \dfrac{1}{3} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \sqrt 2 \\ x = \dfrac{{-1 - 3\sqrt 2 }}{3} \end{array} \right. \end{array}\)
