1. Giải phương trình 3x3 – 7x2 + 6x + 4 = 3
1.
ĐKXĐ: \(\forall \)x Î R
3x3 – 7x2 + 6x + 4 = 3\(\sqrt[3]{{\frac{{16{x^2} + 6x + 2}}{3}}}\)
⇒3x3 + 9x2 + 12x + 6 = 16x2 + 6x + 2 + 3\(\sqrt[3]{{\frac{{16{x^2} + 6x + 2}}{3}}}\)
Đặt \(\sqrt[3]{{\frac{{16{x^2} + 6x + 2}}{3}}}\) = t , ta có:
3x3 + 9x2 + 12x + 6 = 3t3 + 3t
⇔ x3 + 3x2 + 4x + 2 = t3 + t
⇔ (x + 1)3 + (x + 1) = t3 + t
⇔ (x + 1 – t) [(x + 1)2 – (x + 1)t + t2 +1] = 0
Mà (x + 1)2 – (x + 1)t + t2 +1 > 0 Þ x + 1 – t = 0
⇒ x + 1 = t
⇔ x + 1 = \(\sqrt[3]{{\frac{{16{x^2} + 6x + 2}}{3}}}\)
⇔3x3 + 9x2 + 12x + 6 = 16x2 + 6x + 2
⇔3x3 – 7x2 + 3x + 1 = 0
⇔3x3 – 3x2 – 4x2 + 4x – x + 1 = 0
⇔ (x – 1)( 3x2 – 4x + 1) = 0
⇒\(\left[ {\begin{array}{*{20}{c}}{x = 1\;\;\;\;\;}\\{x = \frac{{2 + \sqrt 7 }}{3}}\\{x = \frac{{2 - \sqrt 7 }}{3}}\end{array}} \right.\)
Thử lại thấy cả 3 nghiệm thoả mãn.
Vậy x ∈\(\left\{ {1;\;\frac{{2 + \sqrt 7 }}{3};\;\frac{{2 - \sqrt 7 }}{3}\;} \right\}\)
\(\begin{array}{l}2.\left\{ \begin{array}{l}{x^2} + {y^2} + x + y = 8\left( 1 \right)\\2{x^2} + {y^2} - 3xy + 3x - 2y + 1 = 0\left( 2 \right)\end{array} \right.\\\left( 2 \right) \Leftrightarrow 2{x^2} + 3\left( {1 - y} \right)x + {\left( {y - 1} \right)^2} = 0 \Leftrightarrow \left( {x + 1 - y} \right)\left( {2x + 1 - y} \right) = 0\\ \Rightarrow \left[ \begin{array}{l}x = \frac{{y - 1}}{2}\\x = y - 1\end{array} \right. \Rightarrow \left[ \begin{array}{l}y = 2x + 1 \Rightarrow \left( 1 \right) \Leftrightarrow {x^2} + {\left( {2x + 1} \right)^2} + x + 2x + 1 = 8 \Leftrightarrow \left[ \begin{array}{l}x = - 2 \Rightarrow y = - 3\\x = \frac{3}{5} \Rightarrow y = \frac{{11}}{5}\end{array} \right.\\y = x + 1 \Rightarrow \left( 1 \right) \Leftrightarrow {x^2} + {\left( {x + 1} \right)^2} + x + x + 1 = 8 \Leftrightarrow 2{x^2} + 4x - 6 = 0 \Leftrightarrow \left[ \begin{array}{l}x = 1;y = 2\\x = - 3;y = - 2\end{array} \right.\end{array} \right.\\(x;y) \in \left\{ {\left( { - 2; - 3} \right);\left( {\frac{3}{5};\frac{{11}}{5}} \right);\left( {1;2} \right);\left( { - 3; - 2} \right)} \right\}\end{array}\)