2x `2x+3=x^2+(x-1)`$\sqrt[3]{3x^3+3}$ `nhanh` `nhé` `mik` `cần` `gấp` `lắm` `ạ`

Đáp án:

$S = \left\{ { – 1;2} \right\}$

Giải thích các bước giải:

$\begin{array}{l}
2x + 3 = {x^2} + \left( {x – 1} \right)\sqrt[3]{{3{x^3} + 3}}\\
 \Leftrightarrow {x^2} – 2x – 3 + \left( {x – 1} \right)\sqrt[3]{{3{x^3} + 3}} = 0\\
 \Leftrightarrow \left( {x + 1} \right)\left( {x – 3} \right) + \left( {x – 1} \right)\sqrt[3]{{3\left( {{x^3} + 1} \right)}} = 0\\
 \Leftrightarrow \left( {x + 1} \right)\left( {x – 3} \right) + \left( {x – 1} \right)\sqrt[3]{{3\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}} = 0\\
 \Leftrightarrow \left( {x + 1} \right)\left( {x – 3} \right) + \left( {x – 1} \right)\sqrt[3]{{x + 1}}.\sqrt[3]{{3\left( {{x^2} – x + 1} \right)}} = 0\\
 \Leftrightarrow \sqrt[3]{{x + 1}}\left( {\left( {x – 3} \right){{\left( {\sqrt[3]{{x + 1}}} \right)}^2} + \left( {x – 1} \right)\sqrt[3]{{3\left( {{x^2} – x + 1} \right)}}} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt[3]{{x + 1}} = 0\\
\left( {x – 3} \right){\left( {\sqrt[3]{{x + 1}}} \right)^2} + \left( {x – 1} \right)\sqrt[3]{{3\left( {{x^2} – x + 1} \right)}} = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  – 1\\
\left( {x – 3} \right)\sqrt[3]{{{{\left( {x + 1} \right)}^2}}} = \left( {1 – x} \right)\sqrt[3]{{3\left( {{x^2} – x + 1} \right)}}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  – 1\\
{\left( {x – 3} \right)^3}{\left( {x + 1} \right)^2} = 3{\left( {1 – x} \right)^3}\left( {{x^2} – x + 1} \right)
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  – 1\\
{x^5} – 7{x^4} + 10{x^3} + 18{x^2} – 27x – 27 = 3\left( { – {x^5} + 4{x^4} – 7{x^3} + 7{x^2} – 4x + 1} \right)
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  – 1\\
4{x^5} – 19{x^4} + 31{x^3} – 3{x^2} – 15x – 30 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  – 1\\
\left( {x – 2} \right)\left( {4{x^4} – 11{x^3} + 9{x^2} + 15x + 15} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  – 1\\
x = 2\\
4{x^4} – 11{x^3} + 9{x^2} + 15x + 15 = 0\left( 1 \right)
\end{array} \right.
\end{array}$

Lại có:

$\begin{array}{l}
\left( 1 \right) \Leftrightarrow 16{x^4} – 44{x^3} + 36{x^2} + 60x + 60 = 0\\
 \Leftrightarrow \left( {16{x^4} – 8{x^2} + 1} \right) – 44{x^3} + 44{x^2} + 60x + 59 = 0\\
 \Leftrightarrow {\left( {4{x^2} – 1} \right)^2} – 11x\left( {4{x^2} – 1} \right) + 44{x^2} + 49x + 59 = 0\\
 \Leftrightarrow {\left( {4{x^2} – 1} \right)^2} – 2.\left( {4{x^2} – 1} \right).\dfrac{{11x}}{2} + \dfrac{{121{x^2}}}{4} + \dfrac{{55}}{4}{x^2} + 49x + 59 = 0\\
 \Leftrightarrow {\left( {4{x^2} – 1 – \dfrac{{11x}}{2}} \right)^2} + \dfrac{{55}}{4}\left( {{x^2} + \dfrac{{196}}{{55}}x + {{\left( {\dfrac{{98}}{{55}}} \right)}^2}} \right) + \dfrac{{844}}{{55}} = 0\\
 \Leftrightarrow {\left( {4{x^2} – \dfrac{{15}}{2}x – 1} \right)^2} + \dfrac{{55}}{4}{\left( {x + \dfrac{{98}}{{55}}} \right)^2} + \dfrac{{844}}{{55}} = 0\left( {vn} \right)
\end{array}$

Vậy phương trình có tập nghiệm là: $S = \left\{ { – 1;2} \right\}$