Số nghiệm của phương trình \(\sqrt[3]{{x + 1}} + \...

A

- Hướng dẫn giải

Phương pháp giải:

\(\begin{array}{l}+ \,\,\,\sqrt[3]{{f(x)}} + \sqrt[3]{{g(x)}} = \sqrt[3]{{h(x)}} + \sqrt[3]{{k(x)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\\Leftrightarrow {\left( {\sqrt[3]{{f(x)}} + \sqrt[3]{{g(x)}}} \right)^3} = {\left( {\sqrt[3]{{h(x)}} + \sqrt[3]{{k(x)}}} \right)^3}\\\Leftrightarrow f(x) + g(x) + 3\sqrt[3]{{f(x).g(x)}}\left( {\sqrt[3]{{f(x)}} + \sqrt[3]{{g(x)}}} \right) = h(x) + k(x) + 3\sqrt[3]{{h(x).k(x)}}\left( {\sqrt[3]{{h(x)}} + \sqrt[3]{{k(x)}}} \right)\,\,\,\,\,\,\,\,\,\left( 2 \right)\end{array}\)

+  Thay (1) vào (2) ta giải phương trình ta tìm được x.

Giải chi tiết:

Ta có: 

\[\begin{array}{l}\sqrt[3]{{x + 1}} + \sqrt[3]{{2{x^2} + 1}} = \sqrt[3]{{x + 2}} + \sqrt[3]{{2{x^2}}} \Leftrightarrow {\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{2{x^2} + 1}}} \right)^3} = {\left( {\sqrt[3]{{x + 2}} + \sqrt[3]{{2{x^2}}}} \right)^3}\\ \Leftrightarrow x + 1 + 2{{\rm{x}}^2} + 1 + 3\sqrt[3]{{\left( {x + 1} \right)\left( {2{{\rm{x}}^2} + 1} \right)}}\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{2{{\rm{x}}^2} + 1}}} \right) = x + 2 + 2{{\rm{x}}^2} + 3\sqrt[3]{{\left( {x + 2} \right)2{{\rm{x}}^2}}}\left( {\sqrt[3]{{x + 2}} + \sqrt[3]{{2{{\rm{x}}^2}}}} \right)\\ \Leftrightarrow 3\sqrt[3]{{\left( {x + 1} \right)\left( {2{{\rm{x}}^2} + 1} \right)}}\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{2{{\rm{x}}^2} + 1}}} \right) = 3\sqrt[3]{{\left( {x + 2} \right)2{{\rm{x}}^2}}}\left( {\sqrt[3]{{x + 2}} + \sqrt[3]{{2{{\rm{x}}^2}}}} \right)\\ \Leftrightarrow \sqrt[3]{{\left( {x + 1} \right)\left( {2{{\rm{x}}^2} + 1} \right)}}\left( {\sqrt[3]{{x + 2}} + \sqrt[3]{{2{{\rm{x}}^2}}}} \right) = \sqrt[3]{{\left( {x + 2} \right)2{{\rm{x}}^2}}}\left( {\sqrt[3]{{x + 2}} + \sqrt[3]{{2{{\rm{x}}^2}}}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}\left( {\sqrt[3]{{x + 2}} + \sqrt[3]{{2{{\rm{x}}^2}}}} \right) = 0\\\left( {\sqrt[3]{{\left( {x + 2} \right)2{{\rm{x}}^2}}} - \sqrt[3]{{\left( {x + 1} \right)\left( {2{{\rm{x}}^2} + 1} \right)}}} \right) = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\sqrt[3]{{x + 2}} = - \sqrt[3]{{2{{\rm{x}}^2}}}\\\sqrt[3]{{\left( {x + 2} \right)2{{\rm{x}}^2}}} = \sqrt[3]{{\left( {x + 1} \right)\left( {2{{\rm{x}}^2} + 1} \right)}}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x + 2 = - 2{{\rm{x}}^2}\\2{{\rm{x}}^3} + 4{{\rm{x}}^2} = 2{{\rm{x}}^3} + 2{{\rm{x}}^2} + x + 1\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}2{{\rm{x}}^2} + x + 2 = 0\\2{{\rm{x}}^2} - x - 1 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 1\\x = - \frac{1}{2}\end{array} \right.\end{array}\]

(Vì phương trình:\(2{{\text{x}}^{2}}+x+2=0\) có \(\Delta\) < 0 nên vô nghiệm)

Vậy phương trình có hai nghiệm phân biệt

Chọn A.