Cả nhà ơi giúp e nhé dùng cách nhân liên hợp Giúp e nhanh nhé e đang cần gấp T.T $x(x+5)=3\sqrt[3]{x^2+5x+2}-4$ 31/07/2021 Bởi Ivy Cả nhà ơi giúp e nhé dùng cách nhân liên hợp Giúp e nhanh nhé e đang cần gấp T.T $x(x+5)=3\sqrt[3]{x^2+5x+2}-4$
Đáp án: $\begin{array}{l}x\left( {x + 5} \right) = 3\sqrt[3]{{{x^2} + 5x + 2}} – 4\\ \Rightarrow {x^2} + 5x + 2 = 3\sqrt[3]{{{x^2} + 5x + 2}} + 2 – 4\\ \Rightarrow {\left( {\sqrt[3]{{{x^2} + 5x + 2}}} \right)^3} – 3\sqrt[3]{{{x^2} + 5x + 2}} + 2 = 0\\Dat:\sqrt[3]{{{x^2} + 5x + 2}} = a\\ \Rightarrow {a^3} – 3a + 2 = 0\\ \Rightarrow {a^3} – {a^2} + {a^2} – a – 2a + 2 = 0\\ \Rightarrow \left( {a – 1} \right)\left( {{a^2} + a – 2} \right) = 0\\ \Rightarrow \left( {a – 1} \right)\left( {a – 1} \right)\left( {a + 2} \right) = 0\\ \Rightarrow {\left( {a – 1} \right)^2}\left( {a + 2} \right) = 0\\ \Rightarrow \left[ \begin{array}{l}a = 1\\a = – 2\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}\sqrt[3]{{{x^2} + 5x + 2}} = 1\\\sqrt[3]{{{x^2} + 5x + 2}} = – 2\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}{x^2} + 5x + 2 = 1\\{x^2} + 5x + 2 = – 8\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}{x^2} + 5x + 1 = 0\\{x^2} + 5x + 10 = 0\left( {vo\,nghiem} \right)\end{array} \right.\\ \Rightarrow {x^2} + 2.5x + \dfrac{{25}}{4} – \dfrac{{21}}{4} = 0\\ \Rightarrow {\left( {x + \dfrac{5}{2}} \right)^2} = \dfrac{{21}}{4}\\ \Rightarrow x = \dfrac{{ – 5 \pm \sqrt {21} }}{4}\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
x\left( {x + 5} \right) = 3\sqrt[3]{{{x^2} + 5x + 2}} – 4\\
\Rightarrow {x^2} + 5x + 2 = 3\sqrt[3]{{{x^2} + 5x + 2}} + 2 – 4\\
\Rightarrow {\left( {\sqrt[3]{{{x^2} + 5x + 2}}} \right)^3} – 3\sqrt[3]{{{x^2} + 5x + 2}} + 2 = 0\\
Dat:\sqrt[3]{{{x^2} + 5x + 2}} = a\\
\Rightarrow {a^3} – 3a + 2 = 0\\
\Rightarrow {a^3} – {a^2} + {a^2} – a – 2a + 2 = 0\\
\Rightarrow \left( {a – 1} \right)\left( {{a^2} + a – 2} \right) = 0\\
\Rightarrow \left( {a – 1} \right)\left( {a – 1} \right)\left( {a + 2} \right) = 0\\
\Rightarrow {\left( {a – 1} \right)^2}\left( {a + 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
a = 1\\
a = – 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\sqrt[3]{{{x^2} + 5x + 2}} = 1\\
\sqrt[3]{{{x^2} + 5x + 2}} = – 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
{x^2} + 5x + 2 = 1\\
{x^2} + 5x + 2 = – 8
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
{x^2} + 5x + 1 = 0\\
{x^2} + 5x + 10 = 0\left( {vo\,nghiem} \right)
\end{array} \right.\\
\Rightarrow {x^2} + 2.5x + \dfrac{{25}}{4} – \dfrac{{21}}{4} = 0\\
\Rightarrow {\left( {x + \dfrac{5}{2}} \right)^2} = \dfrac{{21}}{4}\\
\Rightarrow x = \dfrac{{ – 5 \pm \sqrt {21} }}{4}
\end{array}$