Toán Lớp 10: Giúp mik với mọi người ơi
x ²+ √(x+5) -5 =0
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Toán Lớp 10: Giúp mik với mọi người ơi
x ²+ √(x+5) -5 =0
Comments ( 1 )
PT \Leftrightarrow {\left( {{t^2} – 5} \right)^2} + t – 5 = 0\\
\Leftrightarrow {t^4} – 10{t^2} + 25 + t – 5 = 0\\
\Leftrightarrow {t^4} – 10{t^2} + t + 20 = 0\\
\Leftrightarrow {t^2}\left( {{t^2} – t – 4} \right) + t\left( {{t^2} – t – 4} \right) – 5\left( {{t^2} – t – 4} \right) = 0\\
\Leftrightarrow \left( {{t^2} – t – 4} \right)\left( {{t^2} + t – 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
t = \dfrac{{1 + \sqrt {17} }}{2}(tm)\\
t = \dfrac{{1 – \sqrt {17} }}{2}(L)\\
t = \dfrac{{ – 1 + \sqrt {21} }}{2}(tm)\\
t = \dfrac{{ – 1 – \sqrt {21} }}{2}(L)
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\sqrt {x + 5} = \dfrac{{1 + \sqrt {17} }}{2}\\
\sqrt {x + 5} = \dfrac{{\sqrt {21} – 1}}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{{{{\left( {1 + \sqrt {17} } \right)}^2}}}{4} – 5 = \dfrac{{\sqrt {17} – 1}}{2}\\
x = \dfrac{{{{\left( {\sqrt {21} – 1} \right)}^2}}}{4} – 5 = \dfrac{{1 – \sqrt {21} }}{2}
\end{array} \right.\\
\Rightarrow S = \left\{ {\dfrac{{\sqrt {17} – 1}}{2};\dfrac{{1 – \sqrt {21} }}{2}} \right\}
\end{array}$