[Toán 8] Đại số khó

T

trinhminh18

a/Ta có:
$x^3=9+4\sqrt{5}+9-4\sqrt{5}+3\sqrt[3]{(9+4\sqrt{5})(9-4\sqrt{5})}(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}})$
\Leftrightarrow$x^3=18+3x$
\Leftrightarrow$x^3-3x-18=0$
b/$x^3-3x-18=0$ \Leftrightarrow$(x-3)(x^2+3x+6)=0$
\Rightarrow$x-3=0$ (vì $x^2+3x+6=(x+1,5)^2+3,75$>0)
\Rightarrow$x=3$
 
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