Đk:\(x\ge-4\)
\(pt\Leftrightarrow\sqrt[3]{x-4}-1+\sqrt{x+4}-3=0\)
\(\Leftrightarrow\dfrac{x-4-1}{\sqrt[3]{\left(x-4\right)^2}+\sqrt[3]{x-4}+1}+\dfrac{x+4-9}{\sqrt{x+4}+3}=0\)
\(\Leftrightarrow\dfrac{x-5}{\sqrt[3]{\left(x-4\right)^2}+\sqrt[3]{x-4}+1}+\dfrac{x-5}{\sqrt{x+4}+3}=0\)
\(\Leftrightarrow\left(x-5\right)\left(\dfrac{1}{\sqrt[3]{\left(x-4\right)^2}+\sqrt[3]{x-4}+1}+\dfrac{1}{\sqrt{x+4}+3}\right)=0\)
Dễ thấy: \(\dfrac{1}{\sqrt[3]{\left(x-4\right)^2}+\sqrt[3]{x-4}+1}+\dfrac{1}{\sqrt{x+4}+3}=0\) vô nghiệm với \(x\ge-4\)
Nên \(x-5=0\Rightarrow x=5\) (thỏa)
P/s:æ coi hội đúng ko nhé :)
+ 1 cách cho vui nhộn nào:
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x-4}=a\\\sqrt{x+4}=b\left(b\ge0\right)\end{matrix}\right.\)
ta có hệ:\(\left\{{}\begin{matrix}a+b=4\\a^3-b^2=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=3\end{matrix}\right.\)
\(\Leftrightarrow x=5\)(tmđkxđ)
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