Question

\(\sqrt{x+15}+\sqrt{x+8}=4\sqrt[3]{x}-3\)

Answer 1

Lời giải:
ĐKXĐ: $x\geq -8$

$4\sqrt[3]{x}=\sqrt{x+15}+\sqrt{x+8}+3\geq \sqrt{-8+15}+0+3>4$

$\Rightarrow x>1$

$4\sqrt[3]{x}=\sqrt{x+15}+\sqrt{x+8}+3>\sqrt{1+15}+\sqrt{1+8}+3=10$

$\Rightarrow x>15$

$4\sqrt[3]{x}=\sqrt{x+15}+\sqrt{x+8}+3>\sqrt{15+15}+\sqrt{15+8}+3>12$

$\Rightarrow \sqrt[3]{x}>3\Rightarrow x>27$

$4\sqrt[3]{x}=\sqrt{x+15}+\sqrt{x+8}+3> \sqrt{27+15}+\sqrt{27+8}+3>16$

$\Rightarrow \sqrt[3]{x}>4$

$\Rightarrow x>64$

Do đó:

$4\sqrt[3]{x}=\sqrt{x+15}+\sqrt{x+8}+3>2\sqrt{x}$

$=2\sqrt[6]{x}.\sqrt[3]{x}>2\sqrt[6]{64}.\sqrt[3]{x}=4\sqrt[3]{x}$ (vô lý)

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