$\sqrt{x^2+15}-4+3-\sqrt{x^2+8}=3(\sqrt[3]{x}-1)$=> $(x-1)(\frac{x+1}{\sqrt{x^2+15}+4}-\frac{x+1}{3+\sqrt{x^2+8}}-\frac{3}{\sqrt[3]{x^2}+\sqrt[3]{x}+1})$cái khúc đằng sau pt thửu thê :) $\frac{x+1}{\sqrt{x^2+15}+4}-\frac{x+1}{3+\sqrt{x^2+8}}<4$ mà $\frac{3}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}\geq4$ nen pt vô nghiệm :))
$\sqrt{x^2+15}-4+3-\sqrt{x^2+8}=3(\sqrt[3]{x}-1)$=> $(x-1)(\frac{x+1}{\sqrt{x^2+15}+4}-\frac{x+1}{3+\sqrt{x^2+8}}-\frac{3}{\sqrt[3]{x^2}+\sqrt[3]{x}+1})$cái khúc đằng sau pt thửu thê :) $\frac{x+1}{\sqrt{x^2+15}+4}-\frac{x+1}{3+\sqrt{x^2+8}}\leq4$ mà $\frac{3}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}\geq4$ nen pt vô nghiệm :))
$\sqrt{x^2+15}-4+3-\sqrt{x^2+8}=3(\sqrt[3]{x}-1)$=> $(x-1)(\frac{x+1}{\sqrt{x^2+15}+4}-\frac{x+1}{3+\sqrt{x^2+8}}-\frac{3}{\sqrt[3]{x^2}+\sqrt[3]{x}+1})$cái khúc đằng sau pt thửu thê :) $\frac{x+1}{\sqrt{x^2+15}+4}-\frac{x+1}{3+\sqrt{x^2+8}}
&l
t;4$ mà $\frac{3}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}\geq4$ nen pt vô nghiệm :))