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$\Big(\dfrac{8}{\sqrt{x}-1}+\dfrac{\sqrt[4]{x}+1}{\sqrt[4]{x}-1}-\dfrac{\sqrt[4]{x}+3}{\sqrt[4]{x}+1}\Big):\dfrac{3}{\sqrt{x}-1}=\\ \\ \\ =\Big(\dfrac{8}{\sqrt{x}-1}+\dfrac{(\sqrt[4]{x}+1)^2-(\sqrt[4]{x}+3)(\sqrt[4]{x}-1)}{(\sqrt[4]{x}-1)\cdot (\sqrt[4]{x}+1)}\Big)\cdot \dfrac{\sqrt{x}-1}{3}=\\ \\ \\=\Big(\dfrac{8}{\sqrt{x}-1}+\dfrac{\sqrt{x}+2\sqrt[4]{x}+1-(\sqrt{x}-\sqrt[4]{x}+3\sqrt[4]{x}-3)}{\sqrt{x}-1}\Big)\cdot \dfrac{\sqrt{x}-1}{3}=$

$=\Big(\dfrac{8}{\sqrt{x}-1}+\dfrac{\sqrt{x}+2\sqrt[4]{x}+1-\sqrt{x}+\sqrt[4]{x}-3\sqrt[4]{x}+3}{\sqrt{x}-1}\Big)\cdot \dfrac{\sqrt{x}-1}{3}=\\ \\ \\ =\Big(\dfrac{8}{\sqrt{x}-1}+\dfrac{4}{\sqrt{x}-1}\Big)\cdot \dfrac{\sqrt{x}-1}{3}=\\ \\ \\ =\dfrac{12}{\sqrt{x}-1}\cdot \dfrac{\sqrt{x}-1}{3}=4$

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Использованы формулы

$\sqrt[4]{x}^2=\sqrt{x}\\ \\(a+b)^2=a^2+2ab+b^2 \\ \\(a-b)(a+b)=a^2-b^2$