Упростите выражение [tex]( \frac{x \sqrt{x} -8}{x-4} - \frac{6 \sqrt{x} }{ \sqrt{x} +2} ):(1- \frac{4}{ \sqrt{x} +2}) [/tex]. Created by Vampire100. algebra-ru

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June 2022 1 15 Report

Упростите выражение [tex]( \frac{x \sqrt{x} -8}{x-4} - \frac{6 \sqrt{x} }{ \sqrt{x} +2} ):(1- \frac{4}{ \sqrt{x} +2}) [/tex]

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arsenlevadniy

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$(\frac{x\sqrt{x}-8}{x-4}-\frac{6\sqrt{x}}{\sqrt{x}+2}):(1-\frac{4}{\sqrt{x}+2})=(\frac{x\sqrt{x}-8}{(\sqrt{x})^2-4}-\frac{6\sqrt{x}}{\sqrt{x}+2}):(1-\frac{4}{\sqrt{x}+2})=\\=(\frac{x\sqrt{x}-8}{(\sqrt{x}-2)(\sqrt{x}+2)}-\frac{6\sqrt{x}}{\sqrt{x}+2}):\frac{\sqrt{x}+2-4}{\sqrt{x}+2}=\frac{x\sqrt{x}-8-6\sqrt{x}(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}:\frac{\sqrt{x}-2}{\sqrt{x}+2}=$
$=\frac{x\sqrt{x}-8-6x+12\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\cdot\frac{\sqrt{x}+2}{\sqrt{x}-2}=\frac{(\sqrt{x})^3-3\cdot2(\sqrt{x})^2+3\cdot2^2\sqrt{x}-2^3}{(\sqrt{x}-2)^2}=\frac{(\sqrt{x}-2)^3}{(\sqrt{x}-2)^2}=\sqrt{x}-2$

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