Question

Пожалуйста, решите 4,5,6 задание

Answer 1

$4)\ \ \dfrac{\sqrt{a^3}-a}{a-2a^{1/2}+1}=\dfrac{a\sqrt{a}-a}{(\sqrt{a}-1)^2}=\dfrac{a\, (\sqrt{a}-1)}{(\sqrt{a}-1)^2}=\dfrac{a}{\sqrt{a}-1}\\\\\\\star \ \ \sqrt{a^3}=\sqrt{a^2\cdot a}=a\sqrt{a}\ \ \star$

$5)\ \ \Big(2,3\Big)^{\sqrt[7]2}=\Big(\dfrac{23}{10}\Big)^{\sqrt[7]2}\ \ ,\ \ \Big(2\dfrac{2}{9}\Big)^{\sqrt[7]2}=\Big(\dfrac{20}{9}\Big)^{\sqrt[7]2}\\\\\\\dfrac{23}{10}>\dfrac{20}{9}\ \ \ \ \to \ \ \ \ \Big(\dfrac{23}{10}\Big)^{\sqrt[7]2}>\Big(\dfrac{20}{9}\Big)^{\sqrt[7]2}\ \ \ \to \ \ \ \ \Big(2,3\Big)^{\sqrt[7]2}>\Big(2\dfrac{2}{9}\Big)^{\sqrt[7]2}$

$b)\ \ \Big(\dfrac{3}{8}\Big)^{-2\sqrt3}=\Big(\dfrac{8}{3}\Big)^{2\sqrt3}>1\ ,\ \ tak\ kak\ \ \dfrac{8}{3}>1\\\\\\c)\ \ \sqrt[3]{26}=\sqrt[6]{26^2}=\sqrt[6]{676}\ \ ,\ \ \ \sqrt8=\sqrt[6]{8^3}=\sqrt[6]{512}\\\\676>512\ \ \ \to \ \ \ \sqrt[6]{676}>\sqrt[6]{512}\ \ \ \to \ \ \ \sqrt[3]{26}>\sqrt8$

$6)\ \ y=\sqrt[6]{6+0,5x}\\\\OOF:\ 6+0,5x\geq 0\ \ ,\ \ 0,5x\geq -6\ \ ,\ \ \ x\geq -12\\\\x\in [\, -12\, ;\, +\infty )$