1)
x ∈ [3; +∞)
2)
т.к. квадрат любого числа есть число неотрицательное, то y ∈ (-∞; +∞)
3)
y ∈ (-∞; +2)
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Answers & Comments
1)![(x-3)^{\frac{1}{4} } =\sqrt[4]{x-3} \\ \\ x-3\geq 0\\ \\ x\geq 3](https://tex.z-dn.net/?f=%28x-3%29%5E%7B%5Cfrac%7B1%7D%7B4%7D%20%7D%20%3D%5Csqrt%5B4%5D%7Bx-3%7D%20%5C%5C%20%5C%5C%20x-3%5Cgeq%200%5C%5C%20%5C%5C%20x%5Cgeq%203 "(x-3)^{\frac{1}{4} } =\sqrt[4]{x-3} \\ \\ x-3\geq 0\\ \\ x\geq 3")
x ∈ [3; +∞)
2)![(y+3)^{}\frac{2}{5} =\sqrt[5]{(y+3)^{2}}](https://tex.z-dn.net/?f=%28y%2B3%29%5E%7B%7D%5Cfrac%7B2%7D%7B5%7D%20%3D%5Csqrt%5B5%5D%7B%28y%2B3%29%5E%7B2%7D%7D "(y+3)^{}\frac{2}{5} =\sqrt[5]{(y+3)^{2}}")
т.к. квадрат любого числа есть число неотрицательное, то y ∈ (-∞; +∞)
3)![(4-2y)^{-\frac{1}{6}} =\frac{1}{\sqrt[6]{4-2y}} \\ \\ 4-2y>0\\ \\ -2y>-4\\ \\ y](https://tex.z-dn.net/?f=%284-2y%29%5E%7B-%5Cfrac%7B1%7D%7B6%7D%7D%20%3D%5Cfrac%7B1%7D%7B%5Csqrt%5B6%5D%7B4-2y%7D%7D%20%20%5C%5C%20%5C%5C%204-2y%3E0%5C%5C%20%5C%5C%20-2y%3E-4%5C%5C%20%5C%5C%20y%3C2 "(4-2y)^{-\frac{1}{6}} =\frac{1}{\sqrt[6]{4-2y}} \\ \\ 4-2y>0\\ \\ -2y>-4\\ \\ y")
y ∈ (-∞; +2)
Verified answer
1)![(x-3)^ \frac{1}{4} = \sqrt[4]{x-3}](https://tex.z-dn.net/?f=%28x-3%29%5E%20%5Cfrac%7B1%7D%7B4%7D%20%3D%20%5Csqrt%5B4%5D%7Bx-3%7D "(x-3)^ \frac{1}{4} = \sqrt[4]{x-3}")
2)![(y+3)^{ \frac{2}{5} = \sqrt[5]{(y+3)}^{2}](https://tex.z-dn.net/?f=%28y%2B3%29%5E%7B%20%5Cfrac%7B2%7D%7B5%7D%20%3D%20%5Csqrt%5B5%5D%7B%28y%2B3%29%7D%5E%7B2%7D "(y+3)^{ \frac{2}{5} = \sqrt[5]{(y+3)}^{2}")
3)![(4-2y)^{- \frac{1}{6} = \\ \frac{1}{\sqrt[6]{4-2y}}](https://tex.z-dn.net/?f=%284-2y%29%5E%7B-%20%5Cfrac%7B1%7D%7B6%7D%20%3D%20%5C%5C%20%5Cfrac%7B1%7D%7B%5Csqrt%5B6%5D%7B4-2y%7D%7D "(4-2y)^{- \frac{1}{6} = \\ \frac{1}{\sqrt[6]{4-2y}}")