ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 51

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\[1)\ \sqrt[3]{(x - 2)^{3}}\]

\[\textbf{а)}\ \ x \geq 2:\]

\[\sqrt[3]{(x - 2)^{3}} = x - 2.\]

\[\textbf{б)}\ x < 2:\]

\[\sqrt[3]{(x - 2)^{3}} = x - 2.\]

\[2)\ \sqrt{(3 - x)^{6}} = \sqrt{(3 - x)^{2 \bullet 3}} =\]

\[= |3 - x|^{3}\]

\[\textbf{а)}\ x \leq 3:\]

\[|3 - x|^{3} = (3 - x)^{3}.\]

\[\textbf{б)}\ x > 3:\]

\[|3 - x|^{3} = - (3 - x)^{3} = (x - 3)^{3}.\]

\[3)\ \sqrt[4]{(x + 6)^{4}} + \sqrt{(x - 3)^{2}} =\]

\[= |x + 6| + |x - 3|\]

\[\ - 1 < x < 2:\]

\[|x + 6| + |x - 3| =\]

\[= (x + 6) - (x - 3) =\]

\[= x + 6 - x + 3 = 9.\]

\[4)\ \sqrt[6]{(2x + 1)^{6}} - \sqrt[4]{(4 + x)^{4}} =\]

\[= |2x + 1| - |4 + x|\]

\[\ - 3 < x < - 1:\]

\[|2x + 1| - |4 + x| =\]

\[= - (2x + 1) - (4 + x) =\]

\[= - 2x - 1 - 4 - x = - 3x - 5.\]