ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 995

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\[1)\ f(x) = (2x + 1) \bullet \sqrt{x} =\]

\[= 2x\sqrt{x} + \sqrt{x} = 2 \bullet x^{\frac{3}{2}} + x^{\frac{1}{2}};\]

\[F(x) = 2 \bullet x^{\frac{5}{2}}\ :\frac{5}{2} + x^{\frac{3}{2}}\ :\frac{3}{2} =\]

\[= \frac{4}{5}x^{2} \bullet \sqrt{x} + \frac{2}{3}x\sqrt{x} + C =\]

\[= 2x\sqrt{x} \bullet \left( \frac{2}{5}x + \frac{1}{3} \right) + C.\]

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

\[= 3x\sqrt[3]{x} - 2\sqrt[3]{x} = 3 \bullet x^{\frac{4}{3}} - 2 \bullet x^{\frac{1}{3}};\]

\[F(x) = 3 \bullet x^{\frac{7}{3}}\ :\frac{7}{3} - 2 \bullet x^{\frac{4}{3}}\ :\frac{4}{3} =\]

\[= \frac{9}{7}x^{2} \bullet \sqrt[3]{x} - \frac{3}{2}x\sqrt[3]{x} + C =\]

\[= 3x\sqrt[3]{x} \bullet \left( \frac{3}{7}x - \frac{1}{2} \right) + C.\]

\[3)\ f(x) = \frac{x + 4}{\sqrt[3]{x}} = \frac{x}{\sqrt[3]{x}} + \frac{4}{\sqrt[3]{x}} =\]

\[= x^{\frac{2}{3}} + 4 \bullet x^{- \frac{1}{3}};\]

\[F(x) = x^{\frac{5}{3}}\ :\frac{5}{3} + 4 \bullet x^{\frac{2}{3}}\ :\frac{2}{3} =\]

\[= \frac{3}{5}x \bullet \sqrt[3]{x^{2}} + 6\sqrt[3]{x^{2}} + C =\]

\[= 3\sqrt[3]{x^{2}} \bullet \left( \frac{1}{5}x + 2 \right) + C.\]

\[4)\ f(x) = \frac{x - 3}{\sqrt{x}} = \frac{x}{\sqrt{x}} - \frac{3}{\sqrt{x}} =\]

\[= x^{\frac{1}{2}} - 3 \bullet x^{- \frac{1}{2}};\]

\[F(x) = x^{\frac{3}{2}}\ :\frac{3}{2} - 3 \bullet x^{\frac{1}{2}}\ :\frac{1}{2} =\]

\[= \frac{2}{3}x\sqrt{x} - 6\sqrt{x} + C =\]

\[= 2\sqrt{x} \bullet \left( \frac{1}{3}x - 3 \right) + C.\]