ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 821

\[\boxed{\mathbf{821}\mathbf{.}}\]

\[1)\ f(x) = \frac{2x^{2} - 3x + 1}{x + 1}\]

\[= \frac{2x^{2} + 4x - 4}{(x + 1)^{2}}.\]

\[2)\ f(x) = \frac{3x^{2} + 2x - 1}{2x + 1}\]

\[= \frac{6x^{2} + 6x + 4}{(2x + 1)^{2}}.\]

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

\[= \frac{- 2 - x}{2x\sqrt{x}} + \frac{2 + x}{2(2 - x)^{2}\sqrt{x}} =\]

\[= \frac{( - 2 - x)(2 - x)^{2} + (2 + x)x}{2x\sqrt{x} \bullet (2 - x)^{2}} =\]

\[= \frac{( - 2 - x)\left( 4 - 4x + x^{2} \right) + 2x + x^{2}}{2x\sqrt{x} \bullet (2 - x)^{2}} =\]

\[= \frac{- 8 + 8x - 2x^{2} - 4x + 4x^{2} - x^{3} + 2x + x^{2}}{2x\sqrt{x} \bullet (2 - x)^{2}} =\]

\[= \frac{- x^{3} + 3x^{2} + 6x - 8}{2x\sqrt{x} \bullet (2 - x)^{2}} =\]

\[= \frac{5x^{2} - x^{3} - 4x + 10x - 2x^{2} - 8}{2x\sqrt{x} \bullet (2 - x)^{2}} =\]

\[= \frac{(x + 2)(5x - x^{2} - 4)}{2x\sqrt{x} \bullet (2 - x)^{2}}.\]