Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 156

\[\boxed{\mathbf{156.}}\]

\[1)\ \sqrt{2x - 34} = 1 + \sqrt{x}\]

\[2x - 34 = \left( 1 + \sqrt{x} \right)^{2}\]

\[2x - 34 = 1 + 2\sqrt{x} + x\]

\[x - 35 = 2\sqrt{x}\]

\[(x - 35)^{2} = 4x\]

\[x^{2} - 70x + 1225 - 4x = 0\]

\[x^{2} - 74x + 1225 = 0\]

\[D = 74^{2} - 4 \bullet 1225 =\]

\[= 5476 - 4900 = 576\]

\[x_{1} = \frac{74 - 24}{2} = 25;\text{\ \ }\]

\[x_{2} = \frac{74 + 24}{2} = 49.\]

\[Проверим:\]

\[\sqrt{2 \bullet 25 - 34} - \sqrt{25} =\]

\[= \sqrt{50 - 34} - 5 = \sqrt{16} - 5 =\]

\[= 4 - 5 = - 1\]

\[\sqrt{2 \bullet 49 - 34} - \sqrt{49} =\]

\[= \sqrt{98 - 34} - 7 = \sqrt{64} - 7 =\]

\[= 8 - 7 = 1\]

\[Ответ:\ \ x = 49.\]

\[2)\ \sqrt{5x} + \sqrt{14 - x} = 8\]

\[\left( \sqrt{5x} + \sqrt{14 - x} \right)^{2} = 64\]

\[5x + 2\sqrt{5x(14 - x)} + 14 - x =\]

\[= 64\]

\[2\sqrt{5x(14 - x)} = 50 - 4x\]

\[4\left( 70x - 5x^{2} \right) = (50 - 4x)^{2}\]

\[280x - 20x^{2} =\]

\[= 2500 - 400x + 16x^{2}\]

\[36x^{2} - 680x + 2500 = 0\]

\[9x^{2} - 170x + 625 = 0\]

\[D = 170^{2} - 4 \bullet 9 \bullet 625 =\]

\[= 28\ 900 - 22\ 500 = 6400\]

\[x_{1} = \frac{170 - 80}{2 \bullet 9} = \frac{90}{18} = 5;\text{\ \ }\]

\[x_{2} = \frac{170 + 80}{2 \bullet 9} = \frac{250}{18} = 13\frac{8}{9}.\]

\[Проверим:\]

\[\sqrt{5 \bullet 5} + \sqrt{14 - 5} = \sqrt{25} + \sqrt{9} =\]

\[= 5 + 3 = 8;\]

\[\sqrt{5 \bullet \frac{250}{18}} + \sqrt{14 - \frac{250}{18}} =\]

\[= \sqrt{69,(4)} + \sqrt{\frac{1}{9}} > 8.\]

\[Ответ:\ \ x = 5.\]

\[3)\ \sqrt{15 + x} + \sqrt{3 + x} = 6\]

\[\left( \sqrt{15 + x} + \sqrt{3 + x} \right)^{2} = 36\]

\[2\sqrt{45 + 15x + 3x + x^{2}} =\]

\[= 18 - 2x\]

\[4\left( x^{2} + 18x + 45 \right) = (18 - 2x)^{2}\]

\[4x^{2} + 72x + 180 =\]

\[= 324 - 72x + 4x^{2}\]

\[72x + 72x = 324 - 180\]

\[144x = 144\]

\[x = 1.\]

\[Проверим:\]

\[\sqrt{15 + 1} + \sqrt{3 + 1} = \sqrt{16} + \sqrt{4} =\]

\[= 4 + 2 = 6.\]

\[Ответ:\ \ x = 1.\]

\[4)\ \sqrt{3 - 2x} - \sqrt{1 - x} = 1\]

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

\[3 - 3x = 2\sqrt{3 - 3x - 2x + 2x^{2}}\]

\[(3 - 3x)^{2} = 4\left( 2x^{2} - 5x + 3 \right)\]

\[9 - 18x + 9x^{2} = 8x^{2} - 20x + 12\]

\[x^{2} + 2x - 3 = 0\]

\[D = 2^{2} + 4 \bullet 3 = 4 + 12 = 16\]

\[x_{1} = \frac{- 2 - 4}{2} = - 3;\text{\ \ }\]

\[x_{2} = \frac{- 2 + 4}{2} = 1.\]

\[Проверим:\]

\[\sqrt{3 - 2 \bullet ( - 3)} - \sqrt{1 - ( - 3)} =\]

\[= \sqrt{3 + 6} - \sqrt{1 + 3} = \sqrt{9} - \sqrt{4} =\]

\[= 3 - 2 = 1;\]

\[\sqrt{3 - 2 \bullet 1} - \sqrt{1 - 1} =\]

\[= \sqrt{3 - 2} - \sqrt{0} = \sqrt{1} - 0 = 1.\]

\[Отввет:\ \ x_{1} = - 3;\ \ x_{2} = 1.\]