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МатематикаРешебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 32
Задание 32
\[\boxed{\mathbf{32.}}\]
\[\textbf{а)}\ \sqrt{2x + 1} = 2\sqrt{x - 1} + 1\]
\[2x + 1 \geq 0\]
\[2x \geq - 1\]
\[x \geq - 0,5.\]
\[x - 1 \geq 0\]
\[x \geq 1.\]
\[M = \lbrack 1; + \infty).\]
\[\left( \sqrt{2x + 1} \right)^{2} = \left( 2\sqrt{x - 1} + 1 \right)^{2}\]
\[2x + 1 = 4(x - 1) + 4\sqrt{x - 1} + 1\]
\[2x + 1 = 4x - 4 + 1 + 4\sqrt{x - 1}\]
\[- 2x + 4 = 4\sqrt{x - 1}\]
\[2 - x = 2\sqrt{x - 1}\]
\[2 - x \geq 0\]
\[x \leq 2.\]
\[M = \lbrack 1;2\rbrack.\]
\[(2 - x)^{2} = \left( 2\sqrt{x - 1} \right)^{2}\]
\[4 - 4x + x^{2} = 4(x - 1)\]
\[x^{2} - 4x + 4 = 4x - 4\]
\[x^{2} - 8x + 8 = 0\]
\[D_{1} = 16 - 8 = 8\]
\[x_{1} = 4 + 2\sqrt{2} > 2;\]
\[x_{2} = 4 - 2\sqrt{2} > 1.\]
\[Ответ:x = 4 - 2\sqrt{2}.\]
\[\textbf{б)}\ 2\sqrt{3x + 7} = 3\sqrt{2x + 2} + 2\]
\[3x + 7 \geq 0\]
\[x \geq - \frac{7}{3}.\]
\[2x + 2 \geq 0\]
\[2x \geq - 2\]
\[x \geq - 1.\]
\[M = \lbrack - 1; + \infty).\]
\[\left( 2\sqrt{3x + 7} \right)^{2} = \left( 3\sqrt{2x + 2} + 2 \right)^{2}\]
\[4(3x + 7) =\]
\[= 9(2x + 2) + 12\sqrt{2x + 2} + 4\]
\[12x + 28 =\]
\[= 18x + 18 + 12\sqrt{2x + 2} + 4\]
\[- 6x + 6 = 12\sqrt{2x + 2}\]
\[1 - x = 2\sqrt{2x + 2}\]
\[1 - x \geq 0\]
\[x \leq 1.\]
\[M = \lbrack - 1;1\rbrack.\]
\[(1 - x)^{2} = 4(2x + 2)\]
\[x^{2} - 2x + 1 = 8x + 8\]
\[x^{2} - 10x - 7 = 0\]
\[D_{1} = 25 + 7 = 32\]
\[x_{1} = 5 + 4\sqrt{2} > 1;\]
\[x_{2} = 5 - 4\sqrt{2} > - 1.\]
\[Ответ:x = 5 - 4\sqrt{2}.\]
\[\textbf{в)}\ \sqrt{6x - 3} - 2\sqrt{x} = 1\]
\[6x - 3 \geq 0\]
\[6x \geq 3\]
\[x \geq 0,5.\]
\[x \geq 0.\]
\[M = \lbrack 0,5; + \infty).\]
\[\sqrt{6x - 3} = 2\sqrt{x} + 1\]
\[\left( \sqrt{6x - 3} \right)^{2} = \left( 2\sqrt{x} + 1 \right)^{2}\]
\[6x - 3 = 4x + 4\sqrt{x} + 1\]
\[2x - 4 = 4\sqrt{x}\]
\[x - 2 = 2\sqrt{x}\]
\[x - 2 \geq 0\]
\[x \geq 2.\]
\[M = \lbrack 2;\ + \infty).\]
\[(x - 2)^{2} = \left( 2\sqrt{x} \right)^{2}\]
\[x^{2} - 4x + 4 = 4x\]
\[x^{2} - 8x + 4 = 0\]
\[D_{1} = 16 - 4 = 12\]
\[x_{1} = 4 + 2\sqrt{3} > 2;\]
\[x_{2} = 4 - 2\sqrt{3} < 2.\]
\[Ответ:x = 4 + 2\sqrt{3}.\]
\[\textbf{г)}\ \sqrt{2x - 1} - \sqrt{x} = 1\]
\[2x - 1 \geq 0\]
\[2x \geq 1\]
\[x \geq 0,5;\]
\[x \geq 0.\]
\[M = \lbrack 0,5; + \infty).\]
\[\sqrt{2x - 1} = \sqrt{x} + 1\]
\[\left( \sqrt{2x - 1} \right)^{2} = \left( \sqrt{x} + 1 \right)^{2}\]
\[2x - 1 = x + 2\sqrt{x} + 1\]
\[x - 2 = 2\sqrt{x}\]
\[x - 2 \geq 0\]
\[x \geq 2.\]
\[M = \lbrack 2; + \infty).\]
\[(x - 2)^{2} = \left( 2\sqrt{x} \right)^{2}\]
\[x^{2} - 4x + 4 = 4x\]
\[x^{2} - 8x + 4 = 0\]
\[D_{1} = 16 - 4 = 12\]
\[x_{1} = 4 + 2\sqrt{3} > 2;\]
\[x_{2} = 4 - 2\sqrt{3} < 2.\]
\[Ответ:x = 4 + 2\sqrt{3}.\]
