Страница 430 номер 1608, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\begin{cases} 27 \cdot 3^{2x - y} + 3^{x^{2}} = 4\sqrt{3} \\ \lg(y - 4x) = 2\lg(2 + 2x - y) - \lg y \end{cases}\)

Область допустимых значений:

\(\begin{cases} y - 4x > 0 \\ 2 + 2x - y > 0 \\ y > 0 \end{cases},\)
\(4x < y < 2 + 2x, y > 0;\)
\(\lg\dfrac{y - 4x}{2 + 2x - y} = \lg\dfrac{2 + 2x - y}{y},\)
\(\dfrac{y - 4x}{2 + 2x - y} = \dfrac{2 + 2x - y}{y},\)
\(y^{2} - 4xy - (2 + 2x - y)^{2} = 0,\)
\(y^{2} - 4xy - \left( 4 + 8x + 4x^{2} - 2y(2 + 2x) + y^{2} \right) = 0,\)
\(y^{2} - 4xy - 4 - 8x - 4x^{2} + 4y + 4xy - y^{2} = 0,\)
\(- 4x^{2} - 8x + 4y - 4 = 0,\)
\(x^{2} + 2x - y + 1 = 0,\)
\(y = x^{2} + 2x + 1 = (x + 1)^{2},\)
\(3^{3} \cdot 3^{2x - x^{2} - 2x - 1} + 3^{x^{2}} = 4\sqrt{3},\)
\(3^{2 - x^{2}} + 3^{x^{2}} = 4\sqrt{3},\)
\(3^{x^{2}} = z, z > 0\)
\(\dfrac{9}{z} + z - 4\sqrt{3} = 0,\)
\(9 + z^{2} - 4\sqrt{3}z = 0,\)
\(z_{1{,}2} = \dfrac{4\sqrt{3} \pm \sqrt{48 - 36}}{2} = 2\sqrt{3} \pm \sqrt{3},\)
\(z_{1} = 3\sqrt{3}, z_{2} = \sqrt{3}\)

а) \(3^{x^{2}} = 3\sqrt{3} = 3^{\frac{3}{2}}\),
\(x_{1{,}2} = \pm \sqrt{\dfrac{3}{2}},\)
\(y_{1} = \dfrac{3}{2} + \sqrt{6} + 1,\)
\(y_{2} = \dfrac{3}{2} - \sqrt{6} + 1;\)
\(2 + 2x - y = 1 - x^{2} = - \frac{1}{2} < 0\) – не удовлетворяет ОДЗ;

б) \(3^{x^{2}} = \sqrt{3}\),
\(x_{1{,}2} = \pm \dfrac{1}{\sqrt{2}},\)
\(y_{1} = \dfrac{1}{2} + \sqrt{2} + 1,\)
\(y_{2} = \dfrac{1}{2} - \sqrt{2} + 1;\)

Учитывая ОДЗ: \(\left( \frac{1}{\sqrt{2}};\ \frac{3}{2} + \sqrt{2} \right)\), \(\left( - \frac{1}{\sqrt{2}};\ \frac{3}{2} - \sqrt{2} \right)\);

2) \(\begin{cases} 8 \cdot 2^{-x - 2y} + 2^{y^{2}} = 3\sqrt{2} \\ \lg(x + 4y) = 2\lg(2 - x - 2y) - \lg x \end{cases}\)

Область допустимых значений:

\(\begin{cases} x + 4y > 0 \\ 2 - x - 2y > 0 \\ x > 0 \end{cases},\)
\(\begin{cases} x > - 4y \\ x < 2 - 2y \\ x > 0 \end{cases};\)
\(\lg(x + 4y) - \lg(2 - x - 2y) = \lg(2 - x - 2y) - \lg x,\)
\(\dfrac{x + 4y}{2 - x - 2y} = \dfrac{2 - x - 2y}{x},\)
\(x^{2} + 4xy = (2 - x - 2y)^{2},\)
\(x^{2} + 4xy = 4 - 4x + x^{2} - 4y(2 - x) + 4y^{2},\)
\(4y^{2} + 4 - 4x - 8y = 0,\)
\(y^{2} + 1 - x - 2y = 0,\)
\(x = y^{2} - 2y + 1 = (y - 1)^{2},\)
\(2^{3} \cdot 2^{-y^{2} + 2y - 1 - 2y} + 2^{y^{2}} = 3\sqrt{2},\)
\(2^{2} \cdot 2^{-y^{2}} + 2^{y^{2}} = 3\sqrt{2},\)
\(2^{y^{2}} = z, z > 0\)
\(\dfrac{4}{z} + z = 3\sqrt{2},\)
\(4 + z^{2} - 3\sqrt{2}z = 0,\)
\(z_{1{,}2} = \dfrac{3\sqrt{2} \pm \sqrt{18 - 16}}{2} = \dfrac{3\sqrt{2} \pm \sqrt{2}}{2},\)
\(z_{1} = 2\sqrt{2}, z_{2} = \sqrt{2}\)

а) \(2^{y^{2}} = 2\sqrt{2}\),
\(y = \pm \sqrt{\dfrac{3}{2}},\)
\(x_{1} = \dfrac{3}{2} - \sqrt{6} + 1,\)
\(x_{2} = \dfrac{3}{2} + \sqrt{6} + 1;\)
\(2 - x - 2y = 1 - y^{2} = - \frac{1}{2} < 0\) – не удовлетворяет ОДЗ;

б) \(2^{y^{2}} = \sqrt{2}\),
\(y = \pm \dfrac{1}{\sqrt{2}},\)
\(x_{1} = \dfrac{1}{2} - \sqrt{2} + 1,\)
\(x_{2} = \dfrac{1}{2} + \sqrt{2} + 1;\)

Учитывая ОДЗ: \(\left( \frac{3}{2} - \sqrt{2};\ \frac{1}{\sqrt{2}} \right)\), \(\left( \frac{3}{2} + \sqrt{2};\ - \frac{1}{\sqrt{2}} \right)\).