Страница 410 номер 1355, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(x^{\lg x} = 10\)

Область допустимых значений: \(x > 0\);

\(x = 1\) – не корень: \(1^{\lg 1} = 1 \neq 10\);
\(\log_{x}x^{\lg x} = \log_{x}10,\)
\(\lg x = \dfrac{1}{\lg x},\)
\(\lg^{2}x = 1,\)
\(\lg x = \pm 1,\)
\(x_{1} = 10,\)
\(x_{2} = 0{,}1;\)

2) \(x^{\log_{3}x} = 9x\)

Область допустимых значений: \(x > 0\);

\(x = 1\) – не корень: \(1^{\log_{3}1} = 1 \neq 9\);
\(\log_{x}x^{\log_{3}x} = \log_{x}{9x},\)
\(\log_{3}x = \log_{x}x + \log_{x}9,\)
\(\log_{3}x = 1 + 2\log_{x}3,\)
\(\log_{3}x = 1 + \dfrac{2}{\log_{3}x},\)
\(\log_{3}^{2}x - \log_{3}x - 2 = 0,\)
\(\log_{3}x = z, z^{2} - z - 2 = 0\)
\(z_{1{,}2} = \dfrac{1 \pm \sqrt{1 + 8}}{2},\)
\(z_{1} = 2, z_{2} = - 1\)
\(\log_{3}x = 2\) или \(\log_{3}x = - 1\)
\(x_{1} = 9,\)
\(x_{2} = \dfrac{1}{3};\)

3) \(x^{\lg x} - 1 = 10\left( 1 - x^{-\lg x} \right)\)

Область допустимых значений: \(x > 0\);

\(x^{\lg x} - 1 = 10 - 10x^{-\lg x},\)
\(\lg x = z, x = 10^{z},\)
\(\left( 10^{z} \right)^{z} - 11 + 10\left( 10^{z} \right)^{-z} = 0,\)
\(10^{z^{2}} = w, w - 11 + \dfrac{10}{w} = 0\)
\(w^{2} - 11w + 10 = 0,\)
\(w_{1{,}2} = \dfrac{11 \pm \sqrt{121 - 40}}{2},\)
\(w_{1} = 10, w_{2} = 1;\)
\(10^{z^{2}} = 10\) или \(10^{z^{2}} = 1\)
\(z^{2} = 1\) или \(z^{2} = 0\)
\(z_{1} = 1, z_{2} = - 1, z_{3} = 0;\)
\(x_{1} = 10,\)
\(x_{2} = 10^{-1} = 0{,}1,\)
\(x_{3} = 10^{0} = 1;\)

4) \(x^{\sqrt{x}} = \sqrt{x^{x}}\)

Область допустимых значений: \(x > 0\);

\(x_{1} = 1\) – при \(x = 1\) обе части равны 1;
\(\left( x^{\sqrt{x}} \right)^{2} = x^{x},\)
\(x^{2\sqrt{x}} = x^{x},\)
\(\sqrt{x} = z, x = z^{2}\)
\(\left( z^{2} \right)^{2z} = \left( z^{2} \right)^{z^{2}},\)
\(2z = z^{2},\)
\(z(2 - z) = 0,\)
\(z_{1} = 0, z_{2} = 2\)
\(x_{2} = 0\) – не соответствует ОДЗ;
\(x_{3} = 4.\)

Ответ: \(x_{1} = 1\), \(x_{2} = 4\).