Страница 80 номер 223, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(8 \cdot 4^{x} - 6 \cdot 2^{x} + 1 = 0\)
\(2^{x} = z, 8z^{2} - 6z + 1 = 0,\)
\(z_{1{,}2} = \dfrac{6 \pm \sqrt{36-32}}{16},\)
\(z_{1} = \dfrac{1}{2}, z_{2} = \dfrac{1}{4},\)
\(2^{x} = \dfrac{1}{2}, x_{1} = -1;\)
\(2^{x} = \dfrac{1}{4}, x_{2} = -2;\)

2) \(\left(\frac{1}{4}\right)^{x} + \left(\frac{1}{2}\right)^{x} - 6 = 0\)
\(\left(\dfrac{1}{2}\right)^{x} = z, z^{2} + z - 6 = 0,\)
\(z_{1{,}2} = \dfrac{-1 \pm \sqrt{1+24}}{2},\)
\(z_{1} = 2, z_{2} = -3\) – не подходит;
\(\left(\dfrac{1}{2}\right)^{x} = 2, x = -1;\)

3) \(13^{2x+1} - 13^{x} - 12 = 0\)
\(13^{x} = z, 13z^{2} - z - 12 = 0,\)
\(z_{1{,}2} = \dfrac{1 \pm \sqrt{1+624}}{26},\)
\(z_{1} = 1, z_{2} = -\frac{12}{13}\) – не подходит;
\(13^{x} = 1, x = 0;\)

4) \(3^{2x+1} - 10 \cdot 3^{x} + 3 = 0\)
\(3^{x} = z, 3z^{2} - 10z + 3 = 0,\)
\(z_{1{,}2} = \dfrac{10 \pm \sqrt{100-36}}{6},\)
\(z_{1} = 3, z_{2} = \dfrac{1}{3};\)
\(3^{x} = 3, x_{1} = 1;\)
\(3^{x} = \dfrac{1}{3}, x_{2} = -1;\)

5) \(2^{3x} + 8 \cdot 2^{x} - 6 \cdot 2^{2x} = 0\)
\(2^{x}(2^{2x} + 8 - 6 \cdot 2^{x}) = 0,\)
\(2^{x} \neq 0, 2^{x} = z,\)
\(z^{2} + 8 - 6z = 0,\)
\(z_{1{,}2} = \dfrac{6 \pm \sqrt{36-32}}{2},\)
\(z_{1} = 4, z_{2} = 2;\)
\(2^{x} = 4, x_{1} = 2;\)
\(2^{x} = 2, x_{2} = 1;\)

6) \(5^{3x+1} + 34 \cdot 5^{2x} - 7 \cdot 5^{x} = 0\)
\(5^{x}(5^{2x+1} + 34 \cdot 5^{x} - 7) = 0,\)
\(5^{x} \neq 0, 5^{x} = z,\)
\(5z^{2} + 34z - 7 = 0,\)
\(z_{1{,}2} = \dfrac{-34 \pm \sqrt{1156+140}}{10},\)
\(z_{1} = \frac{1}{5}, z_{2} = -7\) – не подходит;
\(5^{x} = \dfrac{1}{5}, x = -1.\)