Question

Решите пожалуйста три этих уравнения. С расчетом.

Answer 1

$5.) \\ x {}^{2} - 3x + 2 = 0 \\ d = 3 {}^{2} - 4 \times 2 = 9 - 8 = 1 \\ x1 = \frac{3 - 1}{2} = \frac{2}{2} = 1 \\ x2 = \frac{3 + 1}{2} = \frac{4}{2} = 2 \\ 6.) \\ 9x {}^{2} - 6x + 1 =0 \\ d = 6 {}^{2} - 4 \times 9 = 36 - 36 = 0 \\ x = \frac{6 - 0}{18} = \frac{6}{18} = \frac{1}{3} \\ 7. \\ 3x {}^{2} - 4x + 1 = 0 \\ d = 4 {}^{2} - 3 \times 4 = 16 - 12 = 4 \\ \sqrt{4} = 2 \\ x1 = \frac{4 - \ 2}{6} = \frac{2}{6} = \frac{1}{3} \\ x2 = \frac{4 + 2}{6} = \frac{6}{6} = 1$