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Ответ дал: сок111213

5.

1) Д

${x}^{2} - 81 = 0 \\ (x - 9)(x + 9) = 0 \\ x_{1} = 9 \\ x_{2} = - 9$

2) В

${x}^{2} - 13x = 0 \\ x(x - 13) = 0 \\ x_{1} =0 \\ x_{2} = 13$

3) А

$(x - 13) {}^{2} = 0 \\ x - 13 = 0 \\ x = 13$

4) Б

$2 {x}^{2} - 9x + 10 = 0 \\ a = 2 \\ b = - 9 \\ c = 10 \\ D = {b}^{2} - 4ac = ( - 9) {}^{2} - 4 \times 2 \times 10 = \\ = 81 - 80 = 1 \\ x_{1} = \frac{9 + 1}{2 \times 2} = \frac{10}{4} = 2.5 \\ x_{2} = \frac{9 - 1}{2 \times 2} = \frac{8}{4} = 2$

6.

$(3x - 8) {}^{2} -( 2x - 6) = 66 - (5x - 2)(x + 2) \\ 9 {x}^{2} - 48x + 64 - 2x + 6 = 66 - (5 {x}^{2} + 10x - 2x - 4) \\ 9 {x}^{2} - 50x + 70 = 66 - 5 {x}^{2} - 8x + 4 \\ 9 {x}^{2} - 50x + 70 - 70 + 5 {x}^{2} + 8x = 0 \\ 14x {}^{2} - 42x = 0 \\ x(x - 3) = 0 \\ x_{1} =0 \\ x_{2} = 3$

7.

${x}^{2} + px + 3 = 0 \\ x_{2} = 3x_{1}$

По теореме Виета:

${x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c$

$x_{1} + x_{2} =x _{1}+ 3x_{1} = 4x_{1} = - p \\ x_{1} x_{2} = x _{1}\times 3x _{1}= 3 {x_{1}}^{2} = 3 \\ \displaystyle\bf\\\left \{ {{4x_{1} = - p} \atop {3 {x_{1}}^{2} = 3 }} \right. \\ \displaystyle\bf\\\left \{ {{4x_{1} = - p} \atop {x_{1} = 1 \: \: ili \: \: x_{1} = - 1 }} \right. \\ \\ 1) \: x_{1} = 1 \\ 4 \times 1 = - p \\ p = - 4 \\ {x}^{2} - 4x + 3 = 0 \\ po \: \: teoreme \: \: vieta \\ x_{1} + x_{2} = 4 \\ x_{1} x_{2} = 3\\ x_{1} = 1 \\ x_{2} = 3 \\ \\ 2) \: x_{1} = - 1 \\ 4 \times ( - 1) = - p \\ p = 4 \\ {x}^{2} + 4x + 3 = 0 \\ po \: \: teoreme \: \: vieta \\ x_{1} + x_{2} = - 4 \\ x_{1} x_{2} = 3 \\ x_{1} = - 1 \\ x_{2} = - 3$