Question

(х+5)^4-6(х+5)^2-7=0
ДАЮ 43 БАЛА

Answer 1

$(x + 5) {}^{4} - 6(x + 5) {}^{2} - 7 = 0 \\ ( x+ 5) {}^{2} = a \: , \: \: a \geqslant 0 \\ {a}^{2} - 6a - 7 = 0 \\ \\ po \: \: \: teoreme \: \: \: vieta \\ {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c \\ \\ a_{1} + a_{2} = 6\\ a_{1} a_{2} = - 7 \\ a_{1} = 7\\ a_{2} = - 1$

Второй корень не подходит

$(x + 5) { }^{2} = 7 \\ (x + 5) { }^{2} - 7 = 0 \\ (x + 5) {}^{2} - ( \sqrt{7} ) {}^{2} = 0 \\ (x + 5 - \sqrt{7} )(x + 5 + \sqrt{7} ) = 0 \\ x_{1} = - 5 + \sqrt{7} \\ x_{2} = - 5 - \sqrt{7}$