Question

Решите пожалуйста, срочно надо

(х+1)^4-4(х+1)^2-21=0

Answer 1

$(x+1)^4-4(x+1)^2-21=0\\\\Zamena:\\(x+1)^2=a\\(x+1)^4=a^2\\\\a^2-4a-21=0\\a^2+3a-7a-21=0\\a(a+3)-7(a+3)=0\\(a+3)(a-7)=0\\a_1=-3 \ \ a_2=7\\\\Obratnaya \ zamena\\(x+1)^2=-3\\x \in \emptyset\\\\(x+1)^2=7\\\\\left \{ {{x+1=\sqrt{7} } \atop {x+1=-\sqrt{7} }} \right. =>\left \{ {{x_1=\sqrt{7}-1 } \atop {x_2=-1-\sqrt{7} }} \right.$