Question

Помогите пж, решит одно уровнение

Answer 1

$\frac{1}{(x-4)^{2} } -\frac{7}{x-4} +10=0\\\\\frac{1-7x+28+10x^{2}-80x+160 }{(x-4)^{2} }=0\\\\\frac{10x^{2}-87x+189 }{(x-4)^{2} }=0\\\\10x^{2}-87x+189=0, \ x\neq4\\\\D=(-87)^{2}-4*10*189=7569-7560=9=3^{2}\\\\x_{1}=\frac{87+3}{20}=4,5\\\\x_{2}=\frac{87-3}{20}=4,2\\\\Otvet:\boxed{4,5 \ ; \ 4,2}$