Question

Решите уравнение \frac{1}{(x-3)^2} =\frac{3}{x-3} +4

Answer 1

$\frac{1}{(x-3)^{2} }=\frac{3}{x-3}+4 \\\\\frac{1}{(x-3)^{2} }-\frac{3}{x-3}-4=0\\\\\frac{1-3x+9-4x^{2}+24x-36 }{(x-3)^{2} }=0\\\\\frac{4x^{2} -21x+26}{(x-3)^{2} } =0\\\\\left \{ {{4x^{2}-21x+26=0 } \atop {x\neq3 }} \right. \\\\4x^{2}-21x+26=0\\\\D=(-21)^{2} -4*4*26=441-416=25=5^{2}\\\\x_{1}=\frac{21-5}{8}=2\\\\x_{2}=\frac{21+5}{8} =3,25\\\\Otvet:\boxed{2 \ ; \ 3,25}$

Answer 2

Ответ:

Объяснение:

Решение дано на фото.