Question

Алгебра. Помогите решить уравнение.

Answer 1

Ответ:

$\displaystyle \tt x_1=6-2\sqrt{2}\\x_2=6+2\sqrt{2}$

Объяснение:

$\displaystyle \tt x+\frac{1}{x-3}=9 \: \: \: \: | \: x\ne3\\x+\frac{1}{x-3}-9=0\\\frac{x(x-3)+1-9(x-3)}{x-3}=0\\\frac{x^2-3x+1-9x+27}{x-3}=0\\\frac{x^2-12x+28}{x-3}=0\\x^2-12x+28=0\\D=(-12)^2-4\cdot1\cdot28=144-112=32\\\sqrt{D}=\sqrt{32}=\sqrt{16\cdot2}=4\sqrt{2}\\x_1=\frac{12+4\sqrt{2}}{2}=\frac{4(3+\sqrt{2})}{2}=2(3+\sqrt{2})=6+2\sqrt{2}\\x_2=\frac{12-4\sqrt{2}}{2}=\frac{4(3-\sqrt{2})}{2}=2(3-\sqrt{2})=6-2\sqrt{2}$