Question

Решите уравнение б) 100БАЛЛОВ ПОМОГИТЕ

Answer 1

$\displaystyle \sqrt{x^2-10x+1}=5\\ x^2-10x+1=25\\x^2-10x+1-25=0\\x^2+2x-12x-24=0\\x(x+2)-12(x+2)=0\\(x+2)(x-12)=0\\x+2=0,x-12=0\\x=-2,x=12\\\sqrt{(-2)^2-10*(-2)+1}=5 \rightarrow 5=5\\ \sqrt{12^2-10*12+1}=5 \rightarrow 5=5\\x_1=-2,x_2=12$

Answer 2

$\sqrt{x^2-10x+1}=5\\\\(\sqrt{x^2-10x+1})^2=5^2\\\\x^2-10x+1=25\\\\x^2-10x+1-25=0\\\\x^2-10x-24=0\\\\D=(-10)^2-4*1*(-24)=100+96=196=14^2\\\\x_1=\frac{10+14}{2}=\frac{24}{2}=12;\; \; \; \; x_2=\frac{10-14}{2}=\frac{-4}{2}=-2$

Проверка:

$x_1=12\\\\\sqrt{12^2-10*12+1}=5\\\sqrt{144-120+1}=5\\\sqrt{25}=5\\5=5$верно

$x_2=-2\\\\\sqrt{(-2)^2-10*(-2)+1}=5\\\sqrt{4+20+1}=5\\\sqrt{25}=5\\5=5$верно

Ответ: -2; 12