Question

помогите пожалуйста ​

Answer 1

$\displaystyle 1)\sqrt{3x-1}=\sqrt{4x+1}\\3x-1=4x+1\\3x-4x=1+1\\-x=2\\x=-2\\\sqrt{3*(-2)-1}=\sqrt{4*(-2)+1}\\x\neq -2\\x \in \phi\\\\ 2)(x-4)\sqrt{x^2-x-20}=6x-24\\(x-4)^2(x^2-x-20)=36x^2-288x+576\\(x-4)^2(x^2-x-20)=36(x^2-8x+16)\\(x-4)^2(x^2-x-20)=36(x-4)^2\\(x-4)^2(x^2-x-20)-36(x-4)^2=0\\(x-4)^2(x^2-x-20-36)=0\\(x-4)^2(x^2-x-56)=0\\(x-4)^2=0,x^2-x-56=0\\x=4,x=-7,x=8\\(4-4)\sqrt{4^2-4-20}=6*4-24, (-7-4)\sqrt{(-7)^2-(-7)-20}=6*(-7)-24,\\ \\ (8-4)\sqrt{8^2-8-20}=6*8-24 \\ x_1=-7,x_2=8$