Question

Помогите пожалуйста! СРОЧНО!!!
125 (2,4)

Answer 1

$sqrt{x+2} =2+ sqrt{x-6} \ left { {{x+2 geq 0} atop {x-6 geq 0}} right. Rightarrow x geq 6 \ ( sqrt{x+2} )^2=(2+ sqrt{x-6} )^2 \ x+2=4+4 sqrt{x-6} +x-6 \ 4 sqrt{x-6} =4 \ sqrt{x-6} =1 \ ( sqrt{x-6} )^2=1^2 \ x-6=1 \ x=7$
Ответ: 7

$sqrt{22-x} - sqrt{10-x} =2 \ left { {{22-x geq 0} atop {10-x geq 0}} right. Rightarrow x leq 10 \ sqrt{22-x} =2+ sqrt{10-x} \ ( sqrt{22-x} )^2=(2+ sqrt{10-x})^2 \ 22-x=4+4 sqrt{10-x} +10-x \ 4 sqrt{10-x} =8 \ sqrt{10-x} =2 \ ( sqrt{10-x} )^2=2^2 \ 10-x=4 \ x=6$
Ответ: 6