Question

Корень х-3=1+уореньх-4

Answer 1

$sqrt{x-3} =1+ sqrt{x-4} \ left { {{x-3 geq 0} atop {x-4 geq 0}} right. \ x geq 4\ ( sqrt{x-3})^2 =(1+ sqrt{x-4})^2 \ x-3=1+2 sqrt{x-4}+x-4\ 2 sqrt{x-4}=0\ x-4=0\ x=4$