√(x-3) > x-5 ОДЗ: х-3>0
(√(x-3) > x-5)² x>3
x-3>(x-5)(x-5)
x-3>x²-10x+25
x-3=x²-10x+25
x²-11x+28=0
x₁+x₂=11
x₁*x₂=28
x₁=4
x₂=7
x>4 ∧ x<7
x∈(4;7)
√(x-3) > x-5 ОДЗ: х-3>0
(√(x-3) > x-5)² x>3
x-3>(x-5)(x-5)
x-3>x²-10x+25
x-3=x²-10x+25
x²-11x+28=0
x₁+x₂=11
x₁*x₂=28
x₁=4
x₂=7
x>4 ∧ x<7
x∈(4;7)