√(x+4)=√(2x-1) , x+4≥0 ∧ 2x-1≥0, x≥-4 ∧ x≥0,5, x∈(0,5, ∞)
x+4=2x-1, x=5 ,
√(5+4)=√9=3, √(2.5-1)=√(10-1)=√9=3
Otvet: x=5
1.a)=(11+44x)-(x²+4x³)=11(1+4x)-x²(1+4x)=(1+4x)(11-x²).
б)=(63mn-49n)-(28m-36m²)=7n(9m-7)-4m(7-9m)=7n(9m-7)+4m(9m-7)=(9m-7)(7n+4m).
D = 4 - 4q.
x_1 =
![x_2 = \frac{-2 - \sqrt{4 - 4q} }{2} = \frac{-2 - 2\sqrt{1 - q} }{2} = -1 - \sqrt{1-q}](https://tex.z-dn.net/?f=x_2+%3D++%5Cfrac%7B-2+-++%5Csqrt%7B4+-+4q%7D+%7D%7B2%7D+%3D++%5Cfrac%7B-2+-++2%5Csqrt%7B1+-+q%7D+%7D%7B2%7D+%3D+-1+-++%5Csqrt%7B1-q%7D)
![x_2 - x_1 = -1 - \sqrt{1-q} - (-1 + \sqrt{1 - q} ) = -2 \sqrt{1 - q}](https://tex.z-dn.net/?f=x_2+-+x_1+%3D+-1+-+%5Csqrt%7B1-q%7D+-+%28-1+%2B+%5Csqrt%7B1+-+q%7D+%29+%3D+-2+%5Csqrt%7B1+-+q%7D+)
, т.е.
![x_2 - x_1 = -\sqrt{D}](https://tex.z-dn.net/?f=x_2+-+x_1+%3D+-%5Csqrt%7BD%7D+)
По обратной теореме Виета:
x₁ + x₂ = -2
x₂² - x₁² = (x₂ - x1)(x₁ + x₂)
![(-2 \sqrt{1-q} )*(-2) = 12](https://tex.z-dn.net/?f=%28-2+%5Csqrt%7B1-q%7D+%29%2A%28-2%29+%3D+12)
![\sqrt{1-q} =3](https://tex.z-dn.net/?f=+%5Csqrt%7B1-q%7D+%3D3)
![1-q = 9](https://tex.z-dn.net/?f=1-q+%3D+9)
![q = -8.](https://tex.z-dn.net/?f=q+%3D+-8.)