Нули функции — это значения аргумента, при которых функция равна нулю.
y = 0;
x^2-6x+8 = 0
D = b^2 - 4ac = 36-32 = 4
x1 = (-b-√d)/2a = (6-2)/2 = 2
x2 = (-b+√d)/2a = (6+2)/2 = 4
Ответ: (2;0) ; (4;0).
Решение на фото......................
2√6 = √(6*4) = √24
4√2 = √(2*16) = √32
√24 < √32
2√6 < 4√2
ОДЗ 2x+8≥0 2x≥-8 x≥-4
![\sqrt{2x+8}=6 \\ (\sqrt{2x+8})^2=6^2 \\ 2x+8=36 \\ 2x=28 \\ x=14](https://tex.z-dn.net/?f=%5Csqrt%7B2x%2B8%7D%3D6+%5C%5C+%28%5Csqrt%7B2x%2B8%7D%29%5E2%3D6%5E2+%5C%5C+2x%2B8%3D36+%5C%5C+2x%3D28+%5C%5C+x%3D14)
ОДЗ x²-4x+13≥0 при любых значениях х
![\sqrt{x^2-4x+13}=5 \\ (\sqrt{x^2-4x+13})^2=5^2 \\ x^2-4x+13=25 \\ x^2-4x-12=0 \\ D=16+48=64 \\ x_1=\frac(4-8}{2}=-2 \ \ \ \ \ x_2=\frac{4+8}{2}=6](https://tex.z-dn.net/?f=%5Csqrt%7Bx%5E2-4x%2B13%7D%3D5+%5C%5C+%28%5Csqrt%7Bx%5E2-4x%2B13%7D%29%5E2%3D5%5E2+%5C%5C+x%5E2-4x%2B13%3D25+%5C%5C+x%5E2-4x-12%3D0+%5C%5C+D%3D16%2B48%3D64+%5C%5C+x_1%3D%5Cfrac%284-8%7D%7B2%7D%3D-2+%5C+%5C+%5C+%5C+%5C+x_2%3D%5Cfrac%7B4%2B8%7D%7B2%7D%3D6)
ОДЗ
![\left \{ {{x^2-4 \geq 0} \atop {8x+5 \geq 0} \right. \left \{ {{(x-2)(x+2) \geq 0} \atop {x \geq -0,625} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bx%5E2-4+%5Cgeq+0%7D+%5Catop+%7B8x%2B5+%5Cgeq+0%7D+%5Cright.+++%5Cleft+%5C%7B+%7B%7B%28x-2%29%28x%2B2%29+%5Cgeq+0%7D+%5Catop+%7Bx+%5Cgeq+-0%2C625%7D+%5Cright.++)
x∈[2; +
![\infty](https://tex.z-dn.net/?f=%5Cinfty)
)
![\sqrt{x^2-4}-\sqrt{8x+5}=0 \\ \sqrt{x^2-4}=\sqrt{8x+5} \\ (\sqrt{x^2-4})^2=(\sqrt{8x+5})^2 \\ x^2-4=8x+5 \\ x^2-8x-9=0 \\ D=64+36=100 \\ x_1=\frac{8-10}{2}=-1 \ \ \ \ \ \ x_2=\frac{8+10}{2}=9 ](https://tex.z-dn.net/?f=%5Csqrt%7Bx%5E2-4%7D-%5Csqrt%7B8x%2B5%7D%3D0+%5C%5C+%5Csqrt%7Bx%5E2-4%7D%3D%5Csqrt%7B8x%2B5%7D+%5C%5C+%28%5Csqrt%7Bx%5E2-4%7D%29%5E2%3D%28%5Csqrt%7B8x%2B5%7D%29%5E2+%5C%5C+x%5E2-4%3D8x%2B5+%5C%5C+x%5E2-8x-9%3D0+%5C%5C+D%3D64%2B36%3D100+%5C%5C+x_1%3D%5Cfrac%7B8-10%7D%7B2%7D%3D-1+%5C+%5C+%5C+%5C+%5C+%5C+x_2%3D%5Cfrac%7B8%2B10%7D%7B2%7D%3D9%0A)
x₁=-1 не удовлетворяет ОДЗ
Можно решать способом проверки корней
![\sqrt{2x^2-5x+1}=x-1 \\ (\sqrt{2x^2-5x+1})^2=(x-1 )^2 \\ 2x^2-5x+1= x^{2} -2x+1 \\ x^2-3x=0 \\ x(x-3)=0 \\ x=0 \ \ \ \ \ x-3=0 \\ . \ \ \ \ \ \ \ \ \ \ \ \ x=3](https://tex.z-dn.net/?f=%5Csqrt%7B2x%5E2-5x%2B1%7D%3Dx-1+%5C%5C+%28%5Csqrt%7B2x%5E2-5x%2B1%7D%29%5E2%3D%28x-1+%29%5E2+%5C%5C+2x%5E2-5x%2B1%3D+x%5E%7B2%7D+-2x%2B1+%5C%5C+x%5E2-3x%3D0+%5C%5C+x%28x-3%29%3D0+%5C%5C+x%3D0+%5C+%5C+%5C+%5C+%5C+x-3%3D0+%5C%5C++.+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C++%5C+x%3D3)
проверим корни уравнения
х=0
![\sqrt{2*0^2-5*0+1}=0-1 \\ 1=-1](https://tex.z-dn.net/?f=%5Csqrt%7B2%2A0%5E2-5%2A0%2B1%7D%3D0-1+%5C%5C+1%3D-1)
значит х=0 посторонний корень
х=3
![\sqrt{2*3^2-5*3+1}=3-1 \\ 2=2](https://tex.z-dn.net/?f=%5Csqrt%7B2%2A3%5E2-5%2A3%2B1%7D%3D3-1+%5C%5C+2%3D2)
ответ 2