Ответ:

1) f(1) = -0.4

f(-3) = -6

2) D(f)={x∈R: x> 5}

D(f) = {x∈R : x≠2; x≠3}

D(f) = {x∈R : x∈ [-2;5) ∪ (5; +∞)}

Объяснение:

1)

[tex]\displaystyle f(x) = \frac{x-3}{x+4} \\\\\\f(1) = \frac{1-3}{1+4} =\frac{-2}{5} =-0.4\\\\\\f(-3) = \frac{-3-3}{-3+4} =\frac{-6}{1} =-6[/tex]

2)

[tex]\displaystyle f(x) = \frac{4}{\sqrt{5-x} } ;\\\\\left \{ {{\sqrt{5-x}\neq 0 } \atop {5-x\geq 0}} \right. \quad \Rightarrow 5-x > 0; \quad x > 5\\\\\\D(f)=\{x \in R: x > 5\}[/tex]

[tex]\displaystyle f(x)=\frac{x+3}{x^2-5x+6}\\\\\\ x^2-5x+6\neq 0\\[/tex]

Разложим на множители

x² - 5x +6 = 0

по теореме Виета

х₁ * х₂ = 6

х₁ + х₂ = 5

х₁ = 3;    х₂ = 2

x² - 5x +6 = (х-2)(х-3)

Тогда

D(f) = {x∈R : x≠2; x≠3}

[tex]\displaystyle f(x) = \sqrt{x+2} +\frac{x-2}{x-5}[/tex]

[tex]\displaystyle \left \{ {{x+2\geq 0} \atop {x\neq 5 \hfill}} \right. \left \{ {{x\geq -2} \atop {x\neq 5}} \right.[/tex]

D(f) = {x∈R : x∈ [-2;5) ∪ (5; +∞)}