Задачии

[Гост]

Моля реши ми тези задачи:
Зад1) х-7=5:(7-х)
Зад2) (х-3)^2=4
Зад3) 3:х=4-х
Зад4) ((х^4-1):(х^2-9)):((х^2+1):(х+3))+(8:(3-х))

[ammornil]

Моля реши ми тези задачи:
Зад1) х-7=5:(7-х)
Зад2) (х-3)^2=4
Зад3) 3:х=4-х
Зад4) ((х^4-1):(х^2-9)):((х^2+1):(х+3))+(8:(3-х))

$\\[12pt] \boxed{1} \quad x-7=\dfrac{5}{7-x} \\[6pt]\quad \text{ДМ: } x\ne{}7 \\[6pt] x-7=\dfrac{5}{-(x-7)} \Leftrightarrow x-7=\dfrac{-5}{x-7} \Leftrightarrow (x-7)^{2}=-5 \\[6pt] \begin{cases} (x -7)^{2}>0, \hspace{0.2em} \forall{x}\in{\text{ДМ}} \\ -5<0, \hspace{0.2em} \forall{x}\in{\text{ДМ}} \end{cases} \Rightarrow \nexists{x} \in\mathbb{R} \\[12pt] \boxed{2} \quad (x-3)^{2}=4 \\[6pt] (x-3)^{2}= 2^{2} \quad \cup \quad (x-3)^{2}= (-2)^{2} \\[6pt] x-3=2 \quad \cup \quad x-3=-2 \\[6pt] x=5 \cup x=1 \\[12pt] \boxed{3} \quad 3\div{}x= 4-x, \quad \text{ДМ: }{}x\ne{}0 \\[6pt] \dfrac{3}{x}= 4-x \quad \Leftrightarrow \quad 3=x(4-x) \quad \Leftrightarrow \quad 3=4x -x^{2} \\[6pt] \quad \Leftrightarrow \quad x^{2} -4x +3=0 \quad \Leftrightarrow \quad x^{2}-x -3x +3=0 \quad \Leftrightarrow \quad x(x-1) -3(x-1)=0 \quad \Leftrightarrow \quad (x-1)(x-3)=0 \\[6pt] x=1 \cup x=3 \\[12pt] \boxed{4} \quad \dfrac{\quad \dfrac{x^{4}-1}{x^{2}-9} \quad}{\dfrac{x^{2}+1}{x+3}}+\dfrac{8}{3-x}=? \\[6pt] \quad \text{ДМ:} x\ne{}\pm{}3 \\[6pt] \dfrac{\quad \dfrac{x^{4}-1}{x^{2}-9} \quad}{\dfrac{x^{2}+1}{x+3}}+\dfrac{8}{3-x}= \dfrac{(x^{2}-1)(x^{2}+1)(x+3)}{(x-3)(x+3)(x^{2}+1)}+\dfrac{8}{-(x-3)}= \\[6pt] \quad =\dfrac{x^{2}-1}{x-3}-\dfrac{8}{x-3}= \dfrac{x^{2}-1-8}{x-3}= \\[6pt] =\dfrac{x^{2}-3^{2}}{x-3}= \dfrac{(x-3)(x+3)}{x-3}= x+3 $