Zadanie :
Podpunkt c)
[tex]x+1=\frac{2-2x}{x-1}\\\\x\neq1\\\\(x+1)(x-1)=2-2x\\\\x^2-1-2+2x=0\\\\x^2+2x-3=0\\\\a=1, \ b=2, \ c=-3\\\\\Delta=2^2-4\cdot1\cdot(-3)=4+12=16\\\\\sqrt{\Delta}=4\\\\x_1=\frac{-2-4}{2}=\boxed{x=-3}\in\mathbb{D}\\\\x_2=\frac{-2+4}{2}=1\notin\mathbb{D}[/tex]
Odp. x = -3.
Podpunkt d)
[tex]\frac{2x-2}{x+2}=\frac{x}{x-1}\\\\x\neq-2, \ x\neq1\\\\(2x-2)(x-1)=x(x+2)\\\\2x^2-2x-2x+2=x^2+2x\\\\2x^2-4x+2-x^2-2x=0\\\\x^2-6x+2=0\\\\a=1, \ b=-6, \ c=2\\\\\Delta=(-6)^2-4\cdot1\cdot2=36-8=28\\\\\sqrt{\Delta}=2\sqrt7\\\\x_1=\frac{-(-6)-2\sqrt7}{2}=\frac{6-2\sqrt7}{2}=\boxed{3-\sqrt7}\in\mathbb{D}\\\\x_2=\frac{-(-6)+2\sqrt7}{2}=\frac{6+2\sqrt7}{2}=\boxed{3+\sqrt7}\in\mathbb{D}[/tex]
Odp. x = 3 - √7 ∨ x = 3 + √7.
Zadanie :
Podpunkt c)
[tex]\frac{x^2+8}{x^2-4}=2\\\\x^2-4\neq0\\\\(x+2)(x-2)\neq0\rightarrow x\neq-2, \ x\neq2\\\\2(x^2-4)=x^2-8\\\\2x^2-8-x^2+8=0\\\\x^2=0\\\\\boxed{x=0}\in\mathbb{D}[/tex]
Odp. x = 0.
Podpunkt d)
[tex]\frac{2x^2-10}{x^2-9}=1\\\\x^2-9\neq0\\\\(x+3)(x-3)\neq0\rightarrow x\neq-3, \ x\neq3\\\\2x^2-10=x^2-9\\\\2x^2-x^2-10+9=0\\\\x^2-1=0\\\\(x+1)(x-1)=0\\\\\boxed{x=-1}\in\mathbb{D} \ \vee \ \boxed{x=1}\in\mathbb{D}[/tex]
Odp. x = -1 ∨ x = 1.
