Zadanie 4 :
Podpunkt a)
[tex]\frac{x}{-3}=\frac{-5}{x+2}\\\\x\neq-2\\\\x(x+2)=15\\\\x^2+2x-15=0\\\\a=1, \ b=2, \ c=-15\\\\\Delta=2^2-4\cdot1\cdot(-15)=4+60=64\\\\\sqrt{\Delta}=8\\\\x_1=\frac{-2-8}{2}=\boxed{-5}\in\mathbb{D}\\\\x_2=\frac{-2+8}{2}=\boxed{3}\in\mathbb{D}[/tex]
Odp. x = -5 ∨ x = 3.
Podpunkt b)
[tex]x=\frac{5x+3}{2x}\\\\x\neq0\\\\2x^2=5x+3\\\\2x^2-5x-3=0\\\\a=2, \ b=-5, \ c=-3\\\\\Delta=(-5)^2-4\cdot2\cdot(-3)=25+24=49\\\\\sqrt{\Delta}=7\\\\x_1=\frac{-(-5)-7}{4}=-\frac{2}{4}=\boxed{-0,5}\in\mathbb{D}\\\\x_2=\frac{-(-5)+7}{4}=\frac{12}{4}=\boxed3\in\mathbb{D}[/tex]
Odp. x = -0,5 ∨ x = 3.
Zadanie 5 :
Podpunkt a)
[tex]\frac{2x^2+1}{x^2}=3\\\\x\neq0\\\\3x^2=2x^2+1\\\\3x^2-2x^2-1=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.
Podpunkt b)
[tex]\frac{6x}{x^2+9}=-1\\\\\mathbb{D}:\mathbb{R}\\\\-(x^2+9)=6x\\\\-x^2-9-6x=0\\\\-x^2-6x-9=0 \ \ |\cdot(-1)\\\\x^2+6x+9=0\\\\(x+3)^2=0\\\\x+3=0\\\\\boxed{x=-3}\in\mathbb{D}[/tex]
Odp. x = -3.
