Question

Rozwiąż równania i wyznacz dziedzinę:

Answer 1

Odpowiedź:

[tex]a)\\\\D=\mathbb {R}-\{0\}\\\\x=-2\sqrt{5} ~~\lor~~x=2\sqrt{5} \\\\\\b)\\\\D=\mathbb {R}-\{-0,5~;~0\}\\\\x=1~~\lor~~x=3[/tex]

Szczegółowe wyjaśnienie:

Korzystamy ze wzorów:

• skróconego mnożenia :[tex]x^{2} -y^{2} =(x-y)(x+y)[/tex]
• [tex]ax^{2} +bx+c=0\\\\\Delta =b^{2}-4ac\\gdy~~\Delta > 0~~\Rightarrow ~~x_{1} =\dfrac{-b-\sqrt{\Delta} }{2a} ~~\lor ~~x_{2} =\dfrac{-b+\sqrt{\Delta} }{2a} \\\\gdy~~\Delta =0~~\Rightarrow ~~x_{o} =\dfrac{-b }{2a} \\\\gdy~~\Delta < 0~~\Rightarrow ~~x\in \varnothing[/tex]

Pamiętamy :

• o wyznaczeniu dziedziny  ⇒ mianownik musi być różny od zera

Rozwiązujemy:

[tex]a)\\\\\dfrac{-x^{2} +20}{x^{2} } =0\\\\zal.~x^{2} \neq 0~~\Rightarrow ~~x\neq 0~~\Rightarrow ~~D=\mathbb {R}-\{0\}\\\\\dfrac{-x^{2} +20}{x^{2} } =0\\\\-x^{2} +20=0\\\\x^{2} -20=0\\\\x^{2} -(\sqrt{20} )^{2}=0\\\\x^{2} -(2\sqrt{5} )^{2}=0\\\\(x-2\sqrt{5} )(x+2\sqrt{5} )=0\\\\x-2\sqrt{5} =0~~\lor ~~x+2\sqrt{5} =0\\\\(~~x=2\sqrt{5}~~\lor ~~x=-2\sqrt{5} ~~)~~\land ~~x\in D\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~\Downarrow \\\\~~~~~~~~~x=2\sqrt{5}~~\lor ~~x=-2\sqrt{5}[/tex]

[tex]b)\\\\\dfrac{x^{2} -4x+3}{2x^{2} +x} =0\\\\zal.~~\\2x^{2} +x\neq 0~~\Rightarrow ~~x(2x+1)\neq 0~~\Rightarrow ~~x\neq 0,~x\neq -0,5~~\Rightarrow ~~D=\mathbb {R}-\{-0,5~;~0\}\\\\\dfrac{x^{2} -4x+3}{2x^{2} +x} =0\\\\x^{2} -4x+3=0\\\\a=1,~b=-4,~c=3\\\\\Delta =(-4)^{2}-4\cdot 1\cdot 3=16-12=4\\\\\sqrt{\Delta} =\sqrt{4} =2\\\\\left (x_{1} =\dfrac{4-2}{2} =1~~\lor ~~x_{2} =\dfrac{4+2}{2} =3 \right)~~\land ~~x\in D\\[/tex]

[tex]~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Downarrow \\\\~~~~~~~~~~~~~~~~~~~~~~~~~~x=1~~\lor ~~x=3[/tex]