Odpowiedź:
[tex]\boxed{D_{f(x)} =R - \{ 3,~~1\dfrac{2}{3} \}}[/tex]
[tex]\boxed{f(x)=1~~dla~~x=4}[/tex]
Szczegółowe wyjaśnienie:
Dane:
- [tex]f(x)=\dfrac{2x^{2} -7x+3}{3x^{2} -14x+15}[/tex]
Wyznaczamy dziedzinę f(x):
[tex]f(x)=\dfrac{2x^{2} -7x+3}{3x^{2} -14x+15}\\\\zal.\\\\3x^{2} -14x+15\neq 0\\\\a=3,~~b=-14,~~c=15\\\\\Delta=b^{2} -4ac\\\\\Delta=(-14)^{2} -4\cdot 3\cdot 15=196-180=16\\\\\sqrt{\Delta} =\sqrt{16} =4\\\\x_{1} \neq \dfrac{-b-\sqrt{x} }{2a} ~~\land~~x_{2} \neq \dfrac{-b+\sqrt{x} }{2a} \\\\\\x_{1} \neq \dfrac{14-4 }{2\cdot 3} ~~\land~~x_{2} \neq \dfrac{14+4 }{2\cdot 3} \\\\x_{1} \neq 1\dfrac{2}{3} ~~\land~~x_{2} \neq 3 \\\\\\\boxed{D_{f(x)} =R - \{ 3,~~1\dfrac{2}{3} \}}[/tex]
Rozwiązujemy równanie:
[tex]f(x)=\dfrac{2x^{2} -7x+3}{3x^{2} -14x+15}~~\land~~f(x)=1\\\\\\\dfrac{2x^{2} -7x+3}{3x^{2} -14x+15}=1\\\\\\\dfrac{2x^{2} -7x+3}{3x^{2} -14x+15}-1=0\\\\\\\dfrac{2x^{2} -7x+3-(3x^{2} -14x+15)}{3x^{2} -14x+15}=0\\\\\\\dfrac{2x^{2} -7x+3-3x^{2} +14x-15}{3x^{2} +14x+15}=0\\\\\\\dfrac{-x^{2} +7x-12}{3x^{2} -14x+15}=0~~\Leftrightarrow~~-x^{2} -21x-12=0\\\\\\-x^{2} +7x-12=0~~\mid \cdot (-1)\\\\x^{2} -7x+12=0\\\\a=1,~~b=-7,~~c=12\\\\\Delta=b^{2} -4ac\\\\[/tex]
[tex]\Delta = (-7)^{2} -4\cdot 1 \cdot 12 = 49-48=1\\\\\sqrt{\Delta} =1\\\\x_{1} =\dfrac{-b-\sqrt{\Delta} }{2a}~~\lor~~x_{2} =\dfrac{-b+\sqrt{\Delta} }{2a}\\\\\\x_{1} =\dfrac{7-1 }{2\cdot 1}=3~~\lor~~x_{2} =\dfrac{7+1 }{2\cdot 1}=4\\\\(~~x_{1} =3~~\lor~~x_{2} =4~~)~~\land~~x\in D_{f(x)} ~~\land~~D_{f(x)} =R - \{ 3,~~1\dfrac{2}{3} \} ~~\Rightarrow~~\boxed{x=4}[/tex]
