[latex]frac {2}{5x+16} = frac {-3}{x^2-4} \\ D_r: 5x
eq 16 ; x^2
eq 4 ; D_r: x in R setminus left { (-2) ; 2; frac{16}{5}
ight } [/latex]
[latex]2(x^2-4)=-3(5x+16)\\ 2x^2-8=-15x-48\\ 2x^2+15x+48-8 =0\\ 2x^2+15x+40=0 [/latex]
[latex]a=2 b=15 c=40 \\ Delta = b^2 - 4ac= 15^2-8cdot 40=225-320 =-95[/latex]
Wyróżnik równania kwadratowego ( delta) jest ujemny, wobec tego nie ma rozwiązań w zbiorze liczb rzeczywistych.
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[latex]frac{2x+5}{x^2-1}=frac{1}{x-1} \\ D_r: xin R setminus {(-1);1}\\ frac{2x+5}{(x+1)(x-1)}=frac{1(x+1)}{(x-1)(x+1)}\\ 2x+5=x+1\\ 2x-x=1-5\\ x=(-4)[/latex]
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[latex]frac 1{1-x}+frac{x}{x-1}=1\\ D_r: xin R setminus {1}\\ (x-1)=x(1-x)\\ x=frac{x-1}{1-x}=frac{-1+x}{1-x}=frac{-1(1-x)}{1(1-x)}=frac {-1}{1}=-1\\ old {x=(-1)}[/latex]
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[latex]frac{2x}{2x+3} -1 = frac {2x}{2x-3}\\ D_r: xin Rsetminus left{-frac 32 ; frac 32
ight }\\ frac{2x}{2x+3}-frac {2x}{2x-3} = 1\\ frac{2x(2x-3)}{(2x+3)(2x-3)}-frac {2x(2x+3)}{(2x-3)(2x+3)} = 1\\ frac{2x(2x-3)-2x(2x+3)}{(2x+3)(2x-3)} = 1\\ frac{4x^2-6x- 4x^2-6x}{4x^2-9} = 1\\ {4x^2-6x- 4x^2-6x}={4x^2-9} \\ {-12x}={4x^2-9} \\ 4x^2+12x-9=0\\ a=4 b=12 c=(-9)\\ Delta = b^2-4ac = 12^2-4cdot4cdot (-9)= 144+144= 288\\ [/latex]
[latex]sqrt {Delta} = sqrt{288}=sqrt{144 cdot 2} = 12sqrt 2 \\ x_1=frac {-b-sqrt Delta}{2a} =frac {-12-12sqrt 2}{8} =frac {-3-3sqrt 2}{2} \\ x_2=frac {-b+sqrt Delta}{2a} =frac {-12+12sqrt 2}{8} =frac {-3+3sqrt 2}{2} [/latex]
Pierwiastki tego równania nie należą do liczb całkowitych
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[latex]frac{6}{x}-1=frac{2}{x-1}\\ D_r:xin R setminus{0;1}\\ frac{6}{x}-frac {2}{x-1} =1\\ frac{6(x-1)}{x(x-1)}-frac {2x}{x(x-1)} =1\\ frac{6x-6-2x}{x(x-1)} =1\\ frac{4x-6}{x^2-x} =1\\ {4x-6}=x^2-x\\ x^2-x-4x+6=0\\ x^2-5x+6 = 0\\ a=1 b=-5 c=6\\ Delta = b^2 - 4ac= 25-24=1\\ sqrt Delta = 1\\ x_1= frac {-b-sqrt Delta}{2a}= frac {5-1}{2}=2\\ x_2= frac {-b+sqrt Delta}{2a}= frac {5+1}{2}=3[/latex]
Pierwiastkami tego równania są liczby 2 i 3.
