potrzebne rozwiązanie do zadań 1,2 i 5. tresc w załączniku

1 a) [latex]frac{x+2}{x^2-5x+6}-frac{1}{x+3}=[/latex] [latex]frac{x+2}{x^2-3x-2x+6}-frac{1}{x+3}=[/latex] [latex]frac{x+2}{x(x-3)-2(x-3)}-frac{1}{x+3}=[/latex] [latex]frac{x+2}{(x-3)(x-2)}-frac{1}{x+3}=[/latex] [latex]frac{(x+2)(x+3)}{(x-3)(x-2)(x+3)}-frac{(x-3)(x-2)}{(x-3)(x-2)(x+3)}=[/latex] [latex]frac{(x+2)(x+3)-(x-3)(x-2)}{(x-3)(x-2)(x+3)}=[/latex] [latex]frac{x^2+3x+2x+6-(x^2-2x-3x+6)}{(x-3)(x-2)(x+3)}=[/latex] [latex]frac{x^2+3x+2x+6-x^2+2x+3x-6}{(x-3)(x-2)(x+3)}=[/latex] [latex]frac{10x}{(x-3)(x-2)(x+3)}[/latex] Założenie: [latex]x eq -3,x eq 2,x eq 3[/latex] b) [latex]frac{2x^2-3x-2}{x^3+3x^2-4x}cdotfrac{x^3-x^2}{2x+1}=[/latex] [latex]frac{2x^2-4x+x-2}{x(x^2+3x-4)}cdotfrac{x^2(x-1)}{2x+1}=[/latex] [latex]frac{2x(x-2)+(x-2)}{x(x^2-x+4x-4)}cdotfrac{x^2(x-1)}{2x+1}=[/latex] [latex]frac{(x-2)(2x+1)}{x[x(x-1)+4(x-1)]}cdotfrac{x^2(x-1)}{2x+1}=[/latex] [latex]frac{(x-2)(2x+1)}{x(x-1)(x+4)}cdotfrac{x^2(x-1)}{2x+1}=[/latex] [latex]frac{x(x-2)}{(x+4)}[/latex] Założenie: [latex]x eq -4,x eq - frac{1}{2},x eq 0,x eq 1[/latex] ================== 2 a) [latex]frac{2x-11}{x-1}=5[/latex] Dziedzina: [latex]x eq 1[/latex] [latex]2x-11=5(x-1)[/latex] [latex]2x-11=5x-5[/latex] [latex]2x-5x=-5+11[/latex] [latex]-3x=6 /:(-3)[/latex] [latex]x=-2[/latex] b) [latex]frac{2x}{x+3}-frac{4}{x+1}=0[/latex] Dziedzina: [latex]x eq -3,x eq -1[/latex] [latex]frac{2x}{x+3}=frac{4}{x+1}[/latex] [latex]2x(x+1)=4(x+3)[/latex] [latex]2x^2+2x=4x+12[/latex] [latex]2x^2+2x-4x-12=0[/latex] [latex]2x^2-2x-12=0 /:2[/latex] [latex]x^2-x-6=0[/latex] [latex]Delta=(-1)^2-4 cdot 1 cdot (-6)=1+24=25[/latex] [latex]sqrt{Delta}= sqrt{25}=5[/latex] [latex]x_1= frac{1-5}{2}=frac{-4}{2}=-2[/latex] [latex]x_2= frac{1+5}{2}=frac{6}{2}=3[/latex] ================== 5 [latex]a_2=frac{-4sqrt3}{3}[/latex] [latex]a_5=-frac{4}{9}[/latex] [latex]a_2=a_1q[/latex] [latex]a_5=a_1q^4[/latex] [latex]egin{cases}a_1q=frac{-4sqrt3}{3}\ a_1q^4=-frac{4}{9}end{cases}[/latex] [latex]egin{cases}a_1q=frac{-4sqrt3}{3}\ a_1q cdot q^3=-frac{4}{9}end{cases}[/latex] [latex]egin{cases}a_1q=frac{-4sqrt3}{3}\ frac{-4sqrt3}{3} cdot q^3=-frac{4}{9} /:left(frac{-4sqrt3}{3} ight) end{cases}[/latex] [latex]egin{cases}a_1q=frac{-4sqrt3}{3}\ q^3=-frac{4}{9} cdot left( - frac{3}{4 sqrt{3} } ight) end{cases}[/latex] [latex]egin{cases}a_1q=frac{-4sqrt3}{3}\ q^3= frac{1}{3sqrt{3} } end{cases}[/latex] [latex]egin{cases}a_1q=frac{-4sqrt3}{3}\ q^3= frac{1}{3 cdot 3^{ frac{1}{2} }} end{cases}[/latex] [latex]egin{cases}a_1q=frac{-4sqrt3}{3}\ q^3= frac{1}{3^{ frac{3}{2} }} end{cases}[/latex] [latex]egin{cases}a_1q=frac{-4sqrt3}{3}\ q^3= 3^{-frac{3}{2} } end{cases}[/latex] [latex]egin{cases}a_1q=frac{-4sqrt3}{3}\ q= sqrt[3]{ 3^{-frac{3}{2} }} end{cases}[/latex] [latex]egin{cases}a_1q=frac{-4sqrt3}{3}\ q=left(3^{-frac{3}{2} } ight)^{ frac{1}{3} } end{cases}[/latex] [latex]egin{cases}a_1q=frac{-4sqrt3}{3}\ q=left(3^{-frac{3}{2} } ight)^{ frac{1}{3} } end{cases}[/latex] [latex]egin{cases}a_1q=frac{-4sqrt3}{3}\ q=3^{-frac{1}{2} } end{cases}[/latex] [latex]egin{cases}a_1q=frac{-4sqrt3}{3}\ q= frac{1}{ sqrt{3} } end{cases}[/latex] [latex]egin{cases}a_1q=frac{-4sqrt3}{3}\ q= frac{ sqrt{3} }{3} end{cases}[/latex] [latex]egin{cases}a_1 cdot frac{ sqrt{3} }{3} =frac{-4sqrt3}{3} /:frac{ sqrt{3} }{3} \ q= frac{ sqrt{3} }{3} end{cases}[/latex] [latex]egin{cases}a_1 =frac{-4sqrt3}{3} cdot frac{3}{ sqrt{3} } \ q= frac{ sqrt{3} }{3} end{cases}[/latex] [latex]egin{cases}a_1 =-4\ q= frac{ sqrt{3} }{3} end{cases}[/latex] Wzór ogólny ciągu jest postaci: [latex]a_n=-4 cdot left( frac{ sqrt{3} }{3} ight) ^{n-1}[/latex]

Treść w załączniku ,Proszę o rozwiązanie zadań od 1-4 ...Potrzebne na teraz daje naj >.<"

Treść w załączniku ,Proszę o rozwiązanie zadań od 1-4 ...Potrzebne na teraz daje naj >.<"...