Řešené příklady
$\huge\frac{12}{1-9x^{2}}=\frac{1-3x}{1+3x}+\frac{1+3x}{3x-1}\\$
$\Large\frac{12}{1-9x^{2}}=\frac{1-3x}{1+3x}+\frac{1+3x}{3x-1}\; \; \normalsize /\times(1-3x)(1+3x)$
$\Large\frac{12}{(1-3x)(1+3x)}=\frac{1-3x}{1+3x}+\frac{1+3x}{-(1-3x)}$
$12=(1-3x)(1-3x)-(1+3x)(1+3x)$
$12=(1-3x)^{2}-(1+3x)^{2}$
$12=1-6x+9x^{2}-(1+6x+9x^{2})$
$12=1-6x+9x^{2}-1-6x-9x^{2})$
$12=-12x \; \; /\div(-12x)$
$x=-1\\$
$x\neq\pm\frac{1}{3}\\$
$K = \left \{ -1 \right \}$
$\huge\frac{12x^{2}+30x-21}{16x^{2}-9}=\frac{3x-7}{3-4x}+\frac{6x+5}{4x+3}\\$
$\Large\frac{12x^{2}+30x-21}{16x^{2}-9}=\frac{3x-7}{-(4x-3)}+\frac{6x+5}{4x+3} \; \; \normalsize /\times(4x-3)(4x+3)$
$12x^{2}+30x-21=-(3x-7)(4x+3)+(6x+5)(4x-3)$
$12x^{2}+30x-21=-(12x^{2}+9x-28x-21)+24x^{2}-18x+20x-15$
$12x^{2}+30x-21=-12x^{2}-9x+28x+21+24x^{2}-18x+20x-15$
$30x-21=21x+6$
$9x=27\; \; /\div 9$
$x=3\\$
$x\neq\pm\frac{3}{4}\\$
$K = \left \{3 \right \}$
$\huge\frac{2x-5}{3x-4}-\frac{4x-5}{6x-1}=0\\$
$\Large\frac{2x-5}{3x-4}-\frac{4x-5}{6x-1}=0 \; \; \normalsize /\times(3x-4)(6x-1)$
$12x^{2}-2x-30x+5-(12x^{2}-16x-15x+20)=0$
$12x^{2}-2x-30x+5-12x^{2}+16x+15x-20=0$
$-x-15=0$
$-x=15\; \; /\times(-1)$
$x=-15\\$
$x\neq\frac{4}{3}\\$
$x\neq\frac{1}{6}\\$
$K = \left \{-15 \right \}$
$\huge\frac{x+1}{x-1}+\frac{2}{x+2}-1=\frac{6}{x^{2}+x-2}\\$
$\Large\frac{x+1}{x-1}+\frac{2}{x+2}-1=\frac{6}{x^{2}+x-2} \; \; \normalsize /\times(x-1)(x+2)$
$(x+1)(x+2)+2(x-1)-1(x-1)(x+2)=6$
$x^{2}+2x+x+2+2x-2-(x^{2}+x-2)=6$
$x^{2}+5x-x^{2}-x+2=6$
$4x=4 \; \; /\div(4)$
$x=1\\$
$x\neq1\\$
$x\neq-2\\$
$K=\left \{\varnothing \right \}$
$\huge\frac{3+4x}{x^{2}+x}-1=\frac{3}{x}-\frac{x}{x+1}\\$
$\Large\frac{3+4x}{x(x+1)}-1=\frac{3}{x}-\frac{x}{x+1} \; \; \normalsize /\times x(x+1)$
$3+4x-x(x+1)=3(x+1)-x^{2}$
$3+4x-x^{2}-x=3x+3-x^{2}$
$3+3x=3x+3$
$0=0$
$x\neq0\\$
$x\neq-1\\$
$K=R - \left \{0;1 \right \}$
$\huge\frac{5}{2x-3}-\frac{3x-8}{4x-6}=\frac{7}{9}-\frac{6x-1}{10x-15}\\$
$\Large\frac{5}{2x-3}-\frac{3x-8}{2(2x-3)}=\frac{7}{9}-\frac{6x-1}{5(2x-3)} \; \; \normalsize /\times 90(2x-3)$
$5\times 90 - 45(3x-8)=70(2x-3)-18(6x-1)$
$450 - 135x+360=140x-210-108x+18$
$810 - 135x=32x-192$
$-167x=-902\; \; \normalsize /\div -167$
$x=6$
$x\neq \frac{3}{2}\\$
$K=\left \{6 \right \}$
$\huge\frac{3}{x-3}+\frac{5}{x-5}-\frac{34}{x^{2}-8x+15}=0\\$
$\Large\frac{3}{x-3}+\frac{5}{x-5}-\frac{34}{(x-3)(x-5)}=0 \; \; \normalsize /\times(x-3)(x-5)$
$3(x-5)+5(x-3)-34=0$
$3x-15+5x-15-34=0$
$8x-64=0$
$8x=64\; \; /\div 8$
$x=8$
$x\neq 3\\$
$x\neq 5\\$
$K=\left \{8 \right \}$
$\huge\frac{11+3x}{x+3}+\frac{5x}{x-4}-\frac{x}{x^{2}-x-12}+2=0\\$
$\Large\frac{11+3x}{x+3}+\frac{5x}{x-4}-\frac{x}{(x+3)(x-4)}+2=0 \; \; \normalsize /\times(x+3)(x-4)$
$(11+3x)(x-4)-5x(x+3)+x+2(x^{2}-x-12)=0$
$11x-44+3x^{2}-12x-5x^{2}-15x+x+2x^{2}-2x-24=0$
$-17x-68=0$
$-17x=68\; \; /\div -17$
$x=-4$
$x\neq -3\\$
$x\neq 4\\$
$K=\left \{-4 \right \}$
$\huge\frac{1}{x-2}-\frac{x-3}{x+4}=\frac{6}{x^{2}+2x-8}+2-1\\$
$\Large\frac{1}{x-2}-\frac{x-3}{x+4}=\frac{6}{(x-2)(x+4)}+2-1 \; \; \normalsize /\times(x-2)(x+4)$
$(x+4)-(x-3)(x-2)=6-(x^{2}+2x-8)$
$x+4-(x^{2}-2x-3x+6)=6-x^{2}-2x+8$
$x+4-x^{2}+2x+3x-6=6-x^{2}-2x+8$
$6x-2=-2x+14$
$8x=16\; \; /\div 8$
$x=2\\$
$x\neq 2$
$x\neq -4\\$
$K=\left \{\varnothing \right \}$
$\huge\frac{1}{x}+\frac{2}{x+1}+\frac{3}{x+2}=\frac{6x^{2}+20x-8}{x^{3}+3x^{2}+2x}+2-1\\$
$\Large\frac{1}{x}+\frac{2}{x+1}+\frac{3}{x+2}=\frac{6x^{2}+20x-8}{x(x+1)(x+2)}+2-1 \; \; \normalsize /\times x(x+1)(x+2)$
$(x+1)(x+2)+2x(x+2)+3x(x+1)=6x^{2}+20x-8$
$x^{2}+2x+x+2+2x^{2}+4x+3x^{2}+3x=6x^{2}+20x-8$
$10x+2=-20x-8$
$-10x=-10\; \; /\div -10$
$x=1\\$
$x\neq 0$
$x\neq -1\\$
$x\neq -2\\$
$K=\left \{1 \right \}$
$\huge\frac{1}{x-3}-\frac{1}{x+2}=\frac{5}{x^{2}+6}\\$
$\Large\frac{1}{x-3}-\frac{1}{x+2}=\frac{5}{x^{2}+6} \; \; \normalsize /\times(x-3)(x+2)(x^{2}+6)$
$(x+2)(x^{2}+6)-(x-3)(x^{2}+6)=5(x-3)(x+2)$
$x^{3}+6x+2x^{2}+12-(x^{3}+6x-3x^{2}-18)=5(x^{2}+2x-3x-6)$
$x^{3}+6x+2x^{2}+12-x^{3}-6x+3x^{2}+18=5x^{2}+10x-15x-30$
$60=-5x\; \; /\div -5$
$x=-12\\$
$x\neq 3$
$x\neq -2\\$
$K=\left \{-12 \right \}$
$\huge\frac{x+3}{x+1}+\frac{x+2}{x-3}=2+\frac{7x-1}{x^{2}-2x-3}\\$
$\Large\frac{x+3}{x+1}+\frac{x+2}{x-3}=2+\frac{7x-1}{(x+1)(x-3)} \; \; \normalsize /\times(x+1)(x-3)$
$(x+3)(x-3)+(x+2)(x+1)=2(x^{2}-2x-3)+7x-1$
$x^{2}-9+x^{2}+3x+2=2x^{2}-4x-6+7x-1$
$3x-7=3x-7\\$
$0=0\\$
$x\neq -1$
$x\neq 3\\$
$K=R-\left \{-1;3 \right \}$
$\huge\frac{3}{(2x+5)^{2}}+\frac{4}{(2x+1)^{2}}=\frac{7}{(2x+5)(2x+1)}\\$
$\Large\frac{3}{(2x+5)^{2}}+\frac{4}{(2x+1)^{2}}=\frac{7}{(2x+5)(2x+1)} \; \; \normalsize /\times(2x+5)^{2}(2x+1)^{2}$
$3(2x+1)^{2}+4(2x+5)^{2}=7(2x+5)(2x+1)$
$12x^{2}+12x+3+16x^{2}+80x+100=28x^{2}+84x+35$
$92x+103=84x+35$
$8x=-68\; \; /\div 8$
$\large x=-\frac{68}{8}=-\frac{17}{2}\\$
$\large x\neq -\frac{5}{2}$
$\large x\neq -\frac{1}{2}\\$
$K=\left \{-\frac{17}{2} \right \}$
$\huge\frac{3}{1-x^{2}}=\frac{2}{(1+x)^{2}}-\frac{5}{(1-x)^{2}}\\$
$\Large\frac{3}{(1-x)(1+x)}=\frac{2}{(1+x)(1+x)}-\frac{5}{(1-x)(1-x)} \; \; \normalsize /\times(1+x)(1+x)(1-x)(1-x)$
$3(1-x)(1+x)=2(1-x)(1-x)-5(1+x)(1+x)$
$3(1-x^{2})=2(1-2x+x^{2})-5(1+2x+x^{2})$
$3-3x^{2}=2-4x+2x^{2}-5-10x-5x^{2}$
$3=-3-14x$
$14x=-6\; \; /\div 14$
$\large x=-\frac{6}{14}=-\frac{3}{7}\\$
$x\neq\pm1\\$
$K=\left \{-\frac{3}{7} \right \}$
$\huge\frac{1}{(3-2x)^{2}}=\frac{3}{9-4x^{2}}+\frac{4}{(3+2x)^{2}}\\$
$\Large\frac{1}{(3-2x)(3-2x)}=\frac{3}{(3-2x)(3+2x)}+\frac{4}{(3+2x)(3+2x)} \; \; \normalsize /\times(3+2x)(3+2x)(3-2x)(3-2x)$
$(3+2x)^{2}=3(9-4x^{2})+4(3-2x)^{2}$
$9+12x+4x^{2}=27-12x^{2}+4(9-12x+4x^{2})$
$9+12x+4x^{2}=27-12x^{2}+36-48x+16x^{2})$
$9+12x=63-48x$
$60x=54\; \; /\div 60$$
$\large x=\frac{54}{60}=\frac{9}{10}\\$
$x\neq\pm\frac{3}{2}\\$
$K=\left \{\frac{9}{10} \right \}$
$\huge\frac{2}{(1-3x)(3x+11)}=\frac{1}{(3x-1)^{2}}-\frac{3}{(3x+11)^{2}}\\$
$\Large-\frac{2}{(3x-1)(3x+11)}=\frac{1}{(3x-1)(3x-1)}-\frac{3}{(3x+11)(3x+11)} \; \; \normalsize /\times(3x-1)(3x-1)(3x+11)(3x+11)$
$-2(3x-1)(3x+11)=(3x+11)^{2}-3(3x-1)^{2}$
$-2(9x^{2}+33x-3x-11)=9x^{2}+66x+121-3(9x^{2}-6x+1)$
$-18x^{2}-60x+22=9x^{2}+66x+121-27x^{2}+18x-3$
$-60x+22=84x+118$
$-144x=96\; \; /\div -144$$
$\large x=-\frac{96}{144}=-\frac{48}{72}=-\frac{6}{9}=-\frac{2}{3}\\$
$x\neq\frac{1}{3}$
$x\neq-\frac{11}{3}$
$K=\left \{-\frac{2}{3} \right \}$
Zdroj: JANEČEK, František. Sbírka úloh z matematiky pro střední školy: výrazy, rovnice, nerovnice a jejich soustavy : sbírka úloh k opakování a procvičování učiva matematiky střední školy. 4. vyd. Praha: Prometheus, 1998. Pomocné knihy pro žáky (Prometheus). ISBN 80-7196-076-4. (strana 68, cvičení 1.3.)