Zadanie 1.
Rozwiąż równania:
a) $$\frac{12}{1-9 x^2}=\frac{1-3 x}{1+3 x}+\frac{1+3 x}{3 x-1}$$;
b) $$5+\frac{96}{x^2-16}=\frac{2 x-1}{x+4}-\frac{3 x-1}{4-x}$$,
c) $$\frac{14}{x^2-9}+\frac{4-x}{3+x}=\frac{7}{x+3}-\frac{1}{3-x}$$;
d) $$\frac{30}{x^2-1}-\frac{13}{x^2+x+1}=\frac{7+18 x}{x^3-1}$$;
e) $$\frac{3}{x-1}-\frac{4 x-1}{x+1}=\frac{x^2+5}{x^2-1}-5$$;
f) $$\frac{x+3}{x+2}-\frac{x-3}{x-2}=\frac{x^2}{x^2-4}+1$$
a)
$$
\begin{gathered}
\frac{12}{1-9 x^2}=\frac{1-3 x}{1+3 x}+\frac{1+3 x}{3 x-1} \\
\frac{12}{(1-3 x)(1+3 x)}=\frac{1-3 x}{1+3 x}-\frac{1+3 x}{1-3 x}
\end{gathered}
$$
Zakładamy, że $$x \in R \backslash\left\{-\frac{1}{3}, \frac{1}{3}\right\}$$ :
$$
\begin{gathered}
12=(1-3 x)^2-(1+3 x)^2 \\
12=1-6 x+9 x^2-\left(1+6 x+9 x^2\right) \\
12=1-6 x+9 x^2-1-6 x-9 x^2 \\
12=-12 x \\
x=-1
\end{gathered}
$$
b)
$$
\begin{gathered}
5+\frac{96}{x^2-16}=\frac{2 x-1}{x+4}-\frac{3 x-1}{4-x} \\
5+\frac{96}{(x-4)(x+4)}=\frac{2 x-1}{x+4}+\frac{3 x-1}{x-4}
\end{gathered}
$$
Zakładamy, że $$x \in R \backslash\{-4,4\}$$.
$$
\begin{gathered}
5\left(x^2-16\right)+96=(2 x-1)(x-4)+(3 x-1)(x+4) \\
5 x^2-80+96=2 x^2-8 x-x+4+3 x^2+12 x-x-4 \\
5 x^2+16=5 x^2+2 x \\
2 x=16 \\
x=8
\end{gathered}
$$
c)
$$
\begin{aligned}
& \frac{14}{x^2-9}+\frac{4-x}{3+x}=\frac{7}{x+3}-\frac{1}{3-x} \\
& \frac{14}{(x-3)(x+3)}+\frac{4-x}{3+x}=\frac{7}{x+3}+\frac{1}{x-3}
\end{aligned}
$$
Zakładamy, że $$x \in R \backslash\{-3,3\}$$.
$$
\begin{gathered}
14+(4-x)(x-3)=7(x-3)+x+3 \\
14+4 x-12-x^2+3 x=7 x-21+x+3 \\
-x^2+7 x+2=8 x-18 \\
-x^2+7 x-8 x+2+18=0 \\
\quad-x^2-x+20=0 \\
\Delta=(-1)^2-4 \cdot(-1) \cdot 20=1+80=81, \quad \sqrt{\Delta}=9 \\
x_1=\frac{1-9}{2 \cdot(-1)}=\frac{-8}{-2}=4, \quad x_2=\frac{1+9}{2 \cdot(-1)}=\frac{10}{-2}=-5
\end{gathered}
$$
Zatem dane równanie posiada dwa rozwiązania: $$x_1=4, x_2=-5$$.
d)
$$
\begin{gathered}
\frac{30}{x^2-1}-\frac{13}{x^2+x+1}=\frac{7+18 x}{x^3-1} \\
\frac{30}{(x-1)(x+1)}-\frac{13}{x^2+x+1}=\frac{7+18 x}{(x-1)\left(x^2+x+1\right)}
\end{gathered}
$$
Zakładamy, że $$x \in R \backslash\{-1,1\}$$ (trójmian $$y=x^2+x+1$$ nie posiada miejsc zerowych).
$$
\begin{gathered}
30\left(x^2+x+1\right)-13(x-1)(x+1)=(7+18 x)(x+1) \\
30 x^2+30 x+30-13\left(x^2-1\right)=7 x+7+18 x^2+18 x \\
30 x^2+30 x+30-13 x^2+13=18 x^2+25 x+7 \\
17 x^2+30 x+43=18 x^2+25 x+7 \\
17 x^2-18 x^2+30 x-25 x+43-7=0 \\
-x^2+5 x+36=0 \\
\Delta=25-4 \cdot(-1) \cdot 36=25+144=169, \quad \sqrt{\Delta}=13 \\
x_1=\frac{-5+13}{2 \cdot(-1)}=\frac{8}{-2}=-4, \quad x_2=\frac{-5-13}{2 \cdot(-1)}=\frac{-18}{-2}=9
\end{gathered}
$$
Zatem dane równanie posiada dwa rozwiązania: $$x_1=-4, x_2=9$$.
e)
$$
\begin{gathered}
\frac{3}{x-1}-\frac{4 x-1}{x+1}=\frac{x^2+5}{x^2-1}-5 \\
\frac{3}{x-1}-\frac{4 x-1}{x+1}=\frac{x^2+5}{(x-1)(x+1)}-5
\end{gathered}
$$
Zakładamy, że $$x \in R \backslash\{-1,1\}$$.
$$
\begin{gathered}
3(x+1)-(4 x-1)(x-1)=x^2+5-5(x-1)(x+1) \\
3 x+3-\left(4 x^2-4 x-x+1\right)=x^2+5-5\left(x^2-1\right) \\
3 x+3-4 x^2+4 x+x-1=x^2+5-5 x^2+5 \\
-4 x^2+8 x+2=-4 x^2+10 \\
8 x=8 \\
x=1
\end{gathered}
$$
Zatem równanie nie posiada rozwiazania (bo z założenia $$x \in R \backslash\{-1,1\}$$.
f)
$$
\begin{gathered}
\frac{x+3}{x+2}-\frac{x-3}{x-2}=\frac{x^2}{x^2-4}+1 \\
\frac{x+3}{x+2}-\frac{x-3}{x-2}=\frac{x^2}{(x-2)(x+2)}+1
\end{gathered}
$$
Zakładamy, że $$x \in R \backslash\{-2,2\}$$.
$$
\begin{gathered}
(x+3)(x-2)-(x-3)(x+2)=x^2+(x-2)(x+2) \\
x^2-2 x+3 x-6-\left(x^2+2 x-3 x-6\right)=x^2+x^2-4 \\
x^2+x-6-x^2-2 x+3 x+6=2 x^2-4 \\
2 x=2 x^2-4 \\
2 x^2-2 x-4=0 \\
x^2-x-2=0 \\
\Delta=(-1)^2-4 \cdot 1 \cdot(-2)=1+8=9, \quad \sqrt{\Delta}=3 \\
x_1=\frac{1+3}{2 \cdot 1}=\frac{4}{2}=2, \quad x_2=\frac{1-3}{2 \cdot 1}=\frac{-2}{2}=-1
\end{gathered}
$$
Liczba 2 nie jest rozwiązaniem danego równania, poniewaz zalłładalísmy, zee $$x \in R \backslash\{-2,2\}$$. Zatem dane rómanie posiada dokładnie jedno rozwiązanie: $$x=-1$$.
Zadanie 2.
Rozwiąż równania:
a) $$\frac{5}{x^2-4}-\frac{8}{x^2-1}=\frac{2}{x^2-3 x+2}-\frac{20}{x^2-3 x+2}$$
b) $$\frac{20+x}{2 x-2}-\frac{9 x^2+x+2}{6 x^2-6}=\frac{5-3 x}{x+1}-\frac{10-4 x}{3 x+3}$$;
c) $$\frac{1}{x^3-x^2+x-1}-\frac{4}{x+1}=\frac{x^2+10 x}{x^4-1}-\frac{4 x^2+21}{x^3+x^2+x+1}$$.
a)
$$
\begin{gathered}
\frac{5}{x^2-4}-\frac{8}{x^2-1}=\frac{2}{x^2-3 x+2}-\frac{2 \theta}{x^2-3 x+2} \\
\frac{5}{(x-2)(x+2)}-\frac{8}{(x-1)(x+1)}=\frac{2}{(x-1)(x-2)}-\frac{20}{(x-1)(x-2)} \\
\text { Zakiadamy, ze } x \in R \backslash\{-2,-1,1,2\} \\
\frac{5}{(x-2)(x+2)}-\frac{8}{(x-1)(x+1)}=\frac{-18}{(x-1)(x-2)} \\
5(x-1)(x+1)-8(x-2)(x+2)=-18(x+1)(x+2) \\
5\left(x^2-1\right)-8\left(x^2-4\right)=-18\left(x^2+2 x+x+2\right) \\
5 x^2-5-8 x^2+32=-18\left(x^2+3 x+2\right) \\
-3 x^2+27=-18 x^2-54 x-36 \\
-3 x^2+18 x^2+54 x+27+36=0 \\
15 x^2+54 x+63=0 \\
5 x^2+18 x+21=0 \\
\Delta=(18)^2-4 \cdot 5 \cdot 21=324-420=-90
\end{gathered}
$$
Stąd wynika, że dane równanie nie posiada rozwiązania.
b)
$$
\begin{aligned}
& \frac{20+x}{2 x-2}-\frac{9 x^2+x+2}{6 x^2-6}=\frac{5-3 x}{x+1}-\frac{10-4 x}{3 x+3} \\
& \frac{20+x}{2(x-1)}-\frac{9 x^2+x+2}{6\left(x^2-1\right)}=\frac{5-3 x}{x+1}-\frac{10-4 x}{3(x+1)} \\
& \frac{20+x}{2(x-1)}-\frac{9 x^2+x+2}{6(x-1)(x+1)}=\frac{5-3 x}{x+1}-\frac{10-4 x}{3(x+1)}
\end{aligned}
$$
Zaktadamy, że $$x \in R \backslash\{-1,1\}$$.
$$
\begin{aligned}
& 3(x+1)(20+x)-\left(9 x^2+x+2\right)=6(x-1)(5-3 x)-2(x-1)(10-4 x) \\
& 3\left(20 x+x^2+20+x\right)-9 x^2-x-2=6\left(5 x-3 x^2-5+3 x\right)-2\left(10 x-4 x^2-10+4 x\right) \\
& 3\left(x^2+21 x+20\right)-9 x^2-x-2=6\left(-3 x^2+8 x-5\right)-2\left(-4 x^2+14 x-10\right)
\end{aligned}
$$
$$
\begin{gathered}
3 x^2+63 x+60-9 x^2-x-2=-18 x^2+48 x-30+8 x^2-28 x+20 \\
-6 x^2+62 x+58=-10 x^2+20 x-10 \\
-6 x^2+10 x^2+62 x-20 x+58+10=0 \\
4 x^2+42 x+68=0 \\
2 x^2+21 x+34=0 \\
\Delta=(21)^2-4 \cdot 2 \cdot 34=441-272=169, \quad \sqrt{\Delta}=13 \\
x_1=\frac{-21-13}{2 \cdot 2}=\frac{-34}{4}=-\frac{17}{2}, \quad x_2=\frac{-21+13}{2 \cdot 2}=\frac{-8}{4}=-2
\end{gathered}
$$
Zatem dane równanie posiada dwa rozwiązania: $$x_1=-\frac{17}{2}, x_2=-2$$.
c)
$$
\begin{gathered}
\frac{1}{x^3-x^2+x-1}-\frac{4}{x+1}=\frac{x^2+10 x}{x^4-1}-\frac{4 x^2+21}{x^3+x^2+x+1} \\
\frac{1}{x^2(x-1)+(x-1)}-\frac{4}{x+1}=\frac{x^2+10 x}{\left(x^2-1\right)\left(x^2+1\right)}-\frac{4 x^2+21}{x^2(x+1)+(x+1)} \\
\frac{1}{(x-1)\left(x^2+1\right)}-\frac{4}{x+1}=\frac{x^2+10 x}{(x-1)(x+1)\left(x^2+1\right)}-\frac{4 x^2+21}{(x+1)\left(x^2+1\right)}
\end{gathered}
$$
Zakładamy, że $$x \in R \backslash\{-1,1\}$$ (trójmian $$y=x^2+1$$ nie posiada miejsc zerowych).
$$
\begin{gathered}
x+1-4(x-1)\left(x^2+1\right)=x^2+10 x-\left(4 x^2+21\right)(x-1) \\
x+1-4\left(x^3+x-x^2-1\right)=x^2+10 x-\left(4 x^3-4 x^2+21 x-21\right) \\
x+1-4 x^3-4 x+4 x^2+4=x^2+10 x-4 x^3+4 x^2-21 x+21 \\
-4 x^3+4 x^2-3 x+5=-4 x^3+5 x^2-11 x+21 \\
-4 x^3+4 x^3+4 x^2-5 x^2-3 x+11 x+5-21=0 \\
-x^2+8 x-16=0 \\
x^2-8 x+16=0 \\
(x-4)^2=0 \\
x=4
\end{gathered}
$$
Zatem równanie posiada dokiadnie jedno rozwiązanie: $$x=4$$.
