Бөлшек-рационал теңдеулерді шешу тәсілдері

$$\{\begin{array}{l}4x^2-7x-2=0,\\ x^2-5x+6\ne 0;\end{array}\{\begin{array}{l}{x}_1=-\frac{1}{4},x_2=2,\\ x\ne 2,x\ne 3\end{array}x=-\frac{1}{4}$$

Жауабы: $\left\{-{\displaystyle \frac{1}{4}}\right\}$.

$$\frac{3(9x-3)}{9x-6}=2+\frac{3x+1}{3x-2};\text{ММЖ: }x\ne \frac{2}{3}$$ $$\frac{9x-3}{3x-2}=2+\frac{3x+1}{3x-2}$$ $$9x-3=6x-4+3x+1$$ $$9x-3=9x-3$$ $$0\cdot x=0$$ $$x\ne \frac{2}{3}.$$

Жауабы: $x\ne {\displaystyle \frac{2}{3}}.$

$${\left(\frac{x^2+35}{x^2-49}\right)}^2={\left(\frac{12x}{x^2-49}\right)}^2,\text{ММЖ: }x\ne \pm 7$$ $$[\begin{array}{l}\frac{x^2+35}{x^2-49}=\frac{12x}{x^2-49},\\ \frac{x^2+35}{x^2-4}=-\frac{12x}{x^2-49};\end{array}\Rightarrow [\begin{array}{l}\frac{x^2-12x+35}{x^2-49}=0\\ \frac{x^2+12x+35}{x^2-49}=0\end{array}\Rightarrow$$ $$\Rightarrow [\begin{array}{l}{x}^2-12x+35=0\\ x^2+12x+35=0\end{array}\Rightarrow [\begin{array}{l}{x}_1=5,x_2=7,\\ x_3=-5,x_4=-7.\end{array}$$ ММЖ-ға сәйкес $x=\pm 5$ түбірін аламыз.

Жауабы: $x=\pm 5$

$$\frac{3}{x}+\frac{33}{x^2-11x}=\frac{x-4}{x-11},x\ne 0,x\ne 11.$$ $$\frac{3}{x}+\frac{33}{x(x-11)}-\frac{x-4}{x-11}=0$$ $$3(x-11)+33-x(x-4)=0$$ $$7x-x^2=0,$$ $$x_1=0-\text{ бөгде түбір}$$ $$x_2=7.$$

Жауабы: $\left\{7\right\}$

$$\frac{x+1}{x+3}+\frac{10x}{\left(x+3\right)\left(x-2\right)}-\frac{4}{x-2}=0\text{ММЖ: }x\ne -3,x\ne 2.$$ $$\left(x+1\right)\left(x-2\right)+10x-4\left(x+3\right)=0$$ $$x^2+5x-14=0$$ $$x_1=2-\text{бөгде түбір}$$ $$x_2=-7.$$

Жауабы: $\{-7\}$

$$\frac{1}{4\left(x+2\right)}-\frac{20x+1}{4\left(x-2\right)\left(x+2\right)}+\frac{7-5x}{{\left(x-2\right)}^2}=0,\text{ММЖ: }x\ne \pm 2$$ $${\left(x-2\right)}^2-\left(20x+1\right)\left(x-2\right)+\left(7-5x\right)\left(4x+8\right)=0$$ $$39x^2-23x-62=0$$ $$b=a+c\left(-23=39-62\right),$$ $$x_1=-1,x_2=\frac{62}{39}.$$

Жауабы: $\left\{-1;{\displaystyle \frac{62}{39}}\right\}$

$$\frac{3-x}{x^2+2x-3}-\frac{3(3-x)}{3x^2-2x-5}=0$$ $$(3-x)\left(\frac{1}{x^2+2x-3}-\frac{3}{3x^2-2x-5}\right)=0$$ $$(3-x)\cdot \frac{3x^2-2x-5-3x^2-6x+9}{\left(x^2+2x-3\right)\left(3x^2-2x-5\right)}=0$$ $$(3-x)\cdot \frac{-8x+4}{(x+3)(x-1)(3x-5)(x+1)}=0,x\ne \pm 1,x\ne -3,x\ne \frac{5}{3}.$$ $$[\begin{array}{l}3-x=0,\\ -8x+4=0;\end{array}[\begin{array}{l}x=3,\\ x=\frac{1}{2}\end{array}$$

Жауабы: $\left\{{\displaystyle \frac{1}{2}};3\right\}$.

$$3\left(x+\frac{1}{x^2}\right)-7\left(1+\frac{1}{x}\right)=0,\text{ММЖ: }x\ne 0.$$ $$3x\left(1+\frac{1}{x^3}\right)-7\left(1+\frac{1}{x}\right)=0$$ $$3x\left(1+\frac{1}{x}\right)\left(1-\frac{1}{x}+\frac{1}{x^2}\right)-7\left(1+\frac{1}{x}\right)=0$$ $$\left(1+\frac{1}{x}\right)\left(3x-3+\frac{3}{x}-7\right)=0$$ $$\left(1+\frac{1}{x}\right)\left(3x+\frac{3}{x}-10\right)=0$$ $$[\begin{array}{l}1+\frac{1}{x}=0,\\ 3x+\frac{3}{x}-10=0;\end{array}\Rightarrow [\begin{array}{l}\frac{1}{x}=-1,\\ 3x^2-10x+3=0;\end{array}\Rightarrow [\begin{array}{l}x=-1,\\ x_1=\frac{1}{3},x_2=3.\end{array}$$

Жауабы: $\left\{-1;{\displaystyle \frac{1}{3}};3\right\}$.

$$\frac{1}{x-1}-\frac{1}{x-2}=\frac{1}{x-3}-\frac{1}{x-4},$$ $$\text{ММЖ:}x\ne 1,x\ne 2,x\ne 3,x\ne 4.$$ $$\frac{1}{x-1}+\frac{1}{x-4}=\frac{1}{x-2}+\frac{1}{x-3}$$ $$\frac{x-4+x-1}{(x-1)(x-4)}=\frac{x-3+x-2}{(x-2)(x-3)}$$ $$\frac{2x-5}{(x-1)(x-4)}=\frac{2x-5}{(x-2)(x-3)}$$ $$(2x-5)\left(\frac{1}{(x-1)(x-4)}-\frac{1}{(x-2)(x-3)}\right)=0$$ $$(2x-5)\cdot \frac{x^2-5x+6-x^2+5x-4}{(x-1)(x-2)(x-3)(x-4)}=0$$ $$(2x-5)\cdot \frac{2}{(x-1)(x-2)(x-3)(x-4)}=0$$ $$2x-5=0$$ $$x=2,5.$$

Жауабы: $\left\{2,5\right\}$.

$$\frac{2}{x-2}+\frac{3}{x-3}=\frac{6}{x+6}-\frac{1}{x-1},$$ $$\text{ММЖ:}x\ne 6;x\ne 1;x\ne 2;x\ne 3.$$ $$\frac{5x-12}{x^2-5x+6}=\frac{5x-12}{x^2+5x-6}$$ $$\left(5x-12\right)\left(\frac{1}{x^2-5x+6}-\frac{1}{x^2+5x-6}\right)=0$$ $$\left(5x-12\right)\cdot \frac{x^2+5x-6-x^2+5x-6}{\left(x^2-5x+6\right)\left(x^2+5x-6\right)}=0$$ $$\left(5x-12\right)\left(10x-12\right)=0$$ $$x_1=2,4;x_2=1,2.$$

Жауабы: $\left\{1,2;2,4\right\}$.

$$\frac{x^2+2}{3x-2}-\frac{3x-2}{x^2+2}=\frac{8}{3},x\ne \frac{2}{3}.$$ $$a=\frac{x^2+2}{3x-2}$$ $$\frac{3x-2}{x^2+2}=\frac{1}{a}$$ $$a-\frac{1}{a}=\frac{8}{3}$$ $$3a^2-8a-3=0$$ $$a_1=3,a_2=-\frac{1}{3}$$ $$[\begin{array}{l}\frac{x^2+2}{3x-2}=3,\\ \frac{x^2+2}{3x-2}=-\frac{1}{3};\end{array}\Rightarrow [\begin{array}{l}{x}^2-9x+8=0,\\ 3x^2+3x+4=0;\end{array}\Rightarrow [\begin{array}{l}{x}_1=1,x_2=8,\\ \text{шешімі жоқ.}\end{array}$$

Жауабы: $\{1;8\}$.

$$\frac{x^2+2x+3}{x}-\frac{6x}{x^2+2x+3}=5,\text{ММЖ: }x\ne 0.$$

Жаңа айнымалы енгіземіз: $a={\displaystyle \frac{x^2+2x+3}{x}}$.

Мынадай теңдеу аламыз:

$$a-\frac{6}{a}=5$$ $$a^2-5a-6=0$$ $$a_1=-1,a_2=6$$ $$[\begin{array}{l}\frac{x^2+2x+3}{x}=-1,\\ \frac{x^2+2x+3}{x}=6\end{array}\Rightarrow [\begin{array}{l}{x}^2+3x+3=0,\\ x^2-4x+3=0\end{array}\Rightarrow [\begin{array}{l}\text{шешімі жоқ},\\ x_1=1,x_2=3.\end{array}$$

Жауабы: $\{1;3\}$.

$$\frac{1}{x(x+6)}-\frac{1}{{(x+3)}^2}=-\frac{9}{20},\text{ММЖ: }x\ne 0,x\ne -6,x\ne -3$$ $$\frac{1}{x^2+6x}-\frac{1}{x^2+6x+9}+\frac{9}{20}=0$$ $$a=x^2+6x$$ $$\frac{1}{a}-\frac{1}{a+9}+\frac{9}{20}=0$$ $$20a+180-20a+9a^2+81a=0$$ $$a^2+9a+20=0$$ $$a_1=-4,a_2=-5$$ $$[\begin{array}{l}{x}^2+6x=-4,\\ x^2+6x=-5;\end{array}\Rightarrow [\begin{array}{l}{x}^2+6x+4=0,\\ x^2+6x+5=0\end{array}\Rightarrow [\begin{array}{l}{x}_{1,2}=-3\pm \sqrt{5}\\ x_3=-1,x_4=-5\end{array}$$

Жауабы: $\left\{-5;-1;-3\pm \sqrt{5}\right\}$.

$$\frac{3}{x^2-4x+1}-\left(x^2-4x\right)-3=0,\text{ММЖ: }x\ne 2+\sqrt{3}$$ $$a=x^2-4x$$ $$\frac{3}{a+1}-a-3=0$$ $$3-a^2-a-3a-3=0$$ $$a^2+4a=0$$ $$a_1=0,a_2=-4$$ $$[\begin{array}{l}{x}^2-4x=0,\\ x^2-4x=-4;\end{array}\Rightarrow [\begin{array}{l}{x}_1=0,x_2=4,\\ x_3=2.\end{array}$$

Жауабы: $\left\{0;24\right\}$.

$$\frac{13x}{2x^2+x+3}+\frac{2x}{2x^2-5x+3}=6,\text{ММЖ: }x\ne 1,x\ne \frac{3}{2}$$ $$\frac{13}{2x+1+\frac{3}{x}}+\frac{2}{2x-5+\frac{3}{x}}=6$$ $$2x+\frac{3}{x}=t$$ $$\frac{13}{t+1}+\frac{2}{t-5}=6$$ $$13(t-5)+2(t+1)=6(t+1)(t-5)$$ $$6t^2-39t+33=0$$ $$2t^2-13t+11=0$$ $$t_1=1,t_2=\frac{11}{2}$$ $$[\begin{array}{l}2x+\frac{3}{x}=1,\\ 2x+\frac{3}{x}=\frac{11}{2}\end{array}\Rightarrow [\begin{array}{l}2x^2-x+3=0\\ 4x^2-11x+6=0\end{array}\Rightarrow [\begin{array}{l}\text{шешімі жоқ,}\\ x_1=2,x_2=\frac{3}{4}.\end{array}$$

Жауабы: $\{\frac{3}{4};2\}$.

$$\frac{15x}{x^2+2x+2}-\frac{8x}{x^2+x+2}=1$$ $$\frac{15}{x+\frac{2}{x}+2}-\frac{8}{x+\frac{2}{x}+1}=1$$ $$a=x+\frac{2}{x}+1$$ $$\frac{15}{a+1}-\frac{8}{a}=1$$ $$7a-8=a^2+a$$ $$a^2-6a+8=0$$ $$a_1=2;a_2=4.$$ $$[\begin{array}{l}x+\frac{2}{x}+1=2,\\ x+\frac{2}{x}+1=4;\end{array}\Rightarrow [\begin{array}{l}{x}^2-x+2=0,\\ x^2-3x+2=0;\end{array}\Rightarrow [\begin{array}{l}\text{шешімі жоқ,}\\ x_1=1,x_2=2.\end{array}$$

Жауабы: $\{1;2\}$.

$$\frac{x^3+2x}{{\left(x^2-x+2\right)}^2}=\frac{3}{4}$$ $$\frac{x+\frac{2}{x}}{{\left(x-1+\frac{2}{x}\right)}^2}=\frac{3}{4}$$ $$a=x+\frac{2}{x}$$ $$\frac{a}{{(a-1)}^2}=\frac{3}{4}$$ $$4a=3\left(a^2-2a+1\right)$$ $$3a^2-10a+3=0$$ $$a_1=\frac{1}{3};a_2=3$$ $$[\begin{array}{l}x+\frac{2}{x}=\frac{1}{3},\\ x+\frac{2}{x}=3\end{array}\Rightarrow [\begin{array}{l}\frac{x^2+2}{x}=\frac{1}{3}\\ \frac{x^2+2}{x}=3\end{array}\Rightarrow$$ $$\Rightarrow [\begin{array}{l}3x^2-x+6=0\\ x^2-3x+2=0\end{array}\Rightarrow [\begin{array}{l}\text{шешімі жоқ,}\\ x_1=1,x_2=2.\end{array}$$

Жауабы: $\{1;2\}$.