Бөлшек-рационал теңдеулерді шешу тәсілдері

$$\left\{ {\begin{array}{*{20}{l}}{4{x^2} – 7x – 2 = 0,}\\{{x^2} – 5x + 6 \ne 0;}\end{array}\quad } \right.\left\{ {\begin{array}{*{20}{l}}{{x_1} = – \frac{1}{4},\quad {x_2} = 2,}\\{x \ne 2,\quad x \ne 3}\end{array}\;} \right.\quad x = – \frac{1}{4}$$

Жауабы: $\left\{ { – \dfrac{1}{4}} \right\}$.

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

Жауабы: $x \ne \dfrac{2}{3}.$

$${\left( {\frac{{{x^2} + 35}}{{{x^2} – 49}}} \right)^2} = {\left( {\frac{{12x}}{{{x^2} – 49}}} \right)^2},\quad \text{ММЖ: } x \ne \pm 7$$ $$\left[ {\begin{array}{*{20}{l}}{\frac{{{x^2} + 35}}{{{x^2} – 49}} = \frac{{12x}}{{{x^2} – 49}},}\\{\frac{{{x^2} + 35}}{{{x^2} – 4}} = – \frac{{12x}}{{{x^2} – 49}};}\end{array} \Rightarrow } \right.\left[ {\begin{array}{*{20}{l}}{\frac{{{x^2} – 12x + 35}}{{{x^2} – 49}} = 0}\\{\frac{{{x^2} + 12x + 35}}{{{x^2} – 49}} = 0}\end{array}\;} \right. \Rightarrow $$ $$ \Rightarrow \left[ {\begin{array}{*{20}{l}}{{x^2} – 12x + 35 = 0}\\{{x^2} + 12x + 35 = 0}\end{array}} \right. \Rightarrow \left[ \begin{array}{l}{x_1} = 5,\quad {x_2} = 7,\\{x_3} = – 5,\quad {x_4} = – 7.\end{array} \right.$$ ММЖ-ға сәйкес $x= \pm 5$ түбірін аламыз.

Жауабы: $x= \pm 5$

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

Жауабы: $\left\{ 7 \right\}$

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

Жауабы: $\left\{ -7 \right\}$

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

Жауабы: $\left\{ { – 1;\,\dfrac{{62}}{{39}}} \right\}$

$$\frac{{3 – x}}{{{x^2} + 2x – 3}} – \frac{{3(3 – x)}}{{3{x^2} – 2x – 5}} = 0$$ $$(3 – x)\left( {\frac{1}{{{x^2} + 2x – 3}} – \frac{3}{{3{x^2} – 2x – 5}}} \right) = 0$$ $$(3 – x) \cdot \frac{{3{x^2} – 2x – 5 – 3{x^2} – 6x + 9}}{{\left( {{x^2} + 2x – 3} \right)\left( {3{x^2} – 2x – 5} \right)}} = 0$$ $$(3 – x) \cdot \frac{{ – 8x + 4}}{{(x + 3)(x – 1)(3x – 5)(x + 1)}} = 0,\quad x \ne \pm 1,\,x \ne – 3,\,x \ne \frac{5}{3}.$$ $$\left[ {\begin{array}{*{20}{l}}{3 – x = 0,}\\{ – 8x + 4 = 0;}\end{array}\quad } \right.\,\left[ {\begin{array}{*{20}{l}}{x = 3,}\\{x = \frac{1}{2}}\end{array}} \right.$$

Жауабы: $\left\{ {\dfrac{1}{2};\,3} \right\}$.

$$3\left( {x + \frac{1}{{{x^2}}}} \right) – 7\left( {1 + \frac{1}{x}} \right) = 0,\quad \text{ММЖ: } x \ne 0.$$ $$3x\left( {1 + \frac{1}{{{x^3}}}} \right) – 7\left( {1 + \frac{1}{x}} \right) = 0$$ $$3x\left( {1 + \frac{1}{x}} \right)\left( {1 – \frac{1}{x} + \frac{1}{{{x^2}}}} \right) – 7\left( {1 + \frac{1}{x}} \right) = 0$$ $$\left( {1 + \frac{1}{x}} \right)\left( {3x – 3 + \frac{3}{x} – 7} \right) = 0$$ $$\left( {1 + \frac{1}{x}} \right)\left( {3x + \frac{3}{x} – 10} \right) = 0$$ $$\left[ {\begin{array}{*{20}{l}}{1 + \frac{1}{x} = 0,}\\{3x + \frac{3}{x} – 10 = 0;}\end{array}} \right. \Rightarrow \left[ {\begin{array}{*{20}{l}}{\frac{1}{x} = – 1,}\\{3{x^2} – 10x + 3 = 0;}\end{array}} \right. \Rightarrow \left[ {\begin{array}{*{20}{l}}{x = – 1,}\\{{x_1} = \frac{1}{3},\,{x_2} = 3.}\end{array}} \right.$$

Жауабы: $\left\{ { – 1;\,\dfrac{1}{3};\,3} \right\}$.

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

Жауабы: $\left\{ {2,5} \right\}$.

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

Жауабы: $\left\{ {1,2;\,2,4} \right\}$.

$$\frac{{{x^2} + 2}}{{3x – 2}} – \frac{{3x – 2}}{{{x^2} + 2}} = \frac{8}{3},\quad x \ne \frac{2}{3}.$$ $$a = \frac{{{x^2} + 2}}{{3x – 2}}$$ $$\frac{{3x – 2}}{{{x^2} + 2}} = \frac{1}{a}$$ $$a – \frac{1}{a} = \frac{8}{3}$$ $$3{a^2} – 8a – 3 = 0$$ $${a_1} = 3,\quad {a_2} = – \frac{1}{3}$$ $$\left[ {\begin{array}{*{20}{l}}{\frac{{{x^2} + 2}}{{3x – 2}} = 3,}\\{\frac{{{x^2} + 2}}{{3x – 2}} = – \frac{1}{3};}\end{array}} \right. \Rightarrow \left[ {\begin{array}{*{20}{l}}{{x^2} – 9x + 8 = 0,}\\{3{x^2} + 3x + 4 = 0;}\end{array}} \right. \Rightarrow \left[ \begin{array}{l}{x_1} = 1,{\quad}{x_2} = 8,\\\text{шешімі жоқ.}\end{array} \right.$$

Жауабы: $\{1 ; 8\}$.

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

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

Мынадай теңдеу аламыз:

$$a – \frac{6}{a} = 5$$ $${a^2} – 5a – 6 = 0$$ $${a_1} = – 1,\quad {a_2} = 6$$ $$\left[ {\begin{array}{*{20}{l}}{\frac{{{x^2} + 2x + 3}}{x} = – 1,}\\{\frac{{{x^2} + 2x + 3}}{x} = 6}\end{array}} \right. \Rightarrow \left[ {\begin{array}{*{20}{l}}{{x^2} + 3x + 3 = 0,}\\{{x^2} – 4x + 3 = 0}\end{array}} \right. \Rightarrow \left[ \begin{array}{l}\text{шешімі жоқ},\\{x_1} = 1,\quad {x_2} = 3.\end{array} \right.$$

Жауабы: $\{1 ; 3\}$.

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

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

$$\frac{3}{{{x^2} – 4x + 1}} – \left( {{x^2} – 4x} \right) – 3 = 0,\quad \text{ММЖ: } x \ne 2 + \sqrt 3 $$ $$a = {x^2} – 4x$$ $$\frac{3}{{a + 1}} – a – 3 = 0$$ $$3 – {a^2} – a – 3a – 3 = 0$$ $${a^2} + 4a = 0$$ $${a_1} = 0,\quad {a_2} = – 4$$ $$\left[ {\begin{array}{*{20}{l}}{{x^2} – 4x = 0,}\\{{x^2} – 4x = – 4;}\end{array}} \right. \Rightarrow \left[ {\begin{array}{*{20}{l}}{{x_1} = 0,{x_2} = 4,}\\{{x_3} = 2.}\end{array}} \right.$$

Жауабы: $\left\{ {0;\,2\,4} \right\}$.

$$\frac{{13x}}{{2{x^2} + x + 3}} + \frac{{2x}}{{2{x^2} – 5x + 3}} = 6,\quad \text{ММЖ: } x \ne 1,\,x \ne \frac{3}{2}$$ $$\frac{{13}}{{2x + 1 + \frac{3}{x}}} + \frac{2}{{2x – 5 + \frac{3}{x}}} = 6$$ $$2x + \frac{3}{x} = t$$ $$\frac{{13}}{{t + 1}} + \frac{2}{{t – 5}} = 6$$ $$13(t – 5) + 2(t + 1) = 6(t + 1)(t – 5)$$ $$6{t^2} – 39t + 33 = 0$$ $$2{t^2} – 13t + 11 = 0$$ $${t_1} = 1,\quad {t_2} = \frac{{11}}{2}$$ $$\left[ {\begin{array}{*{20}{l}}{2x + \frac{3}{x} = 1,}\\{2x + \frac{3}{x} = \frac{{11}}{2}}\end{array}} \right. \Rightarrow \left[ {\begin{array}{*{20}{l}}{2{x^2} – x + 3 = 0}\\{4{x^2} – 11x + 6 = 0}\end{array}} \right. \Rightarrow \left[ \begin{array}{l} \text{шешімі жоқ,}\\{x_1} = 2,\quad {x_2} = \frac{3}{4}.\end{array} \right.$$

Жауабы: $\left\{\frac{3}{4} ; 2\right\}$.

$$\frac{{15x}}{{{x^2} + 2x + 2}} – \frac{{8x}}{{{x^2} + x + 2}} = 1$$ $$\frac{{15}}{{x + \frac{2}{x} + 2}} – \frac{8}{{x + \frac{2}{x} + 1}} = 1$$ $$a = x + \frac{2}{x} + 1$$ $$\frac{{15}}{{a + 1}} – \frac{8}{a} = 1$$ $$7a – 8 = {a^2} + a$$ $${a^2} – 6a + 8 = 0$$ $${a_1} = 2;\quad {a_2} = 4.$$ $$\left[ {\begin{array}{*{20}{l}}{x + \frac{2}{x} + 1 = 2,}\\{x + \frac{2}{x} + 1 = 4;}\end{array}} \right. \Rightarrow \left[ {\begin{array}{*{20}{l}}{{x^2} – x + 2 = 0,}\\{{x^2} – 3x + 2 = 0;}\end{array}} \right. \Rightarrow \left[ \begin{array}{l}\text{шешімі жоқ,}\\{x_1} = 1,\quad {x_2} = 2.\end{array} \right.$$

Жауабы: $\{1 ; 2\}$.

$$\frac{{{x^3} + 2x}}{{{{\left( {{x^2} – x + 2} \right)}^2}}} = \frac{3}{4}$$ $$\frac{{x + \frac{2}{x}}}{{{{\left( {x – 1 + \frac{2}{x}} \right)}^2}}} = \frac{3}{4}$$ $$a = x + \frac{2}{x}$$ $$\frac{a}{{{{(a – 1)}^2}}} = \frac{3}{4}$$ $$4a = 3\left( {{a^2} – 2a + 1} \right)$$ $$3{a^2} – 10a + 3 = 0$$ $${a_1} = \frac{1}{3};\quad {a_2} = 3$$ $$\left[ {\begin{array}{*{20}{l}}{x + \frac{2}{x} = \frac{1}{3},}\\{x + \frac{2}{x} = 3}\end{array}} \right. \Rightarrow \left[ {\begin{array}{*{20}{l}}{\frac{{{x^2} + 2}}{x} = \frac{1}{3}}\\{\frac{{{x^2} + 2}}{x} = 3}\end{array}} \right. \Rightarrow $$ $$ \Rightarrow \left[ {\begin{array}{*{20}{l}}{3{x^2} – x + 6 = 0}\\{{x^2} – 3x + 2 = 0}\end{array}} \right. \Rightarrow \left[ \begin{array}{l}\text{шешімі жоқ,}\\{x_1} = 1,\quad {x_2} = 2.\end{array} \right.$$

Жауабы: $\{1 ; 2\}$.