Рационал теңдеулер (Сканави (А) 6.1-6.12)

Шешуі: ММЖ: $x \ne - 3,x \ne 4$ $${\dfrac{{{x^2} + 1}}{{x - 4}} - \dfrac{{{x^2} - 1}}{{x + 3}} = 23 \Leftrightarrow \dfrac{{16{x^2} - 25x - 275}}{{(x - 4)(x + 3)}} = 0 \Rightarrow }$$ $${ \Rightarrow 16{x^2} - 25x - 275 = 0 \Rightarrow {x_1} = - \dfrac{{55}}{{16}},{x_2} = 5.}$$

Шешуі: ММЖ: $x \ne a, x \ne b$ $${\dfrac{b}{{x - a}} + \dfrac{a}{{x - b}} = 2 \Leftrightarrow \dfrac{{2{x^2} - 3(a + b)x + {{(a + b)}^2}}}{{(x - a)(x - b)}} = 0 \Rightarrow }$$ $${ \Rightarrow 2{x^2} - 3(a + b)x + {{(a + b)}^2} = 0 \Rightarrow {x_1} = \dfrac{{a + b}}{2},{x_2} = a + b}$$

Шешуі: ММЖ: $$x \ne 0,x \ne \dfrac{{ - 1 \pm \sqrt {21} }}{2}$$ Жаңа айнымалы енгізейік: $$\dfrac{{{x^2} + x - 5}}{x} = z$$ Бұдан $${z^2} + 4z + 3 = 0, \Rightarrow {z_1} = - 3,{z_2} = - 1.$$ $$\dfrac{{{x^2} + x - 5}}{x} = - 3{\rm{ }}{\rm{, }}\dfrac{{{x^2} + x - 5}}{x} = - 1.{\rm{ }}$$ Теңдеулердің әрқайсысын шешетін болсақ, $${x_1} = - 5,\,\,{x_2} = 1,\,\,{x_3} = - 1 - \sqrt 6 ,\,\,{x_4} = - 1 + \sqrt 6 $$

Шешуі: ММЖ: $$x \ne \pm \sqrt[4]{{\dfrac{7}{2}}}$$ $$2{x^4} - 7 = z$$ жаңа айнымалысын енгізейік. Онда $$\dfrac{{z + 2}}{2} - \dfrac{{50}}{z} = 14 \Rightarrow {z^2} - 21z - 100 = 0 \Leftrightarrow {z_1} = - 4,{z_2} = 25.$$ Қайтадан орнына апарып қойсақ, $$2{x^4} - 7 = - 4$$ $$2{x^4} - 7 = 25$$ $$x_1 = - \sqrt[4]{{\dfrac{3}{2}}},{x_2} = \sqrt[4]{{\dfrac{3}{2}}},{x_3} = - 2,{x_4} = 2$$

Шешуі: ММЖ: $x \ne 0,x \ne - 1,x \ne - 2$ $$\dfrac{1}{{x(x + 2)}} - \dfrac{1}{{{{(x + 1)}^2}}} = \dfrac{1}{{12}} \Leftrightarrow \dfrac{1}{{{x^2} + 2x}} - \dfrac{1}{{{x^2} + 2x + 1}} = \dfrac{1}{{12}}{\rm{.}}$$ Жаңа айнымалы енгізейік: $${x^2} + 2x = z$$ Онда $$\dfrac{{{z^2} + z + 12}}{{z(z + 1)}} = 0 \Rightarrow {z^2} + z + 12 = 0 \Rightarrow {z_1} = - 4,{z_2} = 3.$$ Апарып орнына қойсақ: $${x^2} + 2x = - 4{\rm{ }}{\rm{, }}{x^2} + 2x = 3.$$ Бұдан $${x_{1,2}} \in \emptyset (D \lt 0),{x_3} = - 3,{x_4} = 1$$

Шешуі: ММЖ: $x \ne 0,m \ne \pm n$ $$x + \dfrac{1}{x} = 2\dfrac{{{m^2} + {n^2}}}{{{m^2} - {n^2}}} \Leftrightarrow {x^2} - 2\dfrac{{{m^2} + {n^2}}}{{{m^2} - {n^2}}}x + 1 = 0$$ $${x_1} = \dfrac{{m + n}}{{m - n}},{x_2} = \dfrac{{m - n}}{{m + n}}$$

Шешуі: ММЖ: $x \ne 0,a \ne 0$ $$\dfrac{{{x^2}}}{{{a^3}}} + \dfrac{{{b^3}}}{{{x^2}}} = \dfrac{b}{a} + \dfrac{{{b^2}}}{{{a^2}}} \Leftrightarrow \dfrac{{{x^4} - \left( {{a^2}b + a{b^2}} \right){x^2} + {a^3}{b^3}}}{{{a^3}{x^2}}} = 0$$ $${x^4} - \left( {{a^2}b + a{b^2}} \right){x^2} + {a^3}{b^3} = 0,ax \ne 0$$ Жаңа айнымалы енгізейік: ${x^2} = y$ $${y^2} - \left( {{a^2}b + a{b^2}} \right)y + {a^3}{b^3} = 0$$ $${{y_{1,2}} = \dfrac{{{a^2}b + a{b^2} \pm \sqrt {{{\left( {{a^2}b + a{b^2}} \right)}^2} - 4{a^3}{b^3}} }}{2} = }$$ $${ = \dfrac{{{a^2}b + a{b^2} \pm \sqrt {{a^4}{b^2} + 2{a^3}{b^3} + {a^2}{b^4} - 4{a^3}{b^3}} }}{2} = }$$ $${ = \dfrac{{{a^2}b + a{b^2} \pm \sqrt {{a^4}{b^2} - 2{a^3}{b^3} + {a^2}{b^4}} }}{2} = \dfrac{{{a^2}b + a{b^2} \pm \sqrt {{{\left( {{a^2}b - a{b^2}} \right)}^2}} }}{2} = }$$ $${ = \dfrac{{{a^2}b + a{b^2} \pm \left( {{a^2}b - a{b^2}} \right)}}{2};{y_1} = a{b^2},{y_2} = {a^2}b}$$ $x$-ті табу үшін мына екі квадрат теңдеуді шешу керек: $${x^2} = a{b^2}{\rm{ }}{\rm{, }}{x^2} = {a^2}b$$ $${{x_{1,2}} = \pm \sqrt {a{b^2}} = \pm b\sqrt a ,{\rm{ }}a \gt 0;}$$ $${{x_{3,4}} = \pm \sqrt {{a^2}b} = \pm a\sqrt b ,{\rm{ }}b \ge 0.}$$

Шешуі: ММЖ: $x \ne \pm 1, x \ne \pm 2 $ $${\dfrac{{x - 3}}{{x - 1}} + \dfrac{{x + 3}}{{x + 1}} = \dfrac{{x + 6}}{{x + 2}} + \dfrac{{x - 6}}{{x - 2}} \Leftrightarrow \dfrac{{2{x^2} - 6}}{{{x^2} - 1}} = \dfrac{{2{x^2} - 24}}{{{x^2} - 4}} \Leftrightarrow }$$ $${ \Leftrightarrow \dfrac{{{x^2} - 3}}{{{x^2} - 1}} = \dfrac{{{x^2} - 12}}{{{x^2} - 4}} \Leftrightarrow \dfrac{{6{x^2}}}{{\left( {{x^2} - 1} \right)\left( {{x^2} - 4} \right)}} = 0}$$ $x=0.$

Шешуі: ММЖ: $y \ne - a,y \ne - 2a,y \ne - 3a$ $$\dfrac{{5a}}{{y + a}} + \dfrac{{4a}}{{y + 2a}} + \dfrac{{3a}}{{y + 3a}} = 8 \Leftrightarrow \dfrac{{8{y^3} + 36a{y^2} + 38{a^2}y}}{{(y + a)(y + 2a)(y + 3a)}} = 0$$ $${\left\{ {\begin{array}{lllllllllllllll}{y\left( {4{y^2} + 18ay + 19{a^2}} \right) = 0,}\\{y \ne - a,}\\{y \ne - 2a,}\\{y \ne - 3a}\end{array} \Rightarrow } \right.}$$ ${ \Rightarrow y = 0}$ немесе ${4{y^2} + 18ay + 19{a^2} = 0.}$ $${{y_{1,2}} = \dfrac{{ - 9a \pm \sqrt {81{a^2} - 76{a^2}} }}{4} = \dfrac{{ - 9a \pm a\sqrt 5 }}{4} = \dfrac{{a( - 9 \pm \sqrt 5 )}}{4}.}$$ Жауабы: $${{y_1} = 0,{y_{2,3}} = \dfrac{{a( - 9 \pm \sqrt 5 )}}{4}}$$

Шешуі: ММЖ: $x \ne - \sqrt[3]{2},x \ne - \sqrt[3]{3}$ $${\dfrac{1}{{{x^3} + 2}} - \dfrac{1}{{{x^3} + 3}} = \dfrac{1}{{12}} \Leftrightarrow \dfrac{{{x^6} + 5{x^3} - 6}}{{12\left( {{x^3} + 2} \right)\left( {{x^3} + 3} \right)}} = 0 \Leftrightarrow }$$ $${ \Leftrightarrow \left\{ {\begin{array}{lllllllllllllll}{{x^6} + 5{x^3} - 6 = 0}\\{\left( {{x^3} + 2} \right)\left( {{x^3} + 3} \right) \ne 0.}\end{array}} \right.}$$ Жаңа айнымалы енгізейік: $x^3=y$. Бұдан мынадай квадрат теңдеу шығарып аламыз: $y^2+5y-6=0$, $y_1=-6, y_2=1$ $$x^3=-6, x^3=1$$ $$x_1=$$ $${x_1} = - \sqrt[3]{6},\,{x_2} = 1$$

Шешуі: ММЖ: $x \ne \pm 1,x \ne \pm 3$ $$\left( {\dfrac{{x - 2}}{{x - 1}} + \dfrac{{x + 2}}{{x + 1}}} \right) - \left( {\dfrac{{x - 4}}{{x - 3}} + \dfrac{{x + 4}}{{x + 3}}} \right) = - \dfrac{{28}}{{15}} \Leftrightarrow $$ $${ \Leftrightarrow \dfrac{{{x^2} - 2}}{{{x^2} - 1}} - \dfrac{{{x^2} - 12}}{{{x^2} - 9}} = - \dfrac{{14}}{{15}} \Leftrightarrow \dfrac{{2{x^2} + 6}}{{\left( {{x^2} - 1} \right)\left( {{x^2} - 9} \right)}} = - \dfrac{{14}}{{15}} \Leftrightarrow }$$ $${ \Leftrightarrow \dfrac{{{x^2} + 3}}{{\left( {{x^2} - 1} \right)\left( {{x^2} - 9} \right)}} = - \dfrac{7}{{15}} \Leftrightarrow \dfrac{{{x^2} + 3}}{{\left( {{x^2} - 1} \right)\left( {{x^2} - 9} \right)}} + \dfrac{7}{{15}} = 0 \Leftrightarrow }$$ $$ \Leftrightarrow \dfrac{{7{x^4} - 55{x^2} + 108}}{{\left( {{x^2} - 1} \right)\left( {{x^2} - 9} \right)}} = 0 \Leftrightarrow \left\{ {\begin{array}{lllllllllllllll}{7{x^4} - 55{x^2} + 108 = 0}\\{\left( {{x^2} - 1} \right)\left( {{x^2} - 9} \right) \ne 0}\end{array}} \right.$$ Жаңа айнымалы енгізейік: $x^2=y$, $$7{y^2} - 55y + 108 = 0;{y_1} = 4,{y_2} = \dfrac{{27}}{7}{\rm{.}}$$ $${x^2} = 4,{x_{1,2}} = \pm 2$$ $${x^2} = \dfrac{{27}}{7},{x_{3,4}} = \pm \sqrt {\dfrac{{27}}{7}} .$$ Жауабы: ${x_{1,2}} = \pm 2,{x_{3,4}} = \pm \sqrt {\dfrac{{27}}{7}} $