Алгебралық теңдеулер жүйесін шешу (Рустюмова 1.4.1)

Шешуі: $${ \Rightarrow \quad \left\{ {\begin{array}{*{20}{l}}{x - y = 6x + 6y}\\{(x - y)(x + y) = 6}\end{array}} \right.}$$ $${(6x + 6y)(x + y) = 6}$$ $${6{{(x + y)}^2} = 6}$$ $${{{(x + y)}^2} = 1}$$ $${x + y = \pm 1}$$ $$1)\left\{ {\begin{array}{*{20}{l}}{x + y = 1}\\{x - y = 6}\end{array} \Rightarrow \left\{ {\begin{array}{*{20}{l}}{x + y = 1}\\{x = y + 6}\end{array}} \right.} \right.$$ $${y + 6 + y = 1}$$ $${y + 6 + y = 1}$$ $$y = - 2,5;\quad x = 3,5$$ $$2)\left\{ {\begin{array}{*{20}{l}}{x + y = - 1}\\{x - y = 6}\end{array} \Rightarrow \left\{ {\begin{array}{*{20}{l}}{x + y = - 1}\\{x = y + 6}\end{array} \Rightarrow } \right.} \right.$$ $${y + 6 + y = - 1}$$ $${2y = - 7}$$ $${y = - 3,5;\quad x = 2,5}$$ Жауабы: $(3,5; - 2,5),\quad (2,5; - 3,5)$

Шешуі: $${\left\{ {\begin{array}{*{20}{l}}{y = 3x + 7}\\{{x^2} + xy = 2}\end{array}} \right.}$$ $${{x^2} + x(3x + 7) = 2}$$ $${{x^2} + 3{x^2} + 7x - 2 = 0}$$ $${4{x^2} + 7x - 2 = 0}$$ $${4{x^2} + 8x - x - 2 = 0}$$ $${4x(x + 2) - (x + 2) = 0}$$ $${(x + 2)(4x - 1) = 0}$$ $${{x_1} = - 2;\quad {x_2} = \dfrac{1}{4}}$$ $${{y_1} = 3 \cdot ( - 2) + 7 = 1;\quad {y_2} = 3 \cdot \dfrac{1}{4} + 7 = 7\dfrac{3}{4}}$$ Жауабы: ${( - 2;1),\quad \left( {\dfrac{1}{4};7\dfrac{3}{4}} \right)}$

Шешуі: $${\left\{ {\begin{array}{*{20}{l}}{{y^2} = 2 - x}\\{2{y^2} + {x^2} = 3}\end{array}} \right.}$$ $${2(2 - x) + {x^2} = 3}$$ $${{x^2} - 2x + 4 - 3 = 0}$$ $${{{(x - 1)}^2} = 0}$$ $${x = 1}$$ $${{y^2} = 2 - 1 = 1}$$ $${y = \pm 1}$$ Жауабы: ${(1;1),\quad (1; - 1)}$

Шешуі: $${\left\{ {\begin{array}{*{20}{l}}{x = {y^2} + 1}\\{{y^2} + {y^3} = xy}\end{array}} \right.}$$ $${{y^3} + {y^2} = \left( {{y^2} + 1} \right)y}$$ $${{y^3} + {y^2} = {y^3} + y}$$ $${{y^2} - y = 0}$$ $${y(y - 1) = 0}$$ $${{y_1} = 0;\quad {y_2} = 1}$$ $${x_1} = {0^2} + 1 = 1;\quad {x_2} = {1^2} + 1 = 2$$ Жауабы: $(1;0),\quad (2;1)$

Шешуі: $${x = 1 - y}$$ $${{{(x + 1)}^2} + {x^2} = 1}$$ $${{x^2} + 2x + 1 + {x^2} - 1 = 0}$$ $${2{x^2} + 2x = 0}$$ $${x(x + 1) = 0}$$ $${{x_1} = 0,\quad {x_2} = - 1}$$ $${1 - y = 0,\quad 1 - {y_2} = - 1}$$ $${{y_1} = 1,\quad {y_2} = 2}$$ Жауабы: ${(0;1),\quad ( - 1;2)}$

Шешуі: $${{{(2(x - y))}^2} + 2{{(x - y)}^2} = 96}$$ $${6{{(x - y)}^2} = 96}$$ $${{{(x - y)}^2} = 16}$$ $${x - y = 4\quad {\text{ немесе}}\quad x - y = - 4}$$ $${\left\{ {\begin{array}{*{20}{l}}{x - y = 4}\\{3x + y = 2(x - y)}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{l}}{x - y = 4}\\{3x + y = 2 \cdot 4}\end{array} \Rightarrow \left\{ \begin{array}{l}x - y = 4\\3x + y = 8\end{array} \right. \Rightarrow \left\{ \begin{array}{l}y = x - 4\\3x + x - 4 = 8\end{array} \right.} \right.}$$ $$4x = 12,\quad {x_1} = 3,\quad {y_1} = - 1$$ $$\left\{ {\begin{array}{*{20}{l}}{x - y = - 4}\\{3x + y = 2(x - y)}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{l}}{x - y = - 4}\\{3x + y = 2 \cdot ( - 4)}\end{array}} \right. \Rightarrow \left\{ \begin{array}{l}y = x + 4\\3x + x + 4 = - 8\end{array} \right.$$ $$4x = - 12,\quad {x_2} = - 3,\quad {y_2} = 1$$ Жауабы: $(3; - 1),\quad ( - 3;1)$

Шешуі: $${x = 1 - 2y}$$ $${{{(1 - 2y)}^2} - 3(1 - 2y) \cdot y - 2{y^2} = 2}$$ $${1 - 4y + 4{y^2} - 3y + 6{y^2} - 2{y^2} - 2 = 0}$$ $${8{y^2} - 7y - 1 = 0}$$ $${8{y^2} - 8y + y - 1 = 0}$$ $${8y(y - 1) + y - 1 = 0}$$ $${(y - 1)(8y + 1) = 0}$$ $${{y_1} = 1,\quad {y_2} = - \dfrac{1}{8}}$$ $${{x_1} = 1 - 2 \cdot 1 = - 1;\quad {y_2} = 1 - 2 \cdot \left( { - \dfrac{1}{8}} \right) = 1\dfrac{1}{4}}$$ Жауабы: ${( - 1;1),\quad \left( {1\dfrac{1}{4}; - \dfrac{1}{8}} \right)}$

Шешуі: $${\left\{ {\begin{array}{*{20}{l}}{x(x + y) = \dfrac{3}{4}}\\{\dfrac{x}{y} = \dfrac{1}{2}}\end{array}} \right.}$$ $${y = 2x}$$ $${x(x + 2x) = \dfrac{3}{4}}$$ $${3{x^2} = \dfrac{3}{4},\quad {x^2} = \dfrac{1}{4}}$$ $${{x_1} = \dfrac{1}{2};\quad {x_2} = - \dfrac{1}{2}}$$ $${{y_1} = 2 \cdot \dfrac{1}{2} = 1;\quad {y_2} = 2 \cdot \left( { - \dfrac{1}{2}} \right) = - 1}$$ Жауабы: ${\left( {\dfrac{1}{2};1} \right),\quad \left( { - \dfrac{1}{2}; - 1} \right)}$

Шешуі: $${x = 3y - 2}$$ $${{{(3y - 2)}^2} - 2(3y - 2) \cdot y = 7}$$ $${9{y^2} - 12y + 4 - 6{y^2} + 4y - 7 = 0}$$ $${3{y^2} - 8y - 3 = 0}$$ $${3{y^2} - 9y + y - 3 = 0}$$ $${3y(y - 3) + (y - 3) = 0}$$ $${(y - 3)(3y + 1) = 0}$$ $${{y_1} = 3,\quad {x_1} = 3 \cdot 3 - 2 = 7}$$ $${{y_2} = - \dfrac{1}{3},\quad {x_2} = 3 \cdot \left( { - \dfrac{1}{3}} \right) - 2 = - 3}$$ Жауабы: $(7;3),\quad \left( { - 3; - \dfrac{1}{3}} \right)$

Шешуі: $${\left\{ {\begin{array}{*{20}{l}}{y - 2 = 2x - 2,\quad x \ne 1}\\{y - 2x = {x^2} - 1}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{l}}{y = 2x,\quad x \ne 1}\\{2x - 2x = {x^2} - 1}\end{array}} \right.}$$ $${{x^2} - 1 = 0}$$ $${{x_1} \ne 1;\quad {x_2} = - 1}$$ $${y = 2 \cdot ( - 1) = - 2}$$ Жауабы: ${( - 1; - 2)}$

Шешуі: $${y = 2x - 1}$$ $${2{x^2} - {{(2x - 1)}^2} + x + 2x - 1 = - 11}$$ $${2{x^2} - 4{x^2} + 4x - 1 + x + 2x - 1 + 11 = 0}$$ $${ - 2{x^2} + 7x + 9 = 0}$$ $${2{x^2} - 7x - 9 = 0}$$ $${2{x^2} - 9x + 2x - 9 = 0}$$ $${x(2x - 9) + (2x - 9) = 0}$$ $${(x + 1)(2x - 9) = 0}$$ $${{x_1} = - 1;\quad {x_2} = 4,5}$$ $${{x_1} = - 1;\quad {x_2} = 4,5}$$ $${y_1} = 2 \cdot ( - 1) - 1 = - 3;\quad {y_2} = 2 \cdot 4,5 - 1 = 8$$ Жауабы: $( - 1;\,\, - 3),\quad (4,5;\,\,8)$

Шешуі: $${{y^2} = 2{x^2} - 4}$$ $${{x^2} + 3\left( {2{x^2} - 4} \right) - 5x = 6}$$ $${7{x^2} - 5x - 12 - 6 = 0}$$ $${7{x^2} - 5x - 18 = 0}$$ $${7{x^2} - 14x + 9x - 18 = 0}$$ $${7x(x - 2) + 9(x - 2) = 0}$$ $${(x - 2)(7x + 9) = 0}$$ $${{x_1} = 2}$$ $${{y^2} = 2 \cdot {2^2} - 4 = 4}$$ $${y = \pm 2}$$ $${{x_2} = - \dfrac{9}{7}}$$ $${{y^2} = 2 \cdot {{\left( { - \dfrac{9}{7}} \right)}^2} - 4 = \dfrac{{162 - 49 \cdot 4}}{{49}} \lt 0}, \quad {y \in \emptyset }$$ Жауабы: ${(2;\, - 2),\quad (2;\,2)}$

Шешуі: $${x = 6 - y}$$ $${1 + 6 - y + {{(6 - y)}^2} = 3 + 3y + 3{y^2}}$$ $${36 - 12y + {y^2} + 7 - y - 3 - 3y - 3{y^2} = 0}$$ $${ - 2{y^2} - 16y + 40 = 0}$$ $${{y^2} + 8y - 20 = 0}$$ $${{y^2} + 10y - 2y - 20 = 0}$$ $${y(y + 10) - 2(y + 10) = 0}$$ $${(y - 2)(y + 10) = 0}$$ $${{y_1} = 2}$$ $${{x_1} = 6 - 2 = 4}$$ $${{y_2} = - 10}$$ $${{x_2} = 6 - ( - 10) = 16}$$ Жауабы: ${(4;\,2);\quad (16;\, - 10)}$

Шешуі: $${x = y + 1}$$ $${\dfrac{{{{(y + 1)}^2} + y + 1}}{{{y^2} + y + 1 + 1}} = \dfrac{3}{2}}$$ $${2 \cdot \left( {{y^2} + 2y + 1 + y + 1} \right) = 3 \cdot \left( {{y^2} + y + 2} \right)}$$ $${2{y^2} + 6y + 4 = 3{y^2} + 3y + 6}$$ $${{y^2} - 3y + 2 = 0}$$ $${{y^2} - y - 2y + 2 = 0}$$ $${y(y - 1) - 2(y - 1) = 0}$$ $${(y - 2)(y - 1) = 0}$$ $${{y_1} = 2;\quad {y_2} = 1}$$ $${{x_1} = 2 + 1 = 3;\quad {x_2} = 1 + 1 = 2}$$ Жауабы: ${(3;2),\quad (2;1)}$

Шешуі: $${x = - 8 - y}$$ $${{{( - 8 - y)}^2} + {y^2} + 6( - 8 - y) + 2y = 0}$$ $${64 + 16y + {y^2} + {y^2} - 48 - 6y + 2y = 0}$$ $${2{y^2} + 12y + 16 = 0}$$ $${{y^2} + 6y + 8 = 0}$$ $${{y^2} + 2y + 4y + 8 = 0}$$ $${y(y + 2) + 4(y + 2) = 0}$$ $${(y + 2)(y + 4) = 0}$$ $${{y_1} = - 2;\quad {x_1} = - 8 - ( - 2) = - 6}$$ $${{y_2} = - 4;\quad {x_2} = - 8 - ( - 4) = - 4}$$ Жауабы: ${( - 6; - 2),\quad ( - 4; - 4)}$

Шешуі: $${x = - \dfrac{1}{{2y}}}$$ $${{{\left( { - \dfrac{1}{{2y}}} \right)}^2} - {y^2} = \dfrac{3}{4}}$$ $${\dfrac{1}{{4{y^2}}} - {y^2} = \dfrac{3}{4}\quad \quad | \times 4{y^2}}$$ $${1 - 4{y^4} = 3{y^2}}$$ $${4{y^4} + 3{y^2} - 1 = 0}$$ $${4{y^4} + 4{y^2} - {y^2} - 1 = 0}$$ $${4{y^2}\left( {{y^2} + 1} \right) - \left( {{y^2} + 1} \right) = 0}$$ $${\left( {{y^2} + 1} \right)\left( {4{y^2} - 1} \right) = 0}$$ $${\left( {{y^2} + 1} \right)(2y - 1)(2y + 1) = 0}$$ $${{y_1} = \dfrac{1}{2};\quad {x_1} = - \dfrac{1}{{2 \cdot \dfrac{1}{2}}} = - 1}$$ $${{y_2} = - \dfrac{1}{2};\quad {x_2} = - \dfrac{1}{{2 \cdot \left( { - \dfrac{1}{2}} \right)}} = 1}$$ Жауабы: ${\left( { - 1;\dfrac{1}{2}} \right),\quad \left( {1; - \dfrac{1}{2}} \right)}$

Шешуі: $${y = 3 - 2x}$$ $${\dfrac{2}{{3 - 2x + x}} - \dfrac{1}{{x - 3}} = \dfrac{2}{{3x - {x^2}}}}$$ $${\dfrac{2}{{3 - x}} + \dfrac{1}{{3 - x}} = \dfrac{2}{{x(3 - x)}}}$$ $${\dfrac{3}{{3 - x}} - \dfrac{2}{{x(3 - x)}} = 0}$$ $${\dfrac{{3x - 2}}{{x(3 - x)}} = 0}$$ $${x = \dfrac{2}{3}}$$ $${y = 3 - 2 \cdot \dfrac{2}{3} = 3 - \dfrac{4}{3} = \dfrac{9}{3} - \dfrac{4}{3} = \dfrac{5}{3}}$$ Жауабы: ${\left( {\dfrac{2}{3};\quad \dfrac{5}{3}} \right)}$

Шешуі: $$x = y + 3$$ $$\dfrac{1}{{y(y + 1)}} + \dfrac{2}{{y + 1}} - \dfrac{1}{{2 - y - 3}} = 0$$ $${\dfrac{1}{{y(y + 1)}} + \dfrac{2}{{y + 1}} + \dfrac{1}{{y + 1}} = 0}$$ $${\dfrac{1}{{y(y + 1)}} + \dfrac{3}{{y + 1}} = 0}$$ $${\dfrac{{1 + 3y}}{{y(y + 1)}} = 0}$$ $${y = - \dfrac{1}{3};\quad x = - \dfrac{1}{3} + 3 = \dfrac{8}{3}}$$ Жауабы: ${\left( {\dfrac{8}{3};\quad - \dfrac{1}{3}} \right)}$

Шешуі: $${x = 2y + 1}$$ $${\dfrac{{2y + 1 - y}}{{y - 1}} + \dfrac{{2y + 1 + 5}}{{y + 1}} = \dfrac{4}{{(y - 1)(y + 1)}}}$$ $${\dfrac{{{{(y + 1)}^2} + (y - 1)(2y + 6)}}{{(y - 1)(y + 1)}} = \dfrac{4}{{(y - 1)(y + 1)}}}$$ $${{y^2} + 2y + 1 + 2{y^2} - 2y + 6y - 6 - 4 = 0,\quad y \ne 1,\,\,y \ne - 1}$$ $${3{y^2} + 6y - 9 = 0}$$ $${{y^2} + 2y - 3 = 0}$$ $${{y^2} - y + 3y - 3 = 0}$$ $${y(y - 1) + 3(y - 1) = 0}$$ $${(y - 1)(y + 3) = 0}$$ $${y \ne 1,\quad y + 3 = 0}$$ $${y = - 3}$$ $${x = 2 \cdot ( - 3) + 1 = - 5}$$ Жауабы: ${( - 5; - 3)}$

Шешуі: $${x - y - 2 = 0,\quad x \ne 3}$$ $${x = y + 2}$$ $${2{{(y + 2)}^2} + {y^2} - 2(y + 2) \cdot y = 13}$$ $${2{y^2} + 8y + 8 + {y^2} - 2{y^2} - 4y - 13 = 0}$$ $${{y^2} + 4y - 5 = 0}$$ $${{y^2} - y + 5y - 5 = 0}$$ $${y(y - 1) + 5(y - 1) = 0}$$ $${(y - 1)(y + 5) = 0}$$ $${y_1} = 1;\quad {x_1} = 3,\quad \text{бөгде түбір}$$ $${y_2} = - 5;\quad {x_2} = - 5 + 2 = - 3$$ Жауабы: ${( - 3; - 5)}$

Шешуі: $$y = x + 1$$ $$\dfrac{{x + 3(x + 1) + 1}}{{x + 1}} - \dfrac{{x + 1 - x + 3}}{{2(x - 2)}} = 2$$ $$\dfrac{{4x + 4}}{{x + 1}} - \dfrac{4}{{2(x - 2)}} = 2$$ $$\dfrac{2}{{x - 2}} = 2;$$ $$x - 2 = 1;\quad x = 3,\quad y = 4$$ Жауабы: $\left( {3;4} \right)$

Шешуі: $$2{y^2} = {x^2} + x + 6$$ $${x^2} - 1,5 \cdot 2{y^2} = - 11$$ $${{x^2} - \left( {1,5{x^2} + 1,5x + 1,5 \cdot 6} \right) = - 11}$$ $${ - 0,5{x^2} - 1,5x - 9 + 11 = 0\quad \quad \mid \times ( - 2)}$$ $${{x^2} + 3x - 4 = 0}$$ $${{x^2} - x + 4x - 4 = 0}$$ $${x(x - 1) + 4(x - 1) = 0}$$ $${(x - 1)(x + 4) = 0}$$ $${{x_1} = 1;\quad 2{y^2} = {1^2} + 1 + 6;\quad 2{y^2} = 8}$$ $${y = \pm 2}$$ $${{x_2} = - 4;\quad 2{y^2} = {{( - 4)}^2} + ( - 4) + 6}$$ $${2{y^2} = 18;\quad {y^2} = 9}$$ $${y = \pm 3}$$ Жауабы: $(1; - 2),\quad (1;2),\quad ( - 4; - 3),\quad ( - 4;3)$