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: $(1; — 2),\quad (1;2),\quad ( — 4; — 3),\quad ( — 4;3)$