Рационал алгебралық өрнектерді теңбе-тең түрлендіру (Рустюмова 1.2.2)

Шешуі: $${3{x^2} - 5x - 2 = 0}$$ $${D = 25 + 24 = 49 = 7^2}$$ $${x_1} = \dfrac{{5 + 7}}{6} = \dfrac{{12}}{6} = 2,$$ $${x_2} = \dfrac{{5 - 7}}{6} = - \dfrac{1}{3}.$$ $$\boxed{a{x^2} + bx + c = a\left( {x - {x_1}} \right)\left( {x - {x_2}} \right)}$$ $$3{x^2} - 5x - 2 = 3\left( {x + \dfrac{1}{3}} \right)(x - 2)$$ $${x^2} - x - 2 = {x^2} + x - 2x - 2 = x\left( {x + 1} \right) - 2\left( {x + 1} \right) = \left( {x + 1} \right)\left( {x - 2} \right)$$ $$\dfrac{{{x^2} - x - 2}}{{3{x^2} - 5x - 2}} = \dfrac{{(x - 2)(x + 1)}}{{3\left( {x + \dfrac{1}{3}} \right)(x - 2)}} = \dfrac{{(x - 2)(x + 1)}}{{(3x + 1)(x - 2)}} = \dfrac{{x + 1}}{{3x + 1}}$$

Шешуі: $$ = \dfrac{{\left( {{x^3} + 4{x^2}} \right) - 9(x + 4)}}{{(x + 4)(x - 3)}} = \dfrac{{{x^2}(x + 4) - 9(x + 4)}}{{(x + 4)(x - 3)}} = $$ $$ = \dfrac{{(x + 4)\left( {{x^2} - 9} \right)}}{{(x + 4)(x - 3)}} = \dfrac{{(x - 3)(x + 3)}}{{(x - 3)}} = x + 3$$

Шешуі: $$ = \dfrac{{\left( {5{x^2} - 5xy} \right) - 7(x - y)}}{{x - y}} = \dfrac{{5x(x - y) - 7(x - y)}}{{x - y}} = $$ $$ = \dfrac{{(x - y)(5x - 7)}}{{x - y}} = 5x - 7$$

Шешуі: $$ = \dfrac{{ - \left( {1 - {a^4}} \right)}}{{\left( {1 - {a^4}} \right)\left( {1 + {a^4}} \right)}} = - \dfrac{1}{{1 + {a^4}}}$$

Шешуі: $$ = \dfrac{{x\left( {{x^3} + {a^3}} \right)}}{{x\left( {{x^2} - ax + {a^2}} \right)}} = \dfrac{{(x + a)\left( {{x^2} - ax + {a^2}} \right)}}{{{x^2} - ax + {a^2}}} = x + a$$

Шешуі: $$ = \dfrac{{(1 - b) + (ab - a)}}{{{{(1 - a)}^2}}} = \dfrac{{(1 - b) - a(1 - b)}}{{{{(1 - a)}^2}}} = \dfrac{{(1 - b)(1 - a)}}{{{{(1 - a)}^2}}} = \dfrac{{1 - b}}{{1 - a}}$$

Шешуі: $$ = \dfrac{{y(x - 1) + 2(x - 1)}}{{x(y + 3) - (y + 3)}} = \dfrac{{(x - 1)(y + 2)}}{{(y + 3)(x - 1)}} = \dfrac{{y + 2}}{{y + 3}}$$

Шешуі: $$ = \dfrac{{(x - 7)(x - 2)}}{{(x - 8)(x - 2)}} = \dfrac{{x - 7}}{{x - 8}}$$

Шешуі: $$ = \dfrac{{7{x^2}{y^2}\left( {{y^2} + {x^2}} \right)}}{{\left( {{x^2} + {y^2}} \right)\left( {{x^4} - {x^2}{y^2} + {y^4}} \right)}} = \dfrac{{7{x^2}{y^2}}}{{{x^4} - {x^2}{y^2} + {y^4}}}$$

Шешуі: 1-жолы: $${x^4} + {x^2} + 1 = \left( {{x^4} + 2{x^2} + 1} \right) - 2{x^2} + {x^2} = $$ $$ = {\left( {{x^2} + 1} \right)^2} - {x^2} = \left( {{x^2} + 1 - x} \right)\left( {{x^2} + 1 + x} \right) = \left( {{x^2} - x + 1} \right)\left( {{x^2} + x + 1} \right)$$ $$\dfrac{{{x^4} + {x^2} + 1}}{{{x^2} + x + 1}} = \dfrac{{\left( {{x^2} + x + 1} \right)\left( {{x^2} - x + 1} \right)}}{{{x^2} + x + 1}} = {x^2} - x + 1$$ 2-жолы:

Шешуі: $${ = \dfrac{{{a^3}\left( {{a^{33}} - 1} \right)}}{{{a^4}\left( {{a^{22}} + {a^{11}} + 1} \right)}} = \dfrac{{\left( {{a^{11}} - 1} \right)\left( {{a^{22}} + {a^{11}} + 1} \right)}}{{a\left( {{a^{22}} + {a^{11}} + 1} \right)}} = \dfrac{{{a^{11}} - 1}}{a}}$$

Шешуі: $${ = \dfrac{{\left( {{a^2} + {b^2} - 2ab} \right) - {c^2}}}{{\left( {{a^2} + {c^2} + 2ac} \right) - {b^2}}} = \dfrac{{{{(a - b)}^2} - {c^2}}}{{{{(a + c)}^2} - {b^2}}} = }$$ $${ = \dfrac{{(a - b - c)(a - b + c)}}{{(a + c - b)(a + c + b)}} = \dfrac{{a - b - c}}{{a + b + c}}}$$

Шешуі: $$ = \dfrac{{\left( {{x^2} - 1} \right)\left( {{x^4} + {x^2} + 1} \right)}}{{3x\left( {{x^2} - 1} \right)}} = \dfrac{{{x^4} + {x^2} + 1}}{{3x}}$$

Шешуі: $$ = \dfrac{{{m^2} - \left( {{n^2} + 2np + {p^2}} \right)}}{{\left( {{m^2} - 2mn + {n^2}} \right) - {p^2}}} = \dfrac{{{m^2} - {{(n + p)}^2}}}{{{{(m - n)}^2} - {p^2}}} = $$ $${ = \dfrac{{(m - n - p)(m + n + p)}}{{(m - n - p)(m - n + p)}} = \dfrac{{m + n + p}}{{m - n + p}}}$$

Шешуі: $${{a^3} - 2{a^2} + 4a - 3 = \left( {{a^3} - 2{a^2} + a} \right) + (3a - 3) = }$$ $${ = a\left( {{a^2} - 2a + 1} \right) + 3(a - 1) = a{{(a - 1)}^2} + 3(a - 1) = }$$ $${ = (a - 1)(a(a - 1) + 3) = (a - 1)\left( {{a^2} - a + 3} \right)}$$ $$\dfrac{{{a^3} - 2{a^2} + 4a - 3}}{{{a^2} - 7a + 6}} = \dfrac{{(a - 1)\left( {{a^2} - a + 3} \right)}}{{(a - 1)(a - 6)}} = \dfrac{{{a^2} - a + 3}}{{a - 6}}$$

Шешуі: $$ = \dfrac{{(x + y) + (x - y)(x + y)}}{{(x - y) + {{(x - y)}^2}}} = \dfrac{{(x + y)(1 + x - y)}}{{(x - y)(1 + x - y)}} = \dfrac{{x + y}}{{x - y}}$$

Шешуі: $$ = \dfrac{{b(a + 1) + z(a + 1)}}{{{{(a + 1)}^3}}} = \dfrac{{(a + 1)(b + z)}}{{{{(a + 1)}^3}}} = \dfrac{{b + z}}{{{{(a + 1)}^2}}}$$

Шешуі:$${x^3} + 2{x^2} + 4x + 3 = \left( {{x^3} + 2{x^2} + x} \right) + (3x + 3) = $$ $$ = x\left( {{x^2} + 2x + 1} \right) + 3(x + 1) = x{(x + 1)^2} + 3(x + 1) = $$ $$ = (x + 1)(x(x + 1) + 3) = (x + 1)\left( {{x^2} + x + 3} \right)$$ $$\dfrac{{{x^3} + 2{x^2} + 4x + 3}}{{{x^2} + 7x + 6}} = \dfrac{{(x + 1)\left( {{x^2} + x + 3} \right)}}{{(x + 1)(x + 6)}} = \dfrac{{{x^2} + x + 3}}{{x + 6}}$$

Шешуі: $${{x^4} + {x^2}{y^2} + {y^4} = \left( {{x^4} + 2{x^2}{y^2} + {y^4}} \right) - 2{x^2}{y^2} + {x^2}{y^2} = }$$ $${ = {{\left( {{x^2} + {y^2}} \right)}^2} - {x^2}{y^2} = \left( {{x^2} + {y^2} - xy} \right)\left( {{x^2} + {y^2} + xy} \right)}$$ $${ = \left( {{x^2} - xy + {y^2}} \right)\left( {{x^2} + xy + {y^2}} \right)}$$ $${\dfrac{{{x^4} + {x^2}{y^2} + {y^4}}}{{{x^2} - xy + {y^2}}} = \dfrac{{\left( {{x^2} - xy + {y^2}} \right)\left( {{x^2} + xy + {y^2}} \right)}}{{{x^2} - xy + {y^2}}} = {x^2} + xy + {y^2}}$$

Шешуі: $${ = \dfrac{{\left( {{x^2} - xy + {y^2}} \right)\left( {{x^2} + xy + {y^2}} \right)}}{{(x - y)(x - y)\left( {{x^2} + xy + {y^2}} \right)}} = \dfrac{{{x^2} - xy + {y^2}}}{{{{(x - y)}^2}}}}$$