Алгебралық бөлшектерге амалдар қолдану

$$1)\quad {\frac{{x - y}}{{x + y}}^{\backslash x - y}} + {\frac{{x + y}}{{x - y}}^{\backslash x + y}} = \frac{{{x^2} - 2xy + {y^2} + {x^2} + 2xy + {y^2}}}{{(x + y)(x - y)}} = \frac{{2\left( {{x^2} + {y^2}} \right)}}{{(x + y)(x - y)}}$$ $$2)\quad \frac{{{x^2} + {y^2}}}{{2xy}} + {1^{\backslash 2xy}} = \frac{{{x^2} + {y^2} + 2xy}}{{2xy}} = \frac{{{{(x + y)}^2}}}{{2xy}}$$ $$3)\quad \frac{{2\left( {{x^2} + {y^2}} \right)}}{{(x + y)(x - y)}} \cdot \frac{{{{(x + y)}^2}}}{{2xy}}:\frac{{{x^2} + {y^2}}}{{xy}} = \frac{{2\left( {{x^2} + {y^2}} \right){{(x + y)}^2} \cdot xy}}{{(x + y)(x - y) \cdot 2xy \cdot \left( {{x^2} + {y^2}} \right)}} = \frac{{x + y}}{{x - y}}$$

Жауабы: $\dfrac{{x + y}}{{x - y}}$

$$1)\quad \frac{2}{{2a - b}} + \frac{{6b}}{{{b^2} - 4{a^2}}} - \frac{4}{{2a + b}} = {\frac{2}{{2a - b}}^{\backslash 2a + b}} - \frac{{6b}}{{(2a - b)(2a + b)}} - {\frac{4}{{2a + b}}^{\backslash 2a - b}} = $$ $$ = \frac{{4a + 2b - 6b - 8a + 4b}}{{(2a - b)(2a + b)}} = - \frac{{4a}}{{4{a^2} - {b^2}}}$$ $$2)\quad {\frac{1}{1}^{\backslash 4{a^2} - {b^2}}} + \frac{{4{a^2} + {b^2}}}{{4{a^2} - {b^2}}} = \frac{{4{a^2} - {b^2} + 4{a^2} + {b^2}}}{{4{a^2} - {b^2}}} = \frac{{8{a^2}}}{{4{a^2} - {b^2}}}$$ $$3)\quad - \frac{{4a}}{{4{a^2} - {b^2}}}:\frac{{8{a^2}}}{{4{a^2} - {b^2}}} = - \frac{{4a \cdot \left( {4{a^2} - {b^2}} \right)}}{{\left( {4{a^2} - {b^2}} \right) \cdot 8{a^2}}} = - \frac{1}{{2a}}$$

Жауабы: $ - \dfrac{1}{{2a}}$

$$\frac{{{x^2}}}{{x - 5}} + \frac{{25}}{{5 - x}} = \frac{{{x^2}}}{{x - 5}} - \frac{{25}}{{x - 5}} = \frac{{{x^2} - 25}}{{x - 5}} = \frac{{(x - 5)(x + 5)}}{{(x - 5)}} = x + 5$$

$$\frac{{{x^2}}}{{{{(x - 2)}^2}}} - \frac{4}{{{{(2 - x)}^2}}} = \frac{{{x^2}}}{{{{(x - 2)}^2}}} - \frac{4}{{{{(x - 2)}^2}}} = \frac{{{x^2} - 4}}{{{{(x - 2)}^2}}} = \frac{{(x - 2)(x + 2)}}{{{{(x - 2)}^2}}} = \frac{{x + 2}}{{x - 2}}$$

$$\frac{{{x^2} + 4}}{{{{(x - 2)}^3}}} + \frac{{4x}}{{{{(2 - x)}^3}}} = \frac{{{x^2} + 4}}{{{{(x - 2)}^3}}} - \frac{{4x}}{{{{(x - 2)}^3}}} = \frac{{{x^2} + 4 - 4x}}{{{{(x - 2)}^3}}} = \frac{{{{(x - 2)}^2}}}{{{{(x - 2)}^3}}} = \frac{1}{{x - 2}}$$

$$1)\quad \frac{3}{{x - 4}} + \frac{{4x - 6}}{{{x^2} - 3x - 4}} + \frac{{2x}}{{x + 1}} = {\frac{3}{{x - 4}}^{\backslash x + 1}} + \frac{{4x - 6}}{{(x + 1)(x - 4)}} + {\frac{{2x}}{{x + 1}}^{\backslash x - 4}} = $$ $$ = \frac{{3x + 3 + 4x - 6 + 2{x^2} - 8x}}{{(x + 1)(x - 4)}} = \frac{{2{x^2} - x - 3}}{{(x + 1)(x - 4)}} = \frac{{(2x - 3)(x + 1)}}{{(x + 1)(x - 4)}} = \frac{{2x - 3}}{{x - 4}}$$ $$2)\quad \frac{{2x - 3}}{{x - 4}} \cdot \frac{x}{{2x - 3}} = \frac{{(2x - 3) \cdot x}}{{(x - 4)(2x - 3)}} = \frac{x}{{x - 4}}$$

Жауабы: $\dfrac{x}{{x - 4}}$