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: $\dfrac{x}{{x – 4}}$