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: $\dfrac{x}{{x — 4}}$