Алгебралық бөлшектерді қысқарту

$$\frac{{3x – xy + 2y – 6}}{{xy – 3x + 2y – 6}} = \frac{{(3x – xy) + (2y – 6)}}{{(xy – 3x) + (2y – 6)}} = $$ $$ = \frac{{ – x(y – 3) + 2(y – 3)}}{{x(y – 3) + 2(y – 3)}} = \frac{{(y – 3)(2 – x)}}{{(y – 3)(x + 2)}} = \frac{{2 – x}}{{x + 2}}$$

$$\frac{{{x^{14}} + {x^7} + 1}}{{{x^{21}} – 1}} = \frac{{{x^{14}} + {x^7} + 1}}{{{{\left( {{x^7}} \right)}^3} – 1}} = \frac{{{x^{14}} + {x^7} + 1}}{{\left( {{x^7} – 1} \right)\left( {{x^{14}} + {x^7} + 1} \right)}} = \frac{1}{{{x^7} – 1}}$$

$$\frac{{{a^{33}} + 1}}{{{a^{11}} – {a^{22}} + {a^{33}}}} = \frac{{{{\left( {{a^{11}}} \right)}^3} + 1}}{{{a^{11}}\left( {1 – {a^{11}} + {a^{22}}} \right)}} = \frac{{\left( {{a^{11}} + 1} \right)\left( {{a^{22}} – {a^{11}} + 1} \right)}}{{{a^{11}}\left( {{a^{22}} – {a^{11}} + 1} \right)}} = \frac{{{a^{11}} + 1}}{{{a^{11}}}}$$

$$\frac{{{a^4} – 16}}{{{a^4} – 4{a^3} + 8{a^2} – 16a + 16}} = \frac{{{{\left( {{a^2}} \right)}^2} – {4^2}}}{{\left( {{a^4} + 8{a^2} + 16} \right) – \left( {4{a^3} + 16a} \right)}} = $$ $$ = \frac{{\left( {{a^2} – 4} \right)\left( {{a^2} + 4} \right)}}{{{{\left( {{a^2} + 4} \right)}^2} – 4a\left( {{a^2} + 4} \right)}} = \frac{{(a – 2)(a + 2)\left( {{a^2} + 4} \right)}}{{\left( {{a^2} + 4} \right)\left( {{a^2} + 4 – 4a} \right)}} = \frac{{(a – 2)(a + 2)}}{{{{(a – 2)}^2}}} = \frac{{a + 2}}{{a – 2}}$$

$${\frac{{{x^4} + {x^2}{y^2} + {y^4}}}{{{x^2} – xy + {y^2}}} = \frac{{{x^4} + 2{x^2}{y^2} + {y^4} – {x^2}{y^2}}}{{{x^2} – xy + {y^2}}} = \frac{{{{\left( {{x^2} + {y^2}} \right)}^2} – {{(xy)}^2}}}{{{x^2} – xy + {y^2}}} = }$$ $${ = \frac{{\left( {{x^2} + {y^2} – xy} \right)\left( {{x^2} + {y^2} + xy} \right)}}{{{x^2} – xy + {y^2}}} = {x^2} + xy + {y^2}}$$

$$\frac{{{x^2} – 9x + 14}}{{ – {x^2} + 10x – 16}} = – \frac{{{x^2} – 9x + 14}}{{{x^2} – 10x + 16}} = \left\| {\begin{array}{*{20}{c}}{{x^2} – 9x + 14 = 0;}&{{x^2} – 10x + 16 = 0}\\{{x_1} = 2,\quad {x_2} = 7;}&{{x_1} = 2,\quad {x_2} = 8}\end{array}} \right\| = $$ $$ = – \frac{{(x – 2)(x – 7)}}{{(x – 2)(x – 8)}} = \frac{{x – 7}}{{8 – x}}$$

$$\frac{{7{x^2}{y^4} + 7{x^4}{y^2}}}{{{x^6} + {y^6}}} = \frac{{7{x^2}{y^2}\left( {{y^2} + {x^2}} \right)}}{{{{\left( {{x^2}} \right)}^3} + {{\left( {{y^2}} \right)}^3}}} = \frac{{7{x^2}{y^2}\left( {{y^2} + {x^2}} \right)}}{{\left( {{x^2} + {y^2}} \right)\left( {{x^4} – {x^2}{y^2} + {y^4}} \right)}} = \frac{{7{x^2}{y^2}}}{{{x^4} – {x^2}{y^2} + {y^4}}}$$

$${\frac{{x(x + 2) – y\left( {y – 2} \right)}}{{x – y – 2}} = \frac{{{x^2} + 2x – {y^2} + 2y}}{{x – y + 2}} = \frac{{\left( {{x^2} – {y^2}} \right) + 2(x + y)}}{{x – y + 2}} = }$$ $${ = \frac{{(x + y)(x – y + 2)}}{{x – y + 2}} = x + y = \frac{4}{{15}} + \frac{2}{5} = \frac{{10}}{{15}} = \frac{2}{3}}$$

$$\frac{{{x^3} + 2{x^2} + 4x + 3}}{{{x^2} + 7x + 6}} = \left\| {\begin{array}{*{20}{l}}{{x^2} + 7x + 6 = 0}\\{{x_1} = – 1,\quad {x_2} = – 6}\end{array}} \right\| = \frac{{{x^3} + \overbrace {{x^2} + {x^2}}^{2{x^2}} + \overbrace {x + 3x}^{4x} + 3}}{{(x + 1)(x + 6)}} = $$ $$ = \frac{{\left( {{x^3} + {x^2}} \right) + \left( {{x^2} + x} \right) + (3x + 3)}}{{(x + 1)(x + 6)}} = \frac{{{x^2}(x + 1) + x(x + 1) + 3(x + 1)}}{{(x + 1)(x + 6)}} = $$ $$ = \frac{{(x + 1)\overbrace {\left( {{x^2} + x + 3} \right)}^{D \lt 0}}}{{(x + 1)(x + 6)}} = \frac{{{x^2} + x + 3}}{{x + 6}}$$

$$\frac{{{{(2x – 1)}^2} + 2\left( {4{x^2} – 1} \right) + {{(2x + 1)}^2}}}{{{{\left( {2{x^2} + 1} \right)}^2} – {{\left( {2{x^2} – 1} \right)}^2}}} = \frac{{{{(2x – 1)}^2} + 2(2x – 1)(2x + 1) + {{(2x + 1)}^2}}}{{\left( {2{x^2} + 1 – 2{x^2} + 1} \right)\left( {2{x^2} + 1 + 2{x^2} – 1} \right)}} = $$ $$ = \frac{{{{(2x – 1 + 2x + 1)}^2}}}{{2 \cdot 4{x^2}}} = \frac{{16{x^2}}}{{8{x^2}}} = 2$$

$${\frac{{9 – 4{a^2} – 4ab – {b^2}}}{{4{a^2} + 2ab + 3b – 9}} = \frac{{3 + 2a + b}}{x}}$$ $${\frac{{9 – {{(2a + b)}^2}}}{{\left( {4{a^2} – 9} \right) + (2ab + 3b)}} = \frac{{3 + 2a + b}}{x}}$$ $${\frac{{(3 – 2a – b)(3 + 2a + b)}}{{(2a – 3)(2a + 3) + b(2a + 3)}} = \frac{{3 + 2a + b}}{x}}$$ $${\frac{{(3 – 2a – b)(3 + 2a + b)}}{{(2a + 3)(2a – 3 + b)}} = \frac{{3 + 2a + b}}{x}}$$ $${\frac{{ – (2a + b – 3)(3 + 2a + b)}}{{(2a + 3)(2a + b – 3)}} = \frac{{3 + 2a + b}}{x}}$$ $${\frac{{ – (3 + 2a + b)}}{{(2a + 3)}} = \frac{{3 + 2a + b}}{x}}$$ $${x = – \frac{{(2a + 3)(3 + 2a + b)}}{{(3 + 2a + b)}}}$$ $${x = – 2a – 3}$$

$a=x^2-x-5$ жаңа айнымалысын енгіземіз. Онда: $${x^2} – x – 2 = {x^2} – x – 5 + 3 = a + 3$$ $${x^2} – x – 1 = {x^2} – x – 5 + 4 = a + 4$$ $$\frac{{\left( {{x^2} – x – 5} \right)\left( {{x^2} – x – 2} \right) + 2}}{{\left( {{x^2} – x – 5} \right)\left( {{x^2} – x – 1} \right) + 4}} = \frac{{a(a + 3) + 2}}{{a(a + 4) + 4}} = \frac{{{a^2} + 3a + 2}}{{{a^2} + 4a + 4}} = $$ $$ = \frac{{(a + 1)(a + 2)}}{{{{(a + 2)}^2}}} = \frac{{a + 1}}{{a + 2}} = \frac{{{x^2} – x – 4}}{{{x^2} – x – 3}}$$