Алгебралық бөлшектерді теңбе-тең түрлендіру

$$\frac{{1 + {a^2} + a}}{{1 – {a^3}}} + \frac{{a – {a^2}}}{{{{(1 – a)}^3}}} = \frac{{1 + {a^2} + a}}{{(1 – a)\left( {1 + {a^2} + a} \right)}} + \frac{{a(1 – a)}}{{{{(1 – a)}^3}}} = $$ $$ = \frac{1}{{1 – a}} + \frac{a}{{{{(1 – a)}^2}}} = \frac{{1 – a + a}}{{{{(1 – a)}^2}}} = \frac{1}{{{{(1 – a)}^2}}}$$

$$\frac{{{a^2}(a – b)}}{{{a^3} – {b^3}}} + \frac{{{b^2} + ab}}{{{a^2} + ab + {b^2}}} = \frac{{{a^2}(a – b)}}{{(a – b)\left( {{a^2} + ab + {b^2}} \right)}} + \frac{{{b^2} + ab}}{{{a^2} + ab + {b^2}}} = $$ $$ = \frac{{{a^2}}}{{{a^2} + ab + {b^2}}} + \frac{{{b^2} + ab}}{{{a^2} + ab + {b^2}}} = \frac{{{a^2} + ab + {b^2}}}{{{a^2} + ab + {b^2}}} = 1$$

$$\frac{{{a^3} + {b^3}}}{{b\left( {{a^2} – ab + {b^2}} \right)}} – \frac{a}{b} = \frac{{(a + b)\left( {{a^2} – ab + {b^2}} \right)}}{{b\left( {{a^2} – ab + {b^2}} \right)}} – \frac{a}{b} = $$ $$ = \frac{{a + b}}{b} – \frac{a}{b} = \frac{{a + b – a}}{b} = 1$$

$$\left( {\frac{1}{{{{(a – b)}^2}}} – \frac{1}{{{{(a + b)}^2}}}} \right) \cdot {\left( {{a^2} – {b^2}} \right)^2} = \left( {\frac{1}{{{{(a – b)}^2}}} – \frac{1}{{{{(a + b)}^2}}}} \right) \cdot {(a – b)^2} \cdot {(a + b)^2} = $$ $$ = \frac{{{{(a – b)}^2} \cdot {{(a + b)}^2}}}{{{{(a – b)}^2}}} – \frac{{{{(a – b)}^2} \cdot {{(a + b)}^2}}}{{{{(a + b)}^2}}} = {(a + b)^2} – {(a – b)^2} = 4ab$$

$$\left( {{a^2} – {x^2}} \right)\left( {\frac{1}{{a + x}} + \frac{1}{{{a^2} – {x^2}}} – \frac{1}{{a – x}}} \right) = \frac{{(a – x)(a + x)}}{{a + x}} + $$ $$ + \frac{{{a^2} – {x^2}}}{{{a^2} – {x^2}}} – \frac{{(a – x)(a + x)}}{{a – x}} = a – x + 1 – a – x = 1 – 2x$$

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

Бөлшектің негізгі қасиетін пайдаланып, алымын да бөлімін де $xy$ өрнегіне көбейтеміз. $$\frac{{\left( {\frac{x}{y} – \frac{y}{x}} \right)xy}}{{\left( {\frac{x}{y} + \frac{y}{x} – 2} \right)xy}} = \frac{{\frac{x}{y} \cdot xy – \frac{y}{x} \cdot xy}}{{\frac{x}{y} \cdot xy + \frac{y}{x} \cdot xy – 2xy}} = $$ $$ = \frac{{{x^2} – {y^2}}}{{{x^2} + {y^2} – 2xy}} = \frac{{(x – y)(x + y)}}{{{{(x – y)}^2}}} = \frac{{x + y}}{{x – y}}$$

Бөлшектің алымы мен бөлімін $x$-ке көбейтеміз. $$\frac{{\left( {1 – \frac{1}{x}} \right) \cdot x}}{{\left( {x + \frac{1}{x} – 2} \right) \cdot x}} = \frac{{x – 1}}{{{x^2} + 1 – 2x}} = \frac{{x – 1}}{{{{(x – 1)}^2}}} = \frac{1}{{x – 1}}$$

$\frac{1}{(x+1)(x+2)}=\frac{1}{x+1}-\frac{1}{x+2}$ теңдігі орындалатынын байқауға болады. Онда $$\frac{1}{{(x + 1)(x + 2)}} + \frac{1}{{(x + 2)(x + 3)}} + \frac{1}{{(x + 3)(x + 4)}} = \frac{1}{{x + 1}} – \frac{1}{{x + 2}} + $$ $$ + \frac{1}{{x + 2}} – \frac{1}{{x + 3}} + \frac{1}{{x + 3}} – \frac{1}{{x + 4}} = \frac{1}{{x + 1}} – \frac{1}{{x + 4}} = \frac{3}{{(x + 1)(x + 4)}}$$

Есепті шығаруды мына қадамдар бойынша орындаймыз:

• Алдымен алғашқы екі бөлшекті қосып аламыз;
• Шыққан өрнекті үшінші бөлшекке қосамыз.
• Алғашқы үш бөлшектің қосындысын соңғы бөлшекке қосамыз.

$$\frac{1}{{x – 2}} + \frac{1}{{x + 2}} + \frac{{2x}}{{{x^2} + 4}} + \frac{{4{x^3}}}{{{x^4} + 16}} = \frac{{2x}}{{{x^2} – 4}} + \frac{{2x}}{{{x^2} + 4}} + \frac{{4{x^3}}}{{{x^4} + 16}} = $$ $$ = \frac{{4{x^3}}}{{{x^4} – 16}} + \frac{{4{x^3}}}{{{x^4} + 16}} = \frac{{8{x^7}}}{{{x^8} – 256}}$$

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

$$\frac{{{x^2} + x – 20}}{{x – 4}} – \frac{{2{x^2} – 5x + 3}}{{2x – 3}} + \frac{{5{x^2} + 8x – 4}}{{x + 2}} = \frac{{(x + 5)(x – 4)}}{{x – 4}} – $$ $$ – \frac{{2(x – 1)\left( {x – \frac{3}{2}} \right)}}{{2x – 3}} + \frac{{5(x + 2)\left( {x – \frac{2}{5}} \right)}}{{x + 2}} = \frac{{(x + 5)(x – 4)}}{{x – 4}} – $$ $$ – \frac{{(x – 1)(2x – 3)}}{{2x – 3}} + \frac{{(x + 2)(5x – 2)}}{{x + 2}} = x + 5 – (x – 1) + 5x – 2 = 5x + 4$$

$$\frac{{(1 – x)(x + 2)}}{{{x^2}{{(x + 1)}^2}}} = \frac{{\left( {1 – \frac{{\sqrt 3 – 1}}{2}} \right)\left( {\frac{{\sqrt 3 – 1}}{2} + 2} \right)}}{{\frac{{{{\left( {\sqrt 3 – 1} \right)}^2}}}{4}{{\left( {\frac{{\sqrt 3 – 1}}{2} + 1} \right)}^2}}} = $$ $$ = \frac{{\frac{{3 – \sqrt 3 }}{2} \cdot \frac{{3 + \sqrt 3 }}{2}}}{{\frac{{{{\left( {\sqrt 3 – 1} \right)}^2}}}{4} \cdot \frac{{{{\left( {\sqrt 3 + 1} \right)}^2}}}{4}}} = \frac{{\frac{{9 – 3}}{4}}}{{\frac{{{{(3 – 1)}^2}}}{{16}}}} = \frac{{6 \cdot 16}}{{4 \cdot 4}} = 6$$