Бүтін көрсеткішті дәрежелі өрнектерді түрлендіру
Алгебралық өрнектердің қосындысынан, айырмасынан, көбейтіндісінен, бөліндісінен және бүтін көрсеткішті дәрежесінен тұратын өрнектерді — рационал өрнектер деп атайды.
Анықтама. Егер $a \ne 0$ және $n$ — натурал сан болса, онда $a^{-n}=\dfrac{1}{a^{n}}$.
$0^{-n}$ өрнегінің мағынасы жоқ.
Мысалдар қарастырайық.
2.33. Амалдарды орындаңыз:
1) $\left(\dfrac{c^4}{6 x^2 y^{-5}}\right)^{-2} \cdot\left(\dfrac{1}{3} c^2 x^3 y^{-2}\right)^4$
Шешуі
$${\left( {\frac{{{c^4}}}{{6{x^2}{y^{ – 5}}}}} \right)^{ – 2}} \cdot {\left( {\frac{1}{3}{c^2}{x^3}{y^{ – 2}}} \right)^4} = \frac{{36{x^4}{y^{ – 10}}}}{{{c^8}}} \cdot \frac{1}{{81}}{c^8}{x^{12}}{y^{ – 8}} = \frac{4}{9}{x^{16}}{y^{ – 18}} = \frac{{4{x^{10}}}}{{9{y^{18}}}}$$
2) $\left(\dfrac{1}{4} m^2 n\right)^3 \cdot\left(-32 m^2 n\right)$
Шешуі
$${\left( {\frac{1}{4}{m^2}n} \right)^3} \cdot \left( { – 32{m^2}n} \right) = – \frac{1}{{64}}{m^6}{n^3} \cdot \left( { – 32{m^2}n} \right) = – \frac{1}{2}{m^8}{n^4}$$
3) $\dfrac{2 a^4 b^{-3}}{3 x^4 y^{-3}} \cdot \dfrac{6 a^{-4} b^4}{5 x^{-5} y^3}$
Шешуі
$$\frac{{2{a^4}{b^{ – 3}}}}{{3{x^4}{y^{ – 3}}}} \cdot \frac{{6{a^{ – 4}}{b^4}}}{{5{x^{ – 5}}{y^3}}} = \frac{{0,8b}}{{{x^{ – 1}}}} = 0,8bx$$
4) $\left(\dfrac{a^{-3} b^2}{9^{-1} c^{-2}}\right)^{-2}:\left(\dfrac{a^2 b^3}{6 c^3}\right)^2$
Шешуі
$${\left( {\frac{{{a^{ – 3}}{b^2}}}{{{9^{ – 1}}{c^{ – 2}}}}} \right)^{ – 2}}:{\left( {\frac{{{a^2}{b^3}}}{{6{c^3}}}} \right)^2} = \frac{{{a^6}{b^{ – 4}}}}{{{9^2}{c^4}}}:\frac{{{a^4}{b^6}}}{{{6^2}{c^6}}} = \frac{{{a^6}{b^{ – 4}} \cdot 36{c^6}}}{{81{c^4} \cdot {a^4}{b^6}}} = \frac{{4{a^2}{c^2}}}{{9{b^{10}}}}$$
2.34. Өрнекті ықшамдаңыз
1) $\left(\dfrac{a^{-1}+b^{-1}}{a+b}\right)^{-1}$
Шешуі
$${\left( {\frac{{{a^{ – 1}} + {b^{ – 1}}}}{{a + b}}} \right)^{ – 1}} = \frac{{a + b}}{{\frac{1}{a} + \frac{1}{b}}} = \frac{{(a + b) \cdot ab}}{{\left( {\frac{1}{a} + \frac{1}{b}} \right) \cdot ab}} = \frac{{(a + b) \cdot ab}}{{a + b}} = ab$$
2) $\left(a b^{-2}+a^{-2} b\right)\left(a^{-1}+b^{-1}\right)^{-1}$
Шешуі
$$\left( {a{b^{ – 2}} + {a^{ – 2}}b} \right){\left( {{a^{ – 1}} + {b^{ – 1}}} \right)^{ – 1}} = \left( {\frac{a}{{{b^2}}} + \frac{b}{{{a^2}}}} \right) \cdot {\left( {\frac{1}{a} + \frac{1}{b}} \right)^{ – 1}} = $$ $$ = \left( {\frac{{{a^3} + {b^3}}}{{{a^2}{b^2}}}} \right) \cdot {\left( {\frac{{a + b}}{{ab}}} \right)^{ – 1}} = \frac{{(a + b)\left( {{a^2} – ab + {b^2}} \right) \cdot ab}}{{{a^2}{b^2} \cdot (a + b)}} = \frac{{{a^2} – ab + {b^2}}}{{ab}}$$
3) $\left(1+\dfrac{x^{-n}+y^{-n}}{x^{-n}-y^{-n}}\right)^{-2}$, мұндағы $x=3, \quad y=0,75, \quad n=1$
Шешуі
$${\left( {1 + \frac{{{x^{ – n}} + {y^{ – n}}}}{{{x^{ – n}} – {y^{ – n}}}}} \right)^{ – 2}} = {\left( {1 + \frac{{\frac{1}{{{x^n}}} + \frac{1}{{{y^n}}}}}{{\frac{1}{{{x^n}}} – \frac{1}{{{y^n}}}}}} \right)^{ – 2}} = {\left( {1 + \frac{{{x^n} + {y^n}}}{{{y^n} – {x^n}}}} \right)^{ – 2}} = $$ $$ = {\left( {\frac{{2{y^n}}}{{{y^n} – {x^n}}}} \right)^{ – 2}} = {\left( {\frac{{{y^n} – {x^n}}}{{2{y^n}}}} \right)^2} = \frac{1}{4}{\left( {1 – {{\left( {\frac{x}{y}} \right)}^n}} \right)^2} = \frac{1}{4}{(1 – 4)^2} = \frac{9}{4} = 2\frac{1}{4}$$
4) $\dfrac{x^{-6}+x^{-4}+x^{-2}}{x^2+x^4+x^6}$
Шешуі
$$\frac{{{x^{ – 6}} + {x^{ – 4}} + {x^{ – 2}}}}{{{x^2} + {x^4} + {x^6}}} = \frac{{{x^{ – 2}}\left( {\frac{1}{{{x^4}}} + \frac{1}{{{x^2}}} + 1} \right)}}{{{x^2}\left( {1 + {x^2} + {x^4}} \right)}} = $$ $$ = {x^{ – 4}} \cdot \frac{{\frac{{1 + {x^2} + {x^4}}}{{{x^4}}}}}{{1 + {x^2} + {x^4}}} = {x^{ – 4}} \cdot \frac{{1 + {x^2} + {x^4}}}{{{x^4}\left( {1 + {x^2} + {x^4}} \right)}} = {x^{ – 8}}$$
5) $\left(x^2-a^{-1} x+a^{-2}\right)\left(x^{-1}+a\right)-x(a x)^{-2}$
Шешуі
$$\left( {{x^2} – {a^{ – 1}}x + {a^{ – 2}}} \right)\left( {{x^{ – 1}} + a} \right) – x{(ax)^{ – 2}} = \left( {{x^2} – \frac{x}{a} + \frac{1}{{{a^2}}}} \right)\left( {\frac{1}{x} + a} \right) – \frac{x}{{{a^2}{x^2}}} = $$ $$ = \frac{{\left( {{a^2}{x^2} – ax + 1} \right)(1 + ax)}}{{{a^2}x}} – \frac{1}{{{a^2}x}} = \frac{{1 + {a^3}{x^3}}}{{{a^2}x}} – \frac{1}{{{a^2}x}} = \frac{{{a^3}{x^3}}}{{{a^2}x}} = a{x^2}$$
6) $\left(\dfrac{x^{-1}+1}{x^{-1}-1}\right)^{-1}-1-2(1+x)^{-1}$
Шешуі
$${\left( {\frac{{{x^{ – 1}} + 1}}{{{x^{ – 1}} – 1}}} \right)^{ – 1}} – 1 – 2{(1 + x)^{ – 1}} = {\left( {\frac{{\frac{1}{x} + 1}}{{\frac{1}{x} – 1}}} \right)^{ – 1}} – 1 – \frac{2}{{x + 1}} = $$ $$ = {\left( {\frac{{1 + x}}{{1 – x}}} \right)^{ – 1}} – 1 – \frac{2}{{x + 1}} = \frac{{1 – x}}{{1 + x}} – 1 – \frac{2}{{x + 1}} = $$ $$ = \frac{{1 – x – x – 1 – 2}}{{x + 1}} = \frac{{ – 2x – 2}}{{x + 1}} = – 2$$
7) $a^{-5}\left(\dfrac{25-4 a^{-4}}{5 a^{-1}+2 a^{-3}}-\dfrac{1-1,5 a^{-2}-a^{-4}}{a^{-1}+0,5 a^{-3}}\right)^5$
Шешуі
$${a^{ – 5}}{\left( {\frac{{25 – 4{a^{ – 4}}}}{{5{a^{ – 1}} + 2{a^{ – 3}}}} – \frac{{1 – 1,5{a^{ – 2}} – {a^{ – 4}}}}{{{a^{ – 1}} + 0,5{a^{ – 3}}}}} \right)^5} = {a^{ – 5}}{\left( {\frac{{25 – \frac{4}{{{a^4}}}}}{{\frac{5}{a} + \frac{2}{{{a^3}}}}} – \frac{{1 – \frac{3}{{2{a^2}}} – \frac{1}{{{a^4}}}}}{{\frac{1}{a} + \frac{1}{{2{a^3}}}}}} \right)^5} = $$ $$ = \frac{1}{{{a^5}}} \cdot {\left( {\frac{{25{a^4} – 4}}{{a\left( {5{a^2} + 2} \right)}} – \frac{{2{a^4} – 3{a^2} – 2}}{{a\left( {2{a^2} + 1} \right)}}} \right)^5} = $$ $$ = \frac{1}{{{a^5}}} \cdot {\left( {\frac{{\left( {5{a^2} + 2} \right)\left( {5{a^2} – 2} \right)}}{{a\left( {5{a^2} + 2} \right)}} \cdot \frac{{\left( {2{a^2} + 1} \right)\left( {{a^2} – 2} \right)}}{{a\left( {2{a^2} + 1} \right)}}} \right)^5} = $$ $$ = \frac{1}{{{a^5}}} \cdot {\left( {\frac{{5{a^2} – 2}}{a} – \frac{{{a^2} – 2}}{a}} \right)^5} = \frac{1}{{{a^5}}} \cdot {\left( {\frac{{4{a^2}}}{a}} \right)^5} = \frac{{{4^5}{a^5}}}{{{a^5}}} = {4^5} = {2^{10}} = 1024$$