Бөлшекті қысқарту
Бөлшекті қысқарту дегеніміз — бөлшектің алымын да, бөлімін деп бірдей санға немесе бірдей өрнекке бөлу.
1.7. Бөлшекті қысқартыңыз:
1) $\dfrac{{{{21}^8} \cdot {4^6}}}{{{3^{21}} \cdot {7^5}}}:\dfrac{{{8^5} \cdot {{49}^3}}}{{{{14}^4} \cdot {9^5}}}$
Шешуі
$$\dfrac{{{{21}^8} \cdot {4^6}}}{{{3^{21}} \cdot {7^5}}}:\dfrac{{{8^5} \cdot {{49}^3}}}{{{{14}^4} \cdot {9^5}}} = \dfrac{{{3^8} \cdot {7^8} \cdot {2^{12}} \cdot {7^4} \cdot {2^4} \cdot {3^{10}}}}{{{3^{21}} \cdot {7^5} \cdot {2^{15}} \cdot {7^6}}} = \dfrac{{{2^{16}} \cdot {3^{18}} \cdot {7^{12}}}}{{{2^{15}} \cdot {3^{21}} \cdot {7^{11}}}} = 2 \cdot {3^{ - 3}} \cdot 7 = \dfrac{{14}}{{27}}$$
2) $\dfrac{{{7^8} \cdot {{49}^{ - 2}} \cdot {5^4} + 49 \cdot 125 \cdot {{\left( {\frac{1}{5}} \right)}^{ - 1}}}}{{{{(7 \cdot 5)}^4} \cdot {7^{ - 3}}}}$
Шешуі
$$\frac{{{7^8} \cdot {{49}^{ - 2}} \cdot {5^4} + 49 \cdot 125 \cdot {{\left( {\frac{1}{5}} \right)}^{ - 1}}}}{{{{(7 \cdot 5)}^4} \cdot {7^{ - 3}}}} = \frac{{{7^8} \cdot {7^{ - 4}} \cdot {5^4} + {7^2} \cdot {5^3} \cdot 5}}{{{7^4} \cdot {5^4} \cdot {7^{ - 3}}}} = \frac{{{7^4} \cdot {5^4} + {7^2} \cdot {5^4}}}{{7 \cdot {5^4}}} = $$ $$ = \frac{{{7^2} \cdot {5^4} \cdot \left( {{7^2} + 1} \right)}}{{7 \cdot {5^4}}} = 7 \cdot 50 = 350$$
3) $\dfrac{\sqrt{21}+\sqrt{14}}{\sqrt{7}}$
Шешуі
$$\frac{{\sqrt {21} + \sqrt {14} }}{{\sqrt 7 }} = \frac{{\sqrt 7 (\sqrt 3 + \sqrt 2 )}}{{\sqrt 7 }} = \sqrt 3 + \sqrt 2 $$
4) $\dfrac{2 \sqrt{10}+4-2 \sqrt{2}}{\sqrt{5}+\sqrt{2}-1}$
Шешуі
$$\frac{{2\sqrt {10} + 4 - 2\sqrt 2 }}{{\sqrt 5 + \sqrt 2 - 1}} = \frac{{2\sqrt 5 \cdot \sqrt 2 + 2\sqrt 2 \cdot \sqrt 2 - 2\sqrt 2 }}{{\sqrt 5 + \sqrt 2 - 1}} = \frac{{2\sqrt 2 (\sqrt 5 + \sqrt 2 - 1)}}{{\sqrt 5 + \sqrt 2 - 1}} = 2\sqrt 2 $$
5) $\dfrac{2 \sqrt{10}-5}{4-\sqrt{10}}$
Шешуі
$$\frac{{2\sqrt {10} - 5}}{{4 - \sqrt {10} }} = \frac{{\sqrt 5 \cdot (2\sqrt 2 - \sqrt 5 )}}{{\sqrt 2 \cdot (2\sqrt 2 - \sqrt 5 )}} = \sqrt {\frac{5}{2}} $$
6) $\dfrac{{9 - 2\sqrt 3 }}{{3\sqrt 6 - 2\sqrt 2 }}$
Шешуі
$$\frac{{9 - 2\sqrt 3 }}{{3\sqrt 6 - 2\sqrt 2 }} = \frac{{\sqrt 3 \cdot (3\sqrt 3 - 2)}}{{\sqrt 2 \cdot (3\sqrt 3 - 2)}} = \sqrt {\frac{3}{2}} $$
7) $\dfrac{(\sqrt{10}-1)^2-3}{\sqrt{10}+\sqrt{3}-1}$
Шешуі
$$\frac{{{{(\sqrt {10} - 1)}^2} - 3}}{{\sqrt {10} + \sqrt 3 - 1}} = \frac{{{{(\sqrt {10} - 1)}^2} - {{(\sqrt 3 )}^2}}}{{\sqrt {10} + \sqrt 3 - 1}} = \frac{{(\sqrt {10} - 1 + \sqrt 3 )(\sqrt {10} - 1 - \sqrt 3 )}}{{\sqrt {10} + \sqrt 3 - 1}} = $$
8) $\dfrac{\sqrt[8]{9} \cdot \sqrt[3]{40} \cdot \sqrt[4]{4}}{\sqrt[6]{25} \cdot \sqrt{2} \cdot \sqrt[4]{3}}$
Шешуі
$$\frac{{\sqrt[8]{9} \cdot \sqrt[3]{{40}} \cdot \sqrt[4]{4}}}{{\sqrt[6]{{25}} \cdot \sqrt 2 \cdot \sqrt[4]{3}}} = \frac{{\sqrt[8]{{{3^2}}} \cdot \sqrt[3]{{{2^3}}} \cdot 5 \cdot \sqrt[4]{{{2^2}}}}}{{\sqrt[6]{{{5^2}}} \cdot \sqrt 2 \cdot \sqrt[4]{3}}} = \frac{{\sqrt[4]{3} \cdot 2 \cdot \sqrt[3]{5} \cdot \sqrt 2 }}{{\sqrt[3]{5} \cdot \sqrt 2 \cdot \sqrt[4]{3}}} = 2$$
9) $\dfrac{(\sqrt{5}-\sqrt{11})(\sqrt{33}+\sqrt{15}-\sqrt{22}-\sqrt{10})}{\sqrt{75}-\sqrt{50}}$
Шешуі
$$\frac{{(\sqrt 5 - \sqrt {11} )(\sqrt {3 \cdot 11} + \sqrt {3 \cdot 5} - \sqrt {2 \cdot 11} - \sqrt {2 \cdot 5} )}}{{\sqrt {25 \cdot 3} - \sqrt {25 \cdot 2} }} = $$ $$ = \frac{{(\sqrt 5 - \sqrt {11} )(\sqrt 3 (\sqrt {11} + \sqrt 5 ) - \sqrt 2 (\sqrt {11} + \sqrt 5 ))}}{{5(\sqrt 3 - \sqrt 2 )}} = $$ $$ = \frac{{(\sqrt 5 - \sqrt {11} )(\sqrt {11} + \sqrt 5 )(\sqrt 3 - \sqrt 2 )}}{{5(\sqrt 3 - \sqrt 2 )}} = \frac{{5 - 11}}{5} = - \frac{6}{5} = - 1,2$$
10)$\dfrac{\sqrt{6}+\sqrt{3}-\sqrt{2}-1}{\sqrt{6}+2 \sqrt{3}-\sqrt{2}-2}$
Шешуі
$$\frac{{\sqrt 6 + \sqrt 3 - \sqrt 2 - 1}}{{\sqrt 6 + 2\sqrt 3 - \sqrt 2 - 2}} = \frac{{\sqrt {3 \cdot 2} + \sqrt 3 - \sqrt 2 - 1}}{{\sqrt {3 \cdot 2} - \sqrt 2 + 2\sqrt 3 - 2}} = \frac{{\sqrt 3 (\sqrt 2 + 1) - (\sqrt 2 + 1)}}{{\sqrt 2 (\sqrt 3 - 1) + 2(\sqrt 3 - 1)}} = $$ $$\frac{{(\sqrt 2 + 1)(\sqrt 3 - 1)}}{{(\sqrt 3 - 1)(\sqrt 2 + 2)}} = \frac{{\sqrt 2 + 1}}{{\sqrt 2 + 2}} = \frac{{\sqrt 2 + 1}}{{\sqrt 2 (\sqrt 2 + 1)}} = \frac{1}{{\sqrt 2 }} = \frac{{\sqrt 2 }}{2}$$
11)$\dfrac{{{6^{3 + \sqrt 5 }}}}{{{3^{2 + \sqrt 3 }} \cdot {2^{1 + \sqrt 5 }}}}$
Шешуі
$$\frac{{{6^{3 + \sqrt 5 }}}}{{{3^{2 + \sqrt 5 }} \cdot {2^{1 + \sqrt 5 }}}} = \frac{{{6^2} \cdot {6^{1 + \sqrt 5 }}}}{{3 \cdot {3^{1 + \sqrt 5 }} \cdot {2^{1 + \sqrt 5 }}}} = \frac{{36 \cdot {6^{1 + \sqrt 5 }}}}{{3 \cdot {6^{1 + \sqrt 5 }}}} = 12$$
12)$\dfrac{{{{100}^n}}}{{{2^{2n + 1}} \cdot {5^{2n - 2}}}},\quad n \in \mathbb{N}$
Шешуі
$$\frac{{{{100}^n}}}{{{2^{2n + 1}} \cdot {5^{2n - 2}}}} = \frac{{{{\left( {{2^2} \cdot {5^2}} \right)}^n}}}{{{2^{2n + 1}} \cdot {5^{2n - 2}}}} = {2^{2n - 2n - 1}} \cdot {5^{2n - 2n + 2}} = {2^{ - 1}} \cdot {5^2} = \frac{{25}}{2}$$
13)$\dfrac{{4 \cdot {{18}^n}}}{{{3^{2n - 1}} \cdot {2^{n + 1}}}},\quad n \in \mathbb{N}$
Шешуі
$$\frac{{4 \cdot {{18}^n}}}{{{3^{2n - 1}} \cdot {2^{n + 1}}}} = \frac{{{2^{n + 2}} \cdot {3^{2n}}}}{{{2^{n + 1}} \cdot {3^{2n - 1}}}} = {2^{n + 2 - n - 1}} \cdot {3^{2n - 2n + 1}} = 2 \cdot 3 = 6$$
14)$\dfrac{{{{\left( {{8^{k + 1}} + {8^k}} \right)}^2}}}{{{{\left( {{4^k} - {4^{k - 1}}} \right)}^3}}},\quad k \in \mathbb{N}$
Шешуі
$$\frac{{{{\left( {{8^{k + 1}} + {8^k}} \right)}^2}}}{{{{\left( {{4^k} - {4^{k - 1}}} \right)}^3}}} = \frac{{{{\left( {{8^k}(8 + 1)} \right)}^2}}}{{{{\left( {{4^{k - 1}}(4 - 1)} \right)}^3}}} = \frac{{{8^{2k}} \cdot {9^2}}}{{{4^{3(k - 1)}} \cdot {3^3}}} = \frac{{3 \cdot {2^{6k}}}}{{{2^{6k}} \cdot {2^{ - 6}}}} = \frac{3}{{{2^{ - 6}}}} = 3 \cdot 64 = 192$$