1. Usuń niewymierność z mianownikaa) b) c) 2. Udowodnij nierównośća) b) c) d).... Question from @marcinasdd

November 2018 1 9 Report

1. Usuń niewymierność z mianownika
a) $\frac {1}{\sqrt {2}+ \sqrt {3} - \sqrt {6}}$
b) $\frac {1}{\sqrt[3]{3} - \sqrt[3]{2}}$
c) $\frac {1}{\sqrt[3]{9} - \sqrt [3]{3} +1}$

2. Udowodnij nierówność
a) $\sqrt {8+4\sqrt3}} - \sqrt {8-4\sqrt{3}} = 2\sqrt{2}$
b) $\sqrt {6+4\sqrt2}} + \sqrt{6-4\sqrt2} = 4$
c) $\frac {1}{5-2\sqrt{6}} + \frac{1}{5+2\sqrt{6}} = 10$
d) $\frac{\sqrt5-2\sqrt6}{\sqrt{2}-\sqrt{3}} + \frac{\sqrt8-2\sqrt{15}}{\sqrt{3}-\sqrt{5}} = -2$

Zgłoś nadużycie!

1.
a)
$\frac {1}{\sqrt {2}+ \sqrt {3} - \sqrt {6}}=\frac {1[(\sqrt {2}+ \sqrt {3})+\sqrt {6}]}{[(\sqrt {2}+ \sqrt {3}) - \sqrt {6}][(\sqrt {2}+ \sqrt {3})+\sqrt {6}]}=$

$\frac {\sqrt {2}+ \sqrt {3}+\sqrt {6}}{(\sqrt {2}+ \sqrt {3})^2 - (\sqrt {6})^2}=\frac {(\sqrt {2}+ \sqrt {3})+\sqrt {6}}{2+2\sqrt {6}+3 - 6}=$

$\frac {\sqrt {2}+ \sqrt {3}+\sqrt {6}}{2\sqrt {6}-1}=\frac {(\sqrt {2}+ \sqrt {3}+\sqrt {6})(2\sqrt {6}+1)}{(2\sqrt {6}-1)(2\sqrt {6}+1)}=$

$\frac {2\sqrt {12}+ \sqrt{2}+2 \sqrt {18}+ \sqrt{3}+12+ \sqrt {6}}{24-1}=$

$\frac {4\sqrt {3}+ \sqrt{2}+6\sqrt {2}+ \sqrt{3}+12+ \sqrt {6}}{23}=\frac {7\sqrt {2}+5 \sqrt{3}+\sqrt {6}+ 12}{23}$

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b)
wzór:
$a^3-b^3=(a-b)(a^2+ab+b^2) \Rightarrow \frac{1}{a-b} = \frac{a^2+ab+b^2}{a^3-b^3}$

$(a=\sqrt[3]{3},b=\sqrt[3]{2})$

$\frac {1}{\sqrt[3]{3} - \sqrt[3]{2}}= \frac{(\sqrt[3]{3})^2+\sqrt[3]{3}\cdot\sqrt[3]{2}+(\sqrt[3]{2})^2 }{(\sqrt[3]{3})^3-(\sqrt[3]{2})^3}=$

$\frac{\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4} }{3-2}=\frac{\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4} }{1}=$

$\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4}$
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c)
wzór:
$a^3+b^3=(a + b)(a^2 - ab + b^2) \Rightarrow \frac{1}{a^2 - ab + b^2}= \frac{a+b}{a^3+b^3}$

$(a=\sqrt [3]{3},b=1)$

$\frac {1}{\sqrt[3]{9} - \sqrt [3]{3} +1}=\frac {1}{\sqrt[3]{3^2} - \sqrt [3]{3} +1}=$

$\frac {1}{(\sqrt[3]{3})^2 - \sqrt [3]{3}\cdot 1 +1^2}= \frac{\sqrt [3]{3}+1}{(\sqrt [3]{3})^3+1^3}=$

$\frac{\sqrt [3]{3}+1}{3+1}=\frac{\sqrt [3]{3}+1}{4}$
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2.
a)
$\sqrt {8+4\sqrt3}} - \sqrt {8-4\sqrt{3}} = 2\sqrt{2}$
$L=\sqrt {8+4\sqrt3}} - \sqrt {8-4\sqrt{3}}=\sqrt {6+2\sqrt{12}+2} - \sqrt {6-2\sqrt{12}+2}=$

$\sqrt {( \sqrt{6} )^2+2\sqrt{12}+( \sqrt{2} )^2} - \sqrt {( \sqrt{6} )^2-2\sqrt{12}+( \sqrt{2} )^2}=$

$\sqrt {( \sqrt{6}+ \sqrt{2} )^2} - \sqrt {( \sqrt{6}- \sqrt{2} )^2}=$

$\sqrt{6}+ \sqrt{2} - ( \sqrt{6}- \sqrt{2})=\sqrt{6}+ \sqrt{2} - \sqrt{6}+ \sqrt{2}=2 \sqrt{2}=P$

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b)
$\sqrt {6+4\sqrt2}} + \sqrt{6-4\sqrt2} = 4$

$L=\sqrt {6+4\sqrt2}} + \sqrt{6-4\sqrt2}=\sqrt {4+4\sqrt2}+2} + \sqrt{4-4\sqrt2+2}=$

$\sqrt {2^2+4\sqrt2+( \sqrt{2} )^2} + \sqrt{2^2-4\sqrt2+( \sqrt{2} )^2}=$

$\sqrt {(2+ \sqrt{2} )^2} + \sqrt{(2-\sqrt{2} )^2}=2+ \sqrt{2}+ 2-\sqrt{2}=4=P$

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c)
c)
$\frac {1}{5-2\sqrt{6}} + \frac{1}{5+2\sqrt{6}}=10$

$L=\frac {1}{5-2\sqrt{6}} + \frac{1}{5+2\sqrt{6}}=$

$\frac {5+2\sqrt{6}}{(5+2\sqrt{6})(5-2\sqrt{6})} + \frac{5-2\sqrt{6}}{(5+2\sqrt{6})(5-2\sqrt{6})}=$

$\frac {5+2\sqrt{6}+5-2\sqrt{6}}{5^2-(2\sqrt{6})^2}= \frac{10}{25-24}= \frac{10}{1}=10=P$
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d)
$\frac{\sqrt5-2\sqrt6}{\sqrt{2}-\sqrt{3}} + \frac{\sqrt8-2\sqrt{15}}{\sqrt{3}-\sqrt{5}}=-2$

$L=\frac{\sqrt5-2\sqrt6}{\sqrt{2}-\sqrt{3}} + \frac{\sqrt8-2\sqrt{15}}{\sqrt{3}-\sqrt{5}}=$

$\frac{(\sqrt5-2\sqrt6)(\sqrt{2}+\sqrt{3})}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})} + \frac{(\sqrt8-2\sqrt{15})(\sqrt{3}+\sqrt{5})}{(\sqrt{3}-\sqrt{5})(\sqrt{3}+\sqrt{5})}=$

$\frac{\sqrt{10}+\sqrt{15}-2\sqrt{12}-2\sqrt{18}}{2-3} + \frac{\sqrt{24}+\sqrt{40}-2\sqrt{45}-2\sqrt{75}}{3-5}=$

$\frac{\sqrt{10}+\sqrt{15}-4\sqrt{3}-6\sqrt{2}}{-1} + \frac{2\sqrt{6}+2\sqrt{10}-6\sqrt{5}-10\sqrt{3}}{-2}=$

$-\sqrt{10}-\sqrt{15}+4\sqrt{3}+6\sqrt{2}-\sqrt{6}-\sqrt{10}+3\sqrt{5}+5\sqrt{3}=$

$-\sqrt{15}-2\sqrt{10}- \sqrt{6} +3\sqrt{5}+9\sqrt{3}+6 \sqrt{2} \neq P$