Doprowadź do najprostrzej ostaci wyrażenia:1) √27 + √48 - √75 2)√125 + √24 - √45

Tamiza17 @Tamiza17

September 2018 1 41 Report

Doprowadź do najprostrzej ostaci wyrażenia:

1) √27 + √48 - √75

2)√125 + √24 - √45 - √150

3) $\frac{\sqrt{108}- \sqrt{180}}{ \sqrt{12}- \sqrt{20}}$

4) $\frac{ \sqrt{112}-\sqrt{32}}{\sqrt{252}- \sqrt{72}}$

5) $\frac{11\sqrt{3}-4\sqrt{7}}{\sqrt{3}-\sqrt{7}}-\frac{13\sqrt{7}+\sqrt{3}}{\sqrt{12-\sqrt{28}}}$

6) $\frac{26\sqrt{5}-23\sqrt{3}}{\sqrt{45}-\sqrt{27}} - \frac{2\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}$

senseir

ad.1.)
$\sqrt{27}+\sqrt{48}+\sqrt{75}=\\=\sqrt{9*3}+\sqrt{16*3}+\sqrt{25*3}=\\=\sqrt{9}\sqrt{3}+\sqrt{16}\sqrt{3}+\sqrt{25}\sqrt{3}=\\=3\sqrt{3}+4\sqrt{3}+5\sqrt{3}=\\=(3+4+5)\sqrt{3}=12\sqrt{3}.$

ad.2.).
$\sqrt{125}+\sqrt{24}-\sqrt{45}-\sqrt{150}=\\\sqrt{25*5}+\sqrt{4*6}-\sqrt{9*5}-\sqrt{25*6}=\\\sqrt{25}\sqrt{5}+\sqrt{4}\sqrt{6}-\sqrt{9}\sqrt{5}-\sqrt{25}\sqrt{6}=\\5\sqrt{5}+2\sqrt{6}-3\sqrt{5}-5\sqrt{6}=\\(5-3)\sqrt{5}+(2-5)\sqrt{6}=2\sqrt{5}-3\sqrt{6}=\\2\sqrt{5}-3\sqrt{2}\sqrt{3}$

ad.3.)
$\frac{\sqrt{108}-\sqrt{180}}{\sqrt{12}-\sqrt{20}}=\frac{\sqrt{36*3}-\sqrt{36*5}}{\sqrt{4*3}-\sqrt{4*5}}=\frac{6\sqrt{3}-6\sqrt{5}}{2\sqrt{3}-6\sqrt{5}}=\frac{6(\sqrt{3}-\sqrt{5})}{2(\sqrt{3}-3\sqrt{5})}=$ $3\frac{\sqrt{3}-\sqrt{5}}{\sqrt{3}-3\sqrt{5}}=3\frac{\sqrt{3}-\sqrt{5}}{\sqrt{3}-3\sqrt{5}}=3\frac{(\sqrt{3}-\sqrt{5})(\sqrt{3}+3\sqrt{5})}{(\sqrt{3}-3\sqrt{5})(\sqrt{3}+3\sqrt{5})}=$ $3\frac{(\sqrt{3}-\sqrt{5})(\sqrt{3}+3\sqrt{5})}{(\sqrt{3}-3\sqrt{5})(\sqrt{3}+3\sqrt{5})}=3\frac{(\sqrt{3}-\sqrt{5})(\sqrt{3}+3\sqrt{5})}{(\sqrt{3}-3\sqrt{5})(\sqrt{3}+3\sqrt{5})}=3\frac{\sqrt{3}\sqrt{3}+3\sqrt{3}\sqrt{5}-\sqrt{3}\sqrt{5}-3\sqrt{5}\sqrt{5}}{\sqrt{3}\sqrt{3}+3\sqrt{3}\sqrt{5}-3\sqrt{3}\sqrt{5}-3\sqrt{5}*3\sqrt{5}}=$ $3\frac{3+2\sqrt{3}\sqrt{5}-3*5}{3-3*3*5}=3\frac{3-15+2\sqrt{3}\sqrt{5}}{(-42)}=3\frac{(-12)+2\sqrt{3}\sqrt{5}}{(-42)}=\frac{12-2\sqrt{3}\sqrt{5}}{14}=$ $\frac{2(6-\sqrt{3}\sqrt{5})}{14}=\frac{6-\sqrt{3}\sqrt{5}}{7}.$

ad.4.)

$\frac{\sqrt{112}-\sqrt{32}}{\sqrt{252}-\sqrt{72}}=\frac{\sqrt{16*7}-\sqrt{16*2}}{\sqrt{7*4*3*3}-\sqrt{9*8}}=\frac{\sqrt{16}\sqrt{7}-\sqrt{16}\sqrt{2}}{\sqrt{7}\sqrt{4}\sqrt{3}\sqrt{3}-\sqrt{9}\sqrt{4}\sqrt{2}}=$ $\frac{4\sqrt{7}-4\sqrt{2}}{6\sqrt{7}-6\sqrt{2}}=\frac{4(\sqrt{7}-\sqrt{2})}{6(\sqrt{7}-\sqrt{2})}=\frac{2}{3}.$

ad.5.)
$\frac{11\sqrt{3}-4\sqrt{7}}{\sqrt{3}-\sqrt{7}}-\frac{13\sqrt{7}+\sqrt{3}}{\sqrt{12}-\sqrt{28}}=\frac{11\sqrt{3}-4\sqrt{7}}{\sqrt{3}-\sqrt{7}}-\frac{13\sqrt{7}+\sqrt{3}}{\sqrt{4*3}-\sqrt{4*7}}=\frac{11\sqrt{3}-4\sqrt{7}}{\sqrt{3}-\sqrt{7}}-\frac{\sqrt{3}+13\sqrt{7}}{2\sqrt{3}-2\sqrt{7}}=$ $\frac{2(11\sqrt{3}-4\sqrt{7})}{2(\sqrt{3}-\sqrt{7})}-\frac{\sqrt{3}+13\sqrt{7}}{2\sqrt{3}-2\sqrt{7}}=\frac{2(11\sqrt{3}-4\sqrt{7})-\sqrt{3}-13\sqrt{7}}{2\sqrt{3}-2\sqrt{7}}=\frac{21\sqrt{3}-21\sqrt{7}}{2\sqrt{3}-2\sqrt{7}}=\frac{21(\sqrt{3}-\sqrt{7})}{2(\sqrt{3}-\sqrt{7})}=\frac{21}{2}=10\frac{1}{2}.$

ad.6.)
$\frac{25\sqrt{5}-23\sqrt{3}}{\sqrt{45}-\sqrt{27}}-\frac{2\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\frac{25\sqrt{5}-23\sqrt{3}}{\sqrt{9*5}-\sqrt{9*3}}-\frac{2\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\frac{25\sqrt{5}-23\sqrt{3}}{\sqrt{9}\sqrt{5}-\sqrt{9}\sqrt{3}}-\frac{2\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}=$ $\frac{25\sqrt{5}-23\sqrt{3}}{3\sqrt{5}-3\sqrt{3}}-\frac{2\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\frac{25\sqrt{5}-23\sqrt{3}}{3(\sqrt{5}-\sqrt{3})}-\frac{3(2\sqrt{5}-\sqrt{3})}{3(\sqrt{5}-\sqrt{3})}=\frac{25\sqrt{5}-23\sqrt{3}-6\sqrt{5}+3\sqrt{3}}{3(\sqrt{5}-\sqrt{3})}=$ $\frac{19\sqrt{5}-20\sqrt{3}}{3(\sqrt{5}-\sqrt{3})}=\frac{(19\sqrt{5}-20\sqrt{3})(\sqrt{5}+\sqrt{3})}{3(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}=\frac{19\sqrt{5}\sqrt{5}+19\sqrt{3}\sqrt{5}-20\sqrt{3}\sqrt{5}-20\sqrt{3}\sqrt{3}}{3(\sqrt{5}\sqrt{5}+\sqrt{3}\sqrt{5}-\sqrt{3}\sqrt{5}-\sqrt{3}\sqrt{3})}=$ $\frac{19\sqrt{5*5}+19\sqrt{3}\sqrt{5}-20\sqrt{3}\sqrt{5}-20\sqrt{3*3}}{3(\sqrt{5*5}-\sqrt{3*3})}=\frac{19*5+19\sqrt{3}\sqrt{5}-20\sqrt{3}\sqrt{5}-20*3}{3(5-3)}=$ $\frac{95-60+(19-20)\sqrt{3}\sqrt{5}}{3(5-3)}=\frac{35-\sqrt{3}\sqrt{5}}{6}=5\frac{5}{6}-\frac{\sqrt{3}\sqrt{5}}{6}.$