Doprowadź wyrażenie do najprostrzej postaci:a) 6/√8 + 10/√32 + 5/√18b) 6/√27 - 2/√48 + 5/√12c) 7/√10.... Question from @ania97712

ania97712 @ania97712

October 2018 1 31 Report

Doprowadź wyrażenie do najprostrzej postaci:

a) 6/√8 + 10/√32 + 5/√18

b) 6/√27 - 2/√48 + 5/√12

c) 7/√10 + 3/√40 - 6/√90

d) 6/√45 + 15/√20 + 1/√125

e) 12/√6 + 4√3/√2 + 3√2/√3

f) 12/√10 + 6/√2 - 15/√5

g) 4/√3 - 2/√15 + 3/√5

h) 11/√2 - 3/√7 - 2/√14

ag94

$\frac{6}{\sqrt8} + \frac{10}{\sqrt{32}} + \frac{5}{\sqrt{18}}=\\\\ =\frac{6}{\sqrt8} \cdot\frac{\sqrt2}{\sqrt2}+ \frac{10}{\sqrt{32}} \cdot\frac{\sqrt2}{\sqrt2}+ \frac{5}{\sqrt{18}}\cdot\frac{\sqrt2}{\sqrt2}=\\\\ =\frac{6\sqrt2}{\sqrt{16}} + \frac{10\sqrt2}{\sqrt{64}}+ \frac{5\sqrt2}{\sqrt{36}}=\\\\ =\frac{6\sqrt2}{4} + \frac{10\sqrt2}{8}+ \frac{5\sqrt2}{6}=$

$=\frac{6\sqrt2}{4} \cdot\frac{6}{6} + \frac{10\sqrt2}{8}\cdot\frac{3}{3} + \frac{5\sqrt2}{6}\cdot\frac{4}{4} =\\\\ =\frac{36\sqrt2}{24} + \frac{30\sqrt2}{24}+ \frac{20\sqrt2}{24}=\\\\ =\frac{36\sqrt2+30\sqrt2+20\sqrt2}{24}=\\\\ =\frac{86\sqrt2}{24}=\frac{43\sqrt2}{12}$

$\frac{6}{\sqrt{27}} - \frac{2}{\sqrt{48}} + \frac{5}{\sqrt{12}}=\\\\ =\frac{6}{3\sqrt{3}} - \frac{2}{4\sqrt{3}} + \frac{5}{2\sqrt{3}}=\\\\ =\frac{6}{3\sqrt{3}} \cdot\frac{\sqrt3}{\sqrt3}- \frac{2}{4\sqrt{3}}\cdot\frac{\sqrt3}{\sqrt3} + \frac{5}{2\sqrt{3}}=\\\\ =\frac{6\sqrt3}{9}:\frac{3}{3}-\frac{2\sqrt3}{12}:\frac{2}{2}+\frac{5\sqrt3}{6}=\\\\ =\frac{2\sqrt3}{3}\cdot\frac{2}{2}-\frac{\sqrt3}{6}+\frac{5\sqrt3}{6}=$

$=\frac{4\sqrt3}{6}-\frac{\sqrt3}{6}+\frac{5\sqrt3}{6}=\\\\ =\frac{4\sqrt3-\sqrt3+5\sqrt3}{6}=\\\\ =\frac{8\sqrt3}{6}:\frac{2}{2}=\frac{4\sqrt3}{3}$

$\frac{7}{\sqrt{10}} + \frac{3}{\sqrt{40}} - \frac{6}{\sqrt{90}}=\\\\ =\frac{7}{\sqrt{10}}\cdot\frac{\sqrt{10}}{\sqrt{10}} + \frac{3}{\sqrt{40}} \cdot\frac{\sqrt{10}}{\sqrt{10}}- \frac{6}{\sqrt{90}}\cdot\frac{\sqrt{10}}{\sqrt{10}}=\\\\ =\frac{7\sqrt{10}}{10}+\frac{3\sqrt{10}}{20}-\frac{6\sqrt{10}}{30}:\frac{3}{3}=\\\\ =\frac{7\sqrt{10}}{10}\cdot\frac{2}{2}+\frac{3\sqrt{10}}{20}-\frac{2\sqrt{10}}{10}\cdot\frac{2}{2}=$

$=\frac{14\sqrt{10}}{20}+\frac{3\sqrt{10}}{20}-\frac{4\sqrt{10}}{20}=\\\\ =\frac{14\sqrt{10}+3\sqrt{10}-4\sqrt{10}}{20}= \\\\ =\frac{13\sqrt{10}}{20}$

d) $\frac{6}{\sqrt{45}} + \frac{15}{\sqrt{20}} + \frac{1}{\sqrt{125}}=\\\\ =\frac{6}{3\sqrt5}\cdot\frac{\sqrt5}{\sqrt5}+\frac{15}{2\sqrt5}\cdot\frac{\sqrt5}{\sqrt5}+\frac{1}{5\sqrt5}\cdot\frac{\sqrt5}{\sqrt5}=\\\\ =\frac{6\sqrt5}{15}\cdot\frac{10}{10}+\frac{15\sqrt5}{10}\cdot\frac{15}{15}+\frac{\sqrt5}{25}\cdot\frac{6}{6}=\\\\ =\frac{60\sqrt5}{150}+\frac{225\sqrt5}{150}+\frac{6\sqrt5}{150}=\\\\ =\frac{60\sqrt5+225\sqrt5+6\sqrt5}{150}=\\\\ =\frac{291\sqrt5}{150}=1\frac{141\sqrt5}{150}$

e) $\frac{12}{\sqrt6} + \frac{4\sqrt3}{\sqrt2} + \frac{3\sqrt2}{\sqrt3}=\\\\ =\frac{12}{\sqrt6} \cdot\frac{\sqrt6}{\sqrt6}+ \frac{4\sqrt3}{\sqrt2}\cdot\frac{\sqrt{18}}{\sqrt{18}} + \frac{3\sqrt2}{\sqrt3}\cdot\frac{\sqrt{12}}{\sqrt{12}}=\\\\ =\frac{12\sqrt6}{6} + \frac{4\sqrt{54}}{6} + \frac{3\sqrt{24}}{6}=\\\\ =\frac{12\sqrt6}{6} + \frac{4\sqrt{3^2\cdot6}}{6} + \frac{3\sqrt{2^2\cdot6}}{6}=\\\\ =\frac{12\sqrt6}{6} + \frac{12\sqrt{6}}{6} + \frac{6\sqrt{6}}{6}=\\\\ =\frac{30\sqrt6}{6}=5\sqrt6$

f) $\frac{12}{ \sqrt{10}} + \frac{6}{ \sqrt2} - \frac{15}{ \sqrt5} =\\\\ =\frac{12}{ \sqrt{10}} \cdot\frac{\sqrt{10}}{\sqrt{10}}+ \frac{6}{ \sqrt2}\cdot\frac{\sqrt{50}}{\sqrt{50}} - \frac{15}{ \sqrt5}\cdot\frac{\sqrt{20}}{\sqrt{20}} =\\\\ =\frac{12\sqrt10}{10}+\frac{6\sqrt{5^2\cdot2}}{10}-\frac{15\sqrt{2^2\cdot5}}{10}=\\\\ =\frac{12\sqrt{10}+30\sqrt2-30\sqrt5}{10}=\\\\ =\frac{2(6\sqrt{10}+15\sqrt2-15\sqrt5)}{10}=$

$=\frac{6\sqrt{10}+15\sqrt2-15\sqrt5}{5}\ [=\frac{6\sqrt{10}}{5}+3\sqrt2-3\sqrt5]$

g) $\frac{4}{\sqrt3} - \frac{2}{\sqrt{15}} + \frac{3}{\sqrt5}=\\\\ =\frac{4}{\sqrt3} \cdot\frac{\sqrt{75}}{\sqrt75} - \frac{2}{\sqrt{15}} \cdot\frac{\sqrt{15}}{\sqrt15} + \frac{3}{\sqrt5} \cdot\frac{\sqrt{45}}{\sqrt{45}}=\\\\ =\frac{4\sqrt{5^2\cdot3}}{15}-\frac{2\sqrt{15}}{15}+\frac{3\sqrt{3^2\cdot5}}{15}=\\\\ =\frac{20\sqrt3-2\sqrt{15}+9\sqrt5}{15}\ [=\frac{4\sqrt3}{3}-\frac{2\sqrt{15}}{15}+\frac{3\sqrt5}{5}]$

h) $\frac{11}{\sqrt2} - \frac{3}{\sqrt7} - \frac{2}{\sqrt{14}}=\\\\ =\frac{11}{\sqrt2} \cdot\frac{\sqrt{98}}{\sqrt{98}}- \frac{3}{\sqrt7}\cdot\frac{\sqrt{28}}{\sqrt{28}} - \frac{2}{\sqrt{14}}\cdot\frac{\sqrt{14}}{\sqrt{14}}=\\\\ =\frac{11\sqrt{7^2\cdot2}}{14}-\frac{3\sqrt{2^2\cdot7}}{14}-\frac{2\sqrt{14}}{14}=\\\\ =\frac{77\sqrt{2}}{14}-\frac{6\sqrt{7}}{14}-\frac{2\sqrt{14}}{14}\ \ \ [=5\frac{1}{2}\sqrt2-\frac{3}{7}\sqrt7-\frac{1}{7}\sqrt14]$