zadania:http://img707.imageshack.us/img707/9207/1001470e.jpghttp://img190.imageshack.us/img190/5762/.... Question from @Inculta

Inculta @Inculta

September 2018 1 24 Report

zadania:

http://img707.imageshack.us/img707/9207/1001470e.jpg

http://img190.imageshack.us/img190/5762/1001471q.jpg

http://img685.imageshack.us/img685/4138/1001472b.jpg

z ostatniego te zaznaczone ;)

trzeba wykorzystywać w niektórych wzory skróconego mnożenia.

D4m13n

i) $\frac{12}{\sqrt7-\sqrt3}=\frac{12(\sqrt7+\sqrt3)}{(\sqrt7-\sqrt3)(\sqrt7+\sqrt3)}=\frac{12\sqrt7+12\sqrt3}{4}=\frac{12\sqrt7}{4}+\frac{12\sqrt3}{4}=3\sqrt7+3\sqrt3$

j) $\frac{3\sqrt2-\sqrt7}{\sqrt7-\sqrt2}=\frac{(3\sqrt2-\sqrt7)(\sqrt7+\sqrt2)}{(\sqrt7-\sqrt2)(\sqrt7+\sqrt2)}=\frac{3\sqrt14+6-7-\sqrt14}{7-2}=\frac{2\sqrt14-1}{5}$

a) $\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}-\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}+\frac{1-16\sqrt6}{4}=$

$\frac{(\sqrt3+\sqrt2)(\sqrt3+\sqrt2)-(\sqrt3-\sqrt2)(\sqrt3-\sqrt2)}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}+\frac{1-16\sqrt6}{4}=\frac{(3+\sqrt6+\sqrt6+2)-(3-\sqrt6-\sqrt6+2)}{3-2}+\frac{1-16\sqrt6}{4}=$

$\frac{3+\sqrt6+\sqrt6+2-3+\sqrt6+\sqrt6-2}{1}+\frac{1-16\sqrt6}{4}=\frac{4\sqrt6}{1}+\frac{1-16\sqrt6}{4}=\frac{16\sqrt6}{4}+\frac{1-16\sqrt6}{4}=\frac{16\sqrt6+1-16\sqrt6}{4}=\frac{1}{4}$

c) $\frac{2}{\sqrt3-1}-\frac{1}{\sqrt3+1}+\frac{5-2\sqrt3}{4}=\frac{2(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}-\frac{\sqrt3-1}{(\sqrt3+1)(\sqrt3-1)}+\frac{5-2\sqrt3}{4}=$

$\frac{2\sqrt3+2}{3-1}-\frac{\sqrt3-1}{3-1}+\frac{5-2\sqrt3}{4}=\frac{2\sqrt3+2-\sqrt3+1}{2}+\frac{5-2\sqrt3}{4}=\frac{\sqrt3+3}{2}+\frac{5-2\sqrt3}{4}=$