Prosze kochani o pomoc! USUWANIE NIEWYMIERNOŚCI Z MIANOWNIKA.Prosze o rozwiazanie przykladu c,d,e,f,.... Question from @mirella

basetla

1/(6+V2) * (6-V2)/(6-V2) = (6-V2)/(36-2) = (6-V2)/34

2V2/(V5+4) *(V5-4)/(V5-4) = 2V2(V5-4)/(5-16) = (2V10-8V2)/(-11) = -2(V10-4V2)/11

2/(4-3V2) *(4+3V2)/(4+3V2) = 2(4+3V2)/(16-18) = 2(4+3V2)/(-2) = -(4+3V2)

(2V3-1)/(2V3+2) *(2V3-2)/(2V3-2) = (2V3-1)(2V3-2)/(12-4) = (12-4V3-2V3+2)/8 =

=(14-6V3)/8 = 2/8 *(7-3V3) = (7-3V3)/4

1/(V3-V2) *(V3+V2)/(V3+V2) = (V3+V2)/(3-2) = V3+V2

V3/(V5+V3) * (V5-V3)/(V5-V3) = V3(V5-V3)/(5-3) = (V15-3)/(5-3) = (V15-3)/2

1 votes Thanks 1

kornel94

$c) \frac{1}{6+\sqrt{2}}=\frac{1}{6+\sqrt{2}}\times\frac{6-\sqrt{2}}{6-\sqrt{2}}=\frac{6-\sqrt{2}}{(36-2)}=\frac{3}{17}-\frac{\sqrt{2}}{34}$

d) $\frac{2\sqrt{2}}{\sqrt{5}+4}\times\frac{\sqrt{5}-4}{\sqrt{5}-4}=\frac{2\sqrt{10}-8\sqrt{2}}{5-16}=\frac{2\sqrt{10}-8\sqrt{2}}{-11}$

e) $\frac{2}{4-3\sqrt{2}}\times\frac{4+3\sqrt{2}}{{4+3\sqrt{2}}}=\frac{8+6\sqrt{2}}{16-18}=-4-3\sqrt{2}$

f) $\frac{2\sqrt{3}-1}{2\sqrt{3}+2}\times\frac{2\sqrt{3}-2}{2\sqrt{3}-2}=\frac{12-2\sqrt{3}-4\sqrt{3}+2}{12-4}=\frac{14-6\sqrt{3}}{8}$

g) $\frac{1}{\sqrt{3}-\sqrt{2}}\times\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3}+\sqrt{2}}{1}$

h) $\frac{\sqrt{3}}{\sqrt{5}+\sqrt{2}}\times\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}=\frac{\sqrt{15}-\sqrt{6}}{3}$

0 votes Thanks 0