Zadanie 3.211.
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)}-\frac{(\sqrt3-\sqrt2)*(\sqrt3-\sqrt2)}{(\sqrt3+\sqrt2)*(\sqrt3-\sqrt2)}+\frac{1-16\sqrt6}{4}=
\frac{(\sqrt3+2)^2}{(\sqrt3)^2-(\sqrt2)^2}-\frac{(\sqrt3-2)^2}{(\sqrt3)^2-(\sqrt2)^2}+\frac{1-16\sqrt6}{4}=
=\frac{3+2\sqrt6+4}{3-2}-\frac{3-2\sqrt6+4}{3-2}+\frac{1-16\sqrt6}{4}=
=7+2\sqrt6-(7-2\sqrt6)+\frac{1-16\sqrt6}{4}=
=7+2\sqrt6-7+2\sqrt6+\frac{1-16\sqrt6}{4}=
=4\sqrt6+\frac{1-16\sqrt6}{4}=
=\frac{16\sqrt6+1-16\sqrt6}{4}=\frac{1}{4}
b)
\frac{\sqrt7-\sqrt6}{\sqrt7+\sqrt6}-\frac{\sqrt7+\sqrt6}{\sqrt7-\sqrt6}+\frac{8\sqrt{42}+3}{2}=
=\frac{(\sqrt7-\sqrt6)*(\sqrt7-\sqrt6)}{(\sqrt7+\sqrt6)*(\sqrt7-\sqrt6)}-\frac{(\sqrt7+\sqrt6)*(\sqrt7+\sqrt6)}{(\sqrt7-\sqrt6)*(\sqrt7+\sqrt6)}+\frac{8\sqrt{42}+3}{2}=
=\frac{(\sqrt7-\sqrt6)^2}{7-6}-\frac{(\sqrt7+\sqrt6)^2}{7-6}+\frac{8\sqrt{42}+3}{2}=
=7-2\sqrt{42}+6-(7+2\sqrt{42}+6)+\frac{8\sqrt{42}+3}{2}=
=13-2\sqrt{42}-13-2\sqrt{42}+\frac{8\sqrt{42}+3}{2}=
=-4\sqrt{42}+\frac{8\sqrt{42}+3}{2}=
=\frac{-8\sqrt{42}+8\sqrt{42}+3}{2}=\frac{3}{2}=1,5
c)
\frac{2}{\sqrt3-1}-\frac{1}{\sqrt3+1}+\frac{5-2\sqrt3}{4}=
=\frac{2*(\sqrt3+1)}{(\sqrt3-1)*(\sqrt3+1)}-\frac{1*(\sqrt3-1)}{(\sqrt3+1)*(\sqrt3-1)}+\frac{5-2\sqrt3}{4}=
=\frac{2\sqrt3+2}{(\sqrt3)^2-1^2}-\frac{\sqrt3-1}{(\sqrt3)^2-1^2}+\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}{2}-\frac{\sqrt3-1}{2}+\frac{5-2\sqrt3}{4}=
=\frac{2(2\sqrt3+2)-2(\sqrt3-1)+5-2\sqrt3}{4}=
=\frac{4\sqrt3+4-2\sqrt3+2+5-2\sqrt3}{4}=\frac{11}{4}=2\frac{3}{4}
Zastosowane wzory skróconego mnożenia
(a-b)^2=a^2+2ab+b^2
(a+b)^2=a^2+2ab+b^2
a^2-b^2=(a-b)(a+b)