Question

Giải giúp bài 5 6 7 8  của bài 15 i

Answer 1

`5)(6-sqrt6)/(1-sqrt6)+(6-sqrt6)/sqrt6=(sqrt6(sqrt6-1))/(1-sqrt6)+(sqrt6(sqrt6-1))/sqrt6=-sqrt6+sqrt6-1=-1` $\\$ `6)1/(sqrt2-sqrt3)-1/(sqrt3+sqrt2)=(sqrt2+sqrt3)/(2-3)-(sqrt3-sqrt2)/(3-2)=-(sqrt2+sqrt3)-sqrt3+sqrt2=-2sqrt3` $\\$ `7)1/(sqrt5+sqrt3)-1/(sqrt5-sqrt3)=(sqrt5-sqrt3)/(5-3)-(sqrt5+sqrt3)/(5-3)=(sqrt5-sqrt3-sqrt5-sqrt3)/2=-sqrt3` $\\$ `8)6/(1-sqrt3)-(3sqrt3-3)/(sqrt3+1)=(6(1+sqrt3))/(1-3)-(3(sqrt3-1)^2)/(3-1)=(-6(sqrt3+1)-3(4-2sqrt3))/2=-9`

Answer 2

thank

Answer 3

16) Ta có: \(\dfrac{\sqrt{2}-\sqrt{3}}{2-\sqrt{6}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}+2}\)

\(=\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}\left(\sqrt{2}-\sqrt{3}\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}+2}\)

\(=\dfrac{1}{\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)}\)

\(=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)}\)

\(=\dfrac{2\sqrt{3}}{\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)}\)

\(=\dfrac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)

\(=\sqrt{18}-\sqrt{12}\)

\(=3\sqrt{2}-2\sqrt{3}\)