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

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

\(=\sqrt{3}+2+4-2\sqrt{2}-3-\sqrt{3}+2\sqrt{2}\)

\(=3\)

Ta có : \(D=\left(\sqrt{57}+3\sqrt{6}+\sqrt{38}+6\right)\left(\sqrt{57}-3\sqrt{6}-\sqrt{38}+6\right)\)

\(=\left(\sqrt{57}+6\right)^2-\left(3\sqrt{6}+\sqrt{38}\right)^2\)

\(=57+12\sqrt{57}+36-\left(54+12\sqrt{57}+38\right)\)

\(=93-92=1\)

Vậy : \(D=1\)

\(D=\left(\sqrt{57}+3\sqrt{6}+\sqrt{38}+6\right)\left(\sqrt{57}-2\sqrt{6}-\sqrt{38}+6\right)\)

\(=\left(\sqrt{57}+6\right)^2-\left(3\sqrt{6}+\sqrt{38}\right)^2\)

\(=\left(93+12\sqrt{57}\right)-\left(92+12\sqrt{57}\right)\)

\(=1\)

A B C D E F

a)

+) Tứ giác AEDF nội tiếp 

=> ^AED = ^DFC (1)

và ^AFD = ^BED ( 2)

+) Ta có: ^EAD = ^FAD ( AD là phân giác ^BAC ) 

^FDC = ^FAD ( cùng chắn cung DF )

^BDE = ^EAD ( cùng chắn cung DE )

=> ^FDC = ^FAD = ^EAD = ^BDE ( 3)

+) Xét \(\Delta\)AED và  \(\Delta\)DFC  có: 

^EAD = ^FDC ( theo (3))

^AED = ^DFC ( theo (1)

=> \(\Delta\)AED ~ \(\Delta\)DFC 

=> \(\frac{AE}{DF}=\frac{ED}{FC}\)=> AE . FC = DF . ED ( 4)

+) Xét \(\Delta\)AFD và \(\Delta\)DEB có:

^DAF = ^BDE ( theo (3))

^AFD = ^DEB ( theo ( 2)

=> \(\Delta\)AFD ~ \(\Delta\)DEB 

=> \(\frac{AF}{ED}=\frac{DF}{BE}\Rightarrow AF.BE=DF.ED\)(5)

Từ (4) ; (5) => AF.BE = AE.FC

=> \(\frac{AF}{FC}=\frac{AE}{BE}\)

=> EF//BC

b) Xét \(\Delta\)AED và \(\Delta\)ADC có:

^EAD = ^DAC 

^ADE = ^ACD ( vì ^ADE = ^AFE ( chắn cung AE ) và ^AFE = ^ACD  (đồng vị ))

=> \(\Delta\)AED ~ \(\Delta\)ADC

=> \(\frac{AE}{AD}=\frac{AD}{AC}\)

=> AD^2 = AE.AC

c) Tương tự cm \(\Delta\)AFD ~ \(\Delta\)ADB 

=> \(\frac{AF}{AD}=\frac{AD}{AB}\)

=> AD^2=AF.AB

kết hợp vs câu b => AB.AF = AE.AC