Australian Open 2025 • Grand Slam

Q1 Q2 Q3 Q4 Q5 Q6

D1 D2 D3 D4 D5 D6 D7 D8 D9 D10 D11 D12 D13 D14 D15

রবিবার 19 জানুয়ারী

R4

Sabalenka A [1]

d.

Andreeva M [14]

6/1, 6/2

R4

Gauff C [3]

d.

Bencic B [PR]

5/7, 6/2, 6/1

R4

Alcaraz C [3]

d.

Draper J [15]

7/5, 6/1

R4

Djokovic N [7]

d.

Lehecka J [24]

6/3, 6/4, 7/6⁴

R4

Badosa P [11]

d.

Danilovic O

6/1, 7/6²

R4

Paul T [12]

d.

Davidovich Fokina A

6/1, 6/1, 6/1

R4

Pavlyuchenkova A [27]

d.

Vekic D [18]

7/6⁷, 6/0

R4

Zverev A [2]

d.

Humbert U [14]

6/1, 2/6, 6/3, 6/2

R1

Kokkinis T [WC]

d.

Dencheva R [10]

6/3, 6/2

R1

Jeong Y

d.

Strydom J [WC]

6/4, 6/1

R1

Thomas F [11]

d.

Host T

6/4, 6/3

R1

Burcescu M

d.

Subasic A

7/6⁵, 6/2

R1

Mosejczuk D

d.

Bilozertsev N

6/4, 2/6, 7/6⁸

R1

Stusek J [16]

d.

Russell T [WC]

6/2, 6/2

R1

Kovackova J [11]

d.

Mildren S [WC]

3/6, 7/5, 6/2

R1

Basile P [Q]

d.

Dembo J [WC]

6/7⁴, 7/6⁵, 6/1

R1

Sawashiro K

d.

Schuman A

6/3, 6/2

R1

Krejcova T [7]

d.

Vujovic L

6/3, 6/2

R1

Black B [Q]

d.

Vladson L

4/6, 6/3, 6/1

R1

Pohankova M

d.

Penickova A

2/6, 7/6³, 6/2

R1

Derepasko T [9]

d.

Haita S

6/2, 6/3

R1

Goto R

d.

Wei Z [Q]

6/3, 5/7, 6/4

R1

Ibraimi Y [WC]

d.

Albuquerque P

7/5, 6/3

R1

Willwerth B

d.

Wazny A [15]

6/3, 6/3

R1

Chen K

d.

Bonding O [6]

6/7⁸, 6/3, 4/1, ab

R1

Djosic N

d.

Tyagi K

6/1, 6/2

R1

Omarkhanov A [2]

d.

Farzam M

7/6³, 7/6⁸

R1

Crivellaro G [Q]

d.

Itsusaki S

6/1, 7/6³

R1

Vandromme J [3]

d.

Nishikawa K [WC]

6/4, 3/6, 6/2

R1

Mpetshi Perricard D

d.

Lam S

7/5, 6/2

R1

Sonobe W [4]

d.

Zhang R [Q]

6/3, 4/6, 7/6⁸

R1

Markovina D

d.

Kawanishi H [Q]

4/6, 6/1, 6/4

R1

Klugman H [14]

d.

Taraba Wallberg N

6/1, 6/1

R1

Leach J [5]

d.

Iutkin I

6/4, 6/3

R1

Alexandrescou Y

d.

McDonald N

6/3, 6/2

R1

Zhenikhova S

d.

An C

6/3, 6/4

R1

Rejchtman Vinciguerra W [14]

d.

Behrmann T

6/4, 3/6, 7/6⁷

R1

Santamarta Roig A [3]

d.

Koyama H [Q]

6/7⁵, 6/3, 6/0

R1

Dilek D

d.

Chen E [WC]

6/3, 6/4

R1

Yaneva E [8]

d.

Phanthala M [Q]

6/1, 6/3