Помогите решить даю 32

1) (3+2√2) (1-2√2) ^2

2) (√3-1) ^2 (4+2√3)

3) (7+4√3) (2-√3) ^2

4) (√2+3) ^2 (11+6√2)

1) (3+2√2) (1-2√2) ²=

(3+2√2) (1-4√2+4*2) =

(3+2√2) (9-4√2) =

27-12√2+18√2-8 (√2) ²=

27+6√2-8*2=

27+6√2-16=

11+6√2

2) (√3-1) ² (4+2√3) =

(3-2√3+1) (4+2√3) =

(4-2√3) (4+2√3) =

16 - (2√3) ²=16 - 4*3=

16-12=4

3) (7+4√3) (2-√3) ²=

(7+4√3) (4-4√3+3) =

(7+4√3) (7-4√3) =

49 - (4√3) ² = 49-16*3=

49-48=1

4) (√2+3) ² (11+6√2) =

(2+6√2+9) (11+6√2) =

(11+6√2) (11+6√2) =

(11+6√2) ²=

121+132√2+36*2=

121+132√2+72=

193+132√2