Question

1)NUmarul b=1,25:(4/5)^-2+1,(6)+0,1(6) este:
2)Daca x=√2+√3 si y=√6-1,calculeaza (x²+y²-11)^100.
3)Daca a=√3-√6 si b=√8-3,atunci(2a²-b²)^2013 este:
Va rog ajutati -ma!

Answer 1

1)1,25=1 intreg 25/100= 1 intreg  1/4= 5/4

(4/5)⁻²= 1/(4/5)²= 1/16/25 = 25/16

1,(6)=1 intreg 6/9= 1 intreg 2/3= 5/3

0,1(6)= (16-1)/90= 15/90= 1/6

5/4 :25/16 +5/3 +1/6= 5/4 *16/25 +5/3 +1/6= 4/5 +5/3 +1/6 = (24+50+5)/30= 79/30

2)(x² +y²-11)¹⁰⁰ =[ (√2 +√3)² +(√6-1)² -11]¹⁰⁰= (2+2√6+3 +6 -2√6 +1 -11)¹⁰⁰=

=(2+3+6+1 -11)¹⁰⁰= (12-11)¹⁰⁰ =1¹⁰⁰ =1

3)(2a²-b²)²⁰¹³= [2(√3-√6)² -(√8-3)² ]²⁰¹³ = [2(3-6√2 +6) -( 8 -12√2 +9)]²⁰¹³ =

=[2(9-6√2) -(17 -12√2)]²⁰¹³ =  (18-12√2 -17 +12√2)²⁰¹³= 1²⁰¹³ =1