Question

imd plssssssssssss ... .

Answer 1

Răspuns:

Explicație pas cu pas:

a = 4/(√20 + √12) + √3 - 1 = 4/(2√5 + 2√3) + √3 - 1

= 2/(√5 + √3) + √3 - 1 = 2(√5 - √3)/(√5 + √3)(√5 - √3) + √3 - 1

=2(√5 - √3)/(5 - 3) + √3 - 1 = √5 - √3 + √3 - 1 = √5 - 1

a = √5 - 1

____________

b = 6/(2√8 - √20) - √8 + 1 = 6/(4√2 - 2√5) - 2√2 + 1

= 3/(2√2 - √5) - 2√2 + 1 = 3(2√2 + √5)/(2√2 - √5)(2√2 + √5) - 2√2 + 1

= 3(2√2 + √5)/(8 - 5) - 2√2 + 1 = 2√2 + √5 - 2√2 + 1 = √5 + 1

b = √5 + 1

_____________

Ma = (a + b) : 2 = (√5 - 1 + √5 + 1) : 2 = 2√5 : 2 = √5

____________

a*b = (√5 - 1)(√5 + 1) = 5 - 1 = 4

Mg = √4 = 2

Answer 2

Răspuns:

a)

a= [4/(rad20+rad12)]+rad3-1

a= [4/(2rad5+2rad3)]+rad3-1

a= [4/2(rad5+rad3)]+rad3-1

Simplificam 4 cu 2

a= [2/(rad5+rad3)]+rad3-1

Rationalizam cu (rad5-rad3):

a=[2(rad5-rad3)/(5-3)]+rad3-1

a= 2(rad5-rad3)/2+rad3 -1

Simplificam 2 cu 2:

a= rad5-rad3+rad3 -1

a= rad5 -1

b=[6/(2rad8-rad20)]-rad8+1

b= [6/(2×2rad2- 2rad5)]- 2rad2+1

b= [6/(4rad2- 2rad5)]-2rad2+1

b= [6/2(2rad2- rad5)] -2rad2+1

b= 3/(2rad2-rad5)-2rad2+1

Rationalizam:

b= [3(2rad2+rad5) /(8-5)]-2rad2+1

b=3(2rad2+rad5)/3- 2rad2+1

b= 2rad2+rad5- 2rad2+1

b= rad5+1

b)

Ma=(a+b)/2

Ma= (rad5+1+ rad5+1)/2

Ma= (2rad5+2)/2

Ma= 2(rad5+1)/2

Ma= rad5+1

Mg=rad(a×b)

Mg= rad [(rad5+1)×(rad5+1)]

Mg= rad (rad5+1)^2

Mg= rad5+1

Explicație pas cu pas: