Răspuns :
2,1∈Q ; 2,1011001 ∈ R ; -2,103..∈Q 1,41... ∈R
0,10..∈Q 0,101100,...∈R
x₁ = 6 ∈ Q
pentru ca 12√2 = 2·3·2√2 si 17 = 9 + 8 = 3² + (2√2)²
17 - 12√2 = 3² - 2 · 3 ·2√2 + (2√2) ² = binom = ( 3 - 2√2)²
si 17 + 12√2 = 3² + 2·3·2√2 + (2√2) ² = binom = ( 3 +2√2)²
x ₁ = √(3 +2√2)² + √( 3 - 2√2)²= 3 + 2√2 + 3 -2√2 = 6 ∈ Q
x₂ =
4 -2√3 = 3 +1 - 2√3 = √3² - 2√3 + 1² = binom = ( √3 -1) ²
x₂ = √11 - 4√6+ 2·(√3 -1) = √11 - 4√(6 +2√3 -2) = √11 -4√( 4 +2√3) =
rad. lung rad. lung
alt binom 4 + 2√3 = 3 + 2√3 +1 = √3² +2√3 +1² = ( √3 +1) ²
= √11-4·(√3 +1) = √11-4√3 -4 = √7 -4√3 =
alt binom 7 - 4√3 = 4 +3 -4√3 = 2² - 2·2√3 + √3² = ( 2 -√3)²
x₂ = √(2- √3)² = 2 - √3 ∈ R
pentru y :
4 +2√3 =3 + 2√3 + 1 = √3 + 2·√3·1 + 1² = binom =( √3 +1)²
y = √11 + 4√6-2·(√3+1) = √11 + 4√(6 -2√3-2) =
rad. lund rad. lung
y= √11+4√(4-2√3)
4 -2√3 = 3 -2√3 + 1 = √3² - 2√3 + 1² = binom =(√3 -1)²
y = √11+4·(√3 -1) =√11+4√3-4 =√7 +4√3
alt binom 7 +4√3= 4 + 3 + 4√3 = 2² + 2·2√3 + √3² = ( 2+√3)²
y= √( 2 +√3)² = 2 + √3 ∈ R \Q
x·y = ( 2 - √3) ·(2 + √3) = 2² - √3² = 4 - 3 = 1 ∈ Q
x√2 + y√7 = ( 3√2 - 2√7) / ( 3√2 +2√7) ·( 3√2 -2√7) =
= ( 3√2 - 2√7) / [ (3√2)² - (2√7)² ] =
= ( 3√2 - 2√7) / ( -10 )
= -3√2 /10 + 2√7 /10
daca x√2 = - 3√2 /10 x = - 3 /10
y√7 = 2√7 /10 y = 2/10
0,10..∈Q 0,101100,...∈R
x₁ = 6 ∈ Q
pentru ca 12√2 = 2·3·2√2 si 17 = 9 + 8 = 3² + (2√2)²
17 - 12√2 = 3² - 2 · 3 ·2√2 + (2√2) ² = binom = ( 3 - 2√2)²
si 17 + 12√2 = 3² + 2·3·2√2 + (2√2) ² = binom = ( 3 +2√2)²
x ₁ = √(3 +2√2)² + √( 3 - 2√2)²= 3 + 2√2 + 3 -2√2 = 6 ∈ Q
x₂ =
4 -2√3 = 3 +1 - 2√3 = √3² - 2√3 + 1² = binom = ( √3 -1) ²
x₂ = √11 - 4√6+ 2·(√3 -1) = √11 - 4√(6 +2√3 -2) = √11 -4√( 4 +2√3) =
rad. lung rad. lung
alt binom 4 + 2√3 = 3 + 2√3 +1 = √3² +2√3 +1² = ( √3 +1) ²
= √11-4·(√3 +1) = √11-4√3 -4 = √7 -4√3 =
alt binom 7 - 4√3 = 4 +3 -4√3 = 2² - 2·2√3 + √3² = ( 2 -√3)²
x₂ = √(2- √3)² = 2 - √3 ∈ R
pentru y :
4 +2√3 =3 + 2√3 + 1 = √3 + 2·√3·1 + 1² = binom =( √3 +1)²
y = √11 + 4√6-2·(√3+1) = √11 + 4√(6 -2√3-2) =
rad. lund rad. lung
y= √11+4√(4-2√3)
4 -2√3 = 3 -2√3 + 1 = √3² - 2√3 + 1² = binom =(√3 -1)²
y = √11+4·(√3 -1) =√11+4√3-4 =√7 +4√3
alt binom 7 +4√3= 4 + 3 + 4√3 = 2² + 2·2√3 + √3² = ( 2+√3)²
y= √( 2 +√3)² = 2 + √3 ∈ R \Q
x·y = ( 2 - √3) ·(2 + √3) = 2² - √3² = 4 - 3 = 1 ∈ Q
x√2 + y√7 = ( 3√2 - 2√7) / ( 3√2 +2√7) ·( 3√2 -2√7) =
= ( 3√2 - 2√7) / [ (3√2)² - (2√7)² ] =
= ( 3√2 - 2√7) / ( -10 )
= -3√2 /10 + 2√7 /10
daca x√2 = - 3√2 /10 x = - 3 /10
y√7 = 2√7 /10 y = 2/10