Question

aflati nr a pt care avem .[(3+a)×3+3×(3+3:a)]:3-3×3+3:3=2

Answer 1

[ ( 3 + a ) × 3 + 3 × ( 3 + 3 : a ) ] : 3 - 3 × 3 + 3 : 3 = 2

( 9 + 3 a + 9 + 9 / a ) : 3 - 9 + 1 = 2

( 3 a + 9 / a + 18 ) : 3 = 2 - 1 + 9

( 3 a + 9 / a + 18 ) : 3 = 10

3 a + 9 / a + 18 = 10 x 3

3 a + 9 / a + 18 = 30

3 a² + 9 = a × ( 30 - 18 )

3 a² - 12 a + 9 = 0 l : 3

a² - 4 a + 3 = 0

Δ = 16 - 12 = 4

a₁ = ( 4 + √4) / 2 = 6 / 2 = 3

a₂ = ( 4 - 2 ) / 2 = 2 / 2 = 1

Verific

[ ( 3 + 1 ) × 3 + 3 × ( 3 + 3 : 1 ) ] : 3 - 3 × 3 + 3 : 3 =2 ( a = 1)

= ( 12 + 18 ) : 3 - 9 + 1 =

= 10 - 9 + 1 =

= 2

[ ( 3 + 3 ) × 3 + 3 × ( 3 + 3 : 3 ) ] : 3 - 3 × 3 + 3 : 3 = 2 ( a = 3)

= ( 18 + 12 ) : 3 - 9 + 1 =

= 10 - 9 + 1 =

= 2