Question

Pe mulțimea numerelor reale se definește legea de compozitie asociativă $x * y=2 x y-3 x-3 y+6$.

$5 p$ a) Arătați că $x * y=2\left(x-\frac{3}{2}\right)\left(y-\frac{3}{2}\right)+\frac{3}{2}$, pentru orice numere reale $x$ şi $y$.

$5 p$b) Determinati numerele reale $x$ pentru care $x * x=14$

$5 p$ c) Determinaţi numărul natural $n$, ştiind că $\left(2^{n}+\frac{3}{2}\right) *\left(2^{n+1}+\frac{3}{2}\right) *\left(2^{n+2}+\frac{3}{2}\right)=2^{20}+\frac{3}{2} .$

Answer 1

Explicație pas cu pas:

legea de compoziție asociativă:

$x * y = 2xy - 3x - 3y + 6$

a)

$2(x - \frac{3}{2})(y - \frac{3}{2}) + \frac{3}{2} = 2(xy - \frac{3x}{2} - \frac{3y}{2} + \frac{9}{4})+ \frac{3}{2} = 2xy - 3x - 3y + \frac{9}{2} + \frac{3}{2} = 2xy - 3x - 3y + 6$

$=>x * y = 2(x - \frac{3}{2})(y - \frac{3}{2}) + \frac{3}{2}$

sau:

$x * y = 2xy - 3x - 3y + 6 = \frac{4xy}{2} - \frac{6x}{2} - \frac{6y}{2} + \frac{9}{2} + \frac{3}{2} = \frac{1}{2}(4xy - 6x - 6y + 9) + \frac{3}{2} = \frac{1}{2}(2x(2y - 3) - 3(2y - 3)) + \frac{3}{2} = \frac{1}{2}(2x - 3)(2y - 3) + \frac{3}{2} = 2(\frac{2x - 3}{2} )(\frac{2y - 3}{2} ) + \frac{3}{2} = 2(x - \frac{3}{2})(y - \frac{3}{2}) + \frac{3}{2}$

$=>x * y = 2(x - \frac{3}{2})(y - \frac{3}{2}) + \frac{3}{2}$

b) x = ?, x ∈ R

$x * x = 2x^{2} - 3x - 3x + 6 = 2{x}^{2} - 6x + 6$

$2 {x}^{2} - 6x + 6 = 14 \\ 2 {x}^{2} - 6x - 8 = 0 \\ {x}^{2} - 3x - 4 = 0 \\ (x + 1)(x - 4) = 0 \\ = > \\ x = - 1 \\ x = 4$

c) n = ?, n ∈ N, a.î.:

$( {2}^{n} + \frac{3}{2} )*( {2}^{n + 1} + \frac{3}{2} )*( {2}^{n + 2} + \frac{3}{2} ) = {2}^{20} + \frac{3}{2}$

=>

$( {2}^{n} + \frac{3}{2} )*( {2}^{n + 1} + \frac{3}{2} ) = 2({2}^{n} + \frac{3}{2} - \frac{3}{2})({2}^{n + 1} + \frac{3}{2} - \frac{3}{2}) + \frac{3}{2} = 2 \times {2}^{n} \times {2}^{n + 1} + \frac{3}{2} = {2}^{2n + 2} + \frac{3}{2}$

legea * este asociativă:

=>

$( {2}^{n} + \frac{3}{2} )*( {2}^{n + 1} + \frac{3}{2} )*( {2}^{n + 2} + \frac{3}{2} )=(( {2}^{n} + \frac{3}{2} )*( {2}^{n + 1} + \frac{3}{2} ))*( {2}^{n + 2} + \frac{3}{2} )= ( {2}^{2n + 2} + \frac{3}{2} )*( {2}^{n + 2} + \frac{3}{2} )$

=>

$( {2}^{2n + 2} + \frac{3}{2} )*( {2}^{n + 2} + \frac{3}{2} ) = 2({2}^{2n + 2} + \frac{3}{2} - \frac{3}{2})({2}^{n + 2} + \frac{3}{2} - \frac{3}{2}) + \frac{3}{2} = 2 \times {2}^{2n + 2} \times {2}^{n + 2} + \frac{3}{2} = {2}^{3n + 5} + \frac{3}{2}$

${2}^{3n + 5} + \frac{3}{2} = {2}^{20} + \frac{3}{2} \\ {2}^{3n + 5} = {2}^{20} \\ 3n + 5 = 20 \\ 3n = 15 = > n = 5$