Question

Se consideră expresia E(x) = (x+3)²-2(3+x)(x-3) + (x - 2)² -26, unde x este un
imar real.
a) Arătaţi că E(x) = 2x + 5, pentru orice număr real x.
b) Determinaţi numărul întreg a, pentru care E(a√2+2)-6=3(a-1)+4√2.

URGENT VA ROGG !!!

Answer 1

$a)~E(x)=(x+3)^2-2(3+x)(x-3)+(x-2)^2-26;~x\in\mathbb{R}\\\\ E(x)=x^2+2 \cdot x \cdot 3+3^2-2\Big(x^2-9\Big)+x^2-2\cdot x \cdot 2+2^2-26\\\\ E(x)=x^2+6x+9-2x^2+18+x^2-4x+4-26\\\\ E(x)=2x+5$

$\displaystyle b)~E\Big(a\sqrt{2} +2\Big)-6=3(a-1)+4\sqrt{2} \\\\ E\Big(a\sqrt{2} +2\Big)=2\Big(a\sqrt{2} +2\Big)+5=2a\sqrt{2} +4+5=2a\sqrt{2} +9\\\\ E\Big(a\sqrt{2} +2\Big)-6=3(a-1)+4\sqrt{2} \Rightarrow 2a\sqrt{2} +9-6=3a-3+4\sqrt{2} \Rightarrow \\\\ \Rightarrow 2a\sqrt{2} -3a=4\sqrt{2} -3-9+6 \Rightarrow a\Big(2\sqrt{2} -3\Big)=4\sqrt{2} -6\Rightarrow$

$\displaystyle \Rightarrow a=\frac{^{2\sqrt{2}+3) }4\sqrt{2}-6 }{2\sqrt{2}-3 } \Rightarrow a=\frac{\Big(2\sqrt{2}+3\Big) \Big(4\sqrt{2} -6\Big)}{\Big(2\sqrt{2}-3\Big)\Big(2\sqrt{2} +3\Big) } \Rightarrow \\\\ \Rightarrow a=\frac{16-12\sqrt{2}+12\sqrt{2}-18 }{8-9} \Rightarrow a=\frac{16-18}{-1} \Rightarrow a=\frac{-2}{-1} \Rightarrow a=2$

Answer 2

Răspuns:

Explicație pas cu pas:

E(x) = (x+3)²-2(3+x)(x-3) + (x-2)² -26 <=>

E(x) = x²+6x+9-2x²+18+x²-4x+4-26

E(x) = x²-2x²+x²+6x-4x+9+18+4-26 =>

E(x) = 2x+31-26 = >

E(x) = 2x+5

-----------------

E(a√2+2)-6=3(a-1)+4√2 <=>

2(a√2+2)+5-6 = 3a-3+4√2 <=>

2a√2+4-1 = 3a-3+4√2 <=>

2a√2+3 = 3a-3 + 4√2 =>

a(2√2-3) = 4√2-6 =>

a = (4√2-6)/(2√2-3) = [(4√2-6)(2√2+3)]/[(2√2+3)(2√2-3)] =>

a = (16+12√2-12√2-18)/(8-9) =>

a = (-2)/(-1) => a = 2