Question

determinati numerele rationale a si b pentru care - se afla in poza
MATEMATICA DE CLASA A 7 A
RADICALI
cat de repede se poate va rog!!!
ofer coroana!!!!!!!!!!!!

Answer 1

Răspuns:

Explicație pas cu pas:

$\sqrt{3} < \sqrt{4} = 2$

$| \sqrt{3} -2| = 2 -\sqrt{3}$

$2a(2 + \sqrt{3}) - b(2- \sqrt{3} ) = 14 - \sqrt{3}$

$4a + 2a\sqrt{3} - 2b + b\sqrt{3} = 14 - \sqrt{3}$

$4a - 2b + 2a\sqrt{3} + b\sqrt{3} = 14 - \sqrt{3}$

$2(2a - b) + \sqrt{3}\cdot (2a + b) = 14 - \sqrt{3}$

Cum a si b sunt numere rationale, atunci egalitatea are loc doar daca  

$2(2a - b) = 14$  

si

$(2a + b) = - 1$

deci

$2(2a - b) = 14\\2a + b = -1$

$2a - b = 7\\2a + b = -1$

$4a = 6\\\\a = \frac{3}{2}$

$2\cdot \frac{3}{2} + b = -1\\\\3 + b = -1\\b = -4$

Answer 2

$\it 2a(2+\sqrt3)-b|\underbrace{\sqrt3-2}_{<0}|=14-\sqrt3 \Rightarrow4a+2a\sqrt3+b\sqrt3-2b=14-\sqrt3 \Rightarrow\\ \\ \\ \Rightarrow (4a-2b)+\sqrt3(2a+b)=14-\sqrt3 \Rightarrow \begin{cases}\ \it 4a-2b=14|_{:2} \Rightarrow2a-b=7\ \ \ (1) \\ \\ \it 2a+b=-1 \Rightarrow b=-1-2a\ \ \ \ \ (2)\end{cases}\\ \\ \\ (1) \Rightarrow b=2a-7\ \ \ \ \ (3) \\ \\ \\ (2),\ (3) \Rightarrow 2a-7=-1-2a \Rightarrow 2a+2a=-1+7 \Rightarrow 4a=6|_{:4} \Rightarrow a=1,5\ \ \ (4)\\ \\ (3),\ (4) \Rightarrow b=3-7=-4$