Question

Daca a²+b²-2a√2-2b√3+5=0, atunci (2/a+3/b)x(b-a) este:

Answer 1

$a ^{2} - 2a \sqrt{2} + 2 + b^{2} - 2b \sqrt{3} + 3 = 0$

$(a - \sqrt{2} )^{2} + (b - \sqrt{3} )^{2} = 0$

=>

$a - \sqrt{2} = 0$

$b - \sqrt{3} = 0$

=>

$a =radical din 2 [tex]b = \sqrt{3}$

=>

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

rationalizam cu radical din 2 si radical din 3 si ne rezultă:

$( \sqrt{2 } + \sqrt{3} ) \times ( \sqrt{3} - \sqrt{2} ) = ( \sqrt{3} + \sqrt{2} ) \times ( \sqrt{3} - \sqrt{2} )$

folosim formula:

$(a + b)(a - b) = a^{2} - b^{2}$

si rezultă

$( \sqrt{3} + \sqrt{2} )( \sqrt{3} - \sqrt{2} ) = ( \sqrt{3} )^{2} - ( \sqrt{2} )^{2} = 3 - 2 = 1$

Answer 2

$\it a^2+b^2-2a\sqrt2-2b\sqrt3+5=0 \Rightarrow (a^2-2a\sqrt2 + 2) + (b^2-2b\sqrt3+3) = 0 \Rightarrow \\ \\ \\ \Rightarrow (a^2-2a\sqrt2 + \sqrt2^2) + (b^2-2b\sqrt3+\sqrt3^2)=0\Rightarrow (a-\sqrt2)^2+(b-\sqrt3)^2=0\Rightarrow\\ \\ \Rightarrow \begin{cases}\it a-\sqrt2=0 \Rightarrow a=\sqrt2\\ \\ \it b-\sqrt3=0\Rightarrow b=\sqrt3\end{cases}$

$\it \Big(\dfrac{2}{a}+\dfrac{3}{b}\Big)(b-a)=(\dfrac{2}{\sqrt2}+\dfrac{3}{\sqrt3})(\sqrt3-\sqrt2)=\Big(\dfrac{\sqrt2\cdot\sqrt2}{\sqrt2}+\dfrac{\sqrt3\cdot\sqrt3}{\sqrt3}\Big)(\sqrt3-\sqrt2)=\\ \\ =(\sqrt2+\sqrt3)(\sqrt3-\sqrt2)=(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)=(\sqrt3)^2-(\sqrt2)^2=3-2=1$