Question

problema 1 punctele a si b va rog !​

Answer 1

Răspuns:

Explicație pas cu pas:

$a)~\frac{2(\sqrt{2}-\sqrt{3})(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})} =\frac{2((\sqrt{2})^{2}-(\sqrt{3})^{2} )}{(\sqrt{3})^{2}-(\sqrt{2})^{2} } =\frac{2*(2-3)}{3-2}=-2.~este~rational.$

$b)~a=(\frac{1}{\sqrt{2} }+\frac{1}{\sqrt{3} })*\sqrt{6}-\sqrt{3}=(\frac{\sqrt{3} }{\sqrt{6} }+\frac{\sqrt{2} }{\sqrt{6} })*\sqrt{6}-\sqrt{3}=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{6} }*\sqrt{6}-\sqrt{3}= \sqrt{3}+\sqrt{2}-\sqrt{3}=\sqrt{2}.~deci~a= \sqrt{2}.$

Aici am aplicat calculele din subpunctul a).$b=\frac{2\sqrt{2} -2\sqrt{3} }{\sqrt{3}-\sqrt{2}}+2+\sqrt{3}=-2+ 2+\sqrt{3}=\sqrt{3}.~deci~b=\sqrt{3}.$\\

$\frac{b}{a} =\frac{\sqrt{3} }{\sqrt{2} }=\sqrt{\frac{3}{2} } \\Sa~verificam~\frac{6}{5}<\frac{b}{a}.\\ \frac{6}{5}<\sqrt{\frac{3}{2}}~ |^2, ~\frac{36}{25}<\frac{3}{2} ,~36*2<25*3,~72<75,~adevarat,~deci~ \frac{6}{5}<\frac{b}{a}.\\Sa~verificam~\frac{b}{a}<\frac{5}{4}.\\\sqrt{\frac{3}{2}} <\frac{5}{4} ~|^2,~\frac{3}{2}<\frac{25}{16},~3*16<2*25,~48<50,~adevarat,~deci~ \frac{b}{a}<\frac{5}{4}.\\Rezulta~ca~\frac{6}{5}<\frac{b}{a}<\frac{5}{4}.$