Question

ma putrti ajuta va rog dau coroana ​

Answer 1

mg (media geometrică) = $\sqrt{xy}$

a) x = |$4\sqrt{3} -7$|($\sqrt{64} +3\sqrt{7}$)

$4\sqrt{3} < 7$

$\sqrt{48} < \sqrt{49}$

⇒ |$4\sqrt{3} -7$| = $7 - 4\sqrt{3}$

⇒ x = ($7 - 4\sqrt{3}$)*($8 + 3\sqrt{7}$)

y = |$3\sqrt{7} -8$|($\sqrt{49} +4\sqrt{3}$)

$3\sqrt{7} <8$

$\sqrt{63} <\sqrt{64}$

⇒ |$3\sqrt{7} -8$| = $8-3\sqrt{7}$

⇒ y = ($8-3\sqrt{7}$)*($7+4\sqrt{3}$)

mg = $\sqrt{(7 - 4\sqrt{3})(8 + 3\sqrt{7})(8-3\sqrt{7})(7+4\sqrt{3})}$

= $\sqrt{(7 - 4\sqrt{3})(7+4\sqrt{3})(8 + 3\sqrt{7})(8-3\sqrt{7})}$

= $\sqrt{1*1} = \sqrt{1} =1$

b) x = |$4\sqrt{5}-9$|($5\sqrt{2} +\sqrt{49}$)

$4\sqrt{5} <9$

$\sqrt{80} <\sqrt{81}$

⇒ |$4\sqrt{5}-9$| = $9-4\sqrt{5}$

⇒ x = ($9-4\sqrt{5}$)*($5\sqrt{2} +7$)

y = |$7-5\sqrt{2}$|($\sqrt{81} +4\sqrt{5}$)

$7<5\sqrt{2}$

$\sqrt{49} <\sqrt{50}$

⇒|$7-5\sqrt{2}$| = $5\sqrt{2} -7$

⇒ y = ($5\sqrt{2} -7$)*($9+4\sqrt{5}$)

mg = $\sqrt{(9-4\sqrt{5})(5\sqrt{2} +7)(5\sqrt{2} -7)(9+4\sqrt{5})}$

= $\sqrt{(9-4\sqrt{5})(9+4\sqrt{5})(5\sqrt{2} +7)(5\sqrt{2} -7)}$

= $\sqrt{1*1}=\sqrt{1} =1$