Question

exercițiul 4, va rog :) ​

Answer 1

Răspuns:

Am atașat rezolvarea în poze! Sper că este bine :))

Answer 2

a)

$\it a=100+\dfrac{^{6)}1}{\ 2}+100+\dfrac{^{3)}1}{\ 4}+100+\dfrac{^{2)}1}{\ 6}+100+\dfrac{1}{12}=400+\dfrac{6+3+2+1}{12}=\\ \\ \\ =400+\dfrac{12}{12}=400+1=401$

b)

$\it b= \dfrac{(\sqrt2+\sqrt3)+4(\sqrt2+\sqrt3)}{(\sqrt2-\sqrt3)+4(\sqrt2-\sqrt3)}:\dfrac{\sqrt2+\sqrt3}{\sqrt2-\sqrt3}=\dfrac{5(\sqrt2+\sqrt3)}{5(\sqrt2-\sqrt3)}:\dfrac{\sqrt2+\sqrt3}{\sqrt2-\sqrt3}=1$

$\it m_a=\dfrac{401+1}{2}=\dfrac{402}{2}=201$