Question

Exercitiul 17 si 19

Multumesc

Answer 1

17)

[tex]\it 1\ \textless \ 2\ \textless \ 4 \Rightarrow \sqrt1\ \textless \ \sqrt2\ \textless \ \sqrt4 \Rightarrow 1\ \textless \ \sqrt2\ \textless \ 2 \Rightarrow [ \sqrt2 ] =1 \\ \\ 1\ \textless \ 3\ \textless \ 4 \Rightarrow \sqrt1\ \textless \ \sqrt3\ \textless \ \sqrt4 \Rightarrow 1\ \textless \ \sqrt3\ \textless \ 2 \Rightarrow [ \sqrt3 ] =1 [/tex]

[tex]\it (\sqrt2+\sqrt3)^2 = 2+2\sqrt6+3 =5+2\sqrt6 \ \ \ \ (*) \\ \\ 4\ \textless \ 6\ \textless \ 9 \Rightarrow \sqrt4\ \textless \ \sqrt6\ \textless \ \sqrt9 \Rightarrow 2\ \textless \ \sqrt6 \ \textless \ 3|_{\cdot2} \Rightarrow 4\ \textless \ 2\sqrt6\ \textless \ 6|_{+5} \\ \\ \Rightarrow 9\ \textless \ 5+2\sqrt6\ \textless \ 11 \stackrel{(*)}{\Longrightarrow} 9\ \textless \ (\sqrt2+\sqrt3)^2\ \textless \ 11 \Rightarrow [/tex]

[tex]\it \Rightarrow \sqrt9\ \textless \ \sqrt{(\sqrt2+\sqrt3)^2} \ \textless \ \sqrt{11}\ \textless \ \sqrt{16} \Rightarrow 3\ \textless \ \sqrt2+\sqrt3\ \textless \ 4 \Rightarrow \\ \\ \Rightarrow [ \sqrt2+\sqrt3] = 3 \\ \\ a = 1+1+3=5[/tex]

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[tex]\it 45^2= 2025,\ \ \ 44^2= 1936 \\ \\ 1936\ \textless \ 2008\ \textless \ 2025 \Rightarrow \sqrt{1936} \ \textless \ \sqrt{2008}\ \textless \ \sqrt{2025} \Rightarrow \\ \\ \Rightarrow 44\ \textless \ \sqrt{2008} \ \textless \ 45 \Rightarrow [ \sqrt{2008}] = 44 \\ \\ \left\{ -\dfrac{1}{3} \right\} = -\dfrac{1}{3} - \left[ -\dfrac{1}{3} \right] =-\dfrac{1}{3}-(-1) = 1-\dfrac{1}{3} = \dfrac{2}{3}[/tex]

$\it b = 44+3\cdot\dfrac{2}{3} = 44+2=46$

19)

[tex]\it \left[\dfrac{2d}{3} \right] =2 \\ \\ \\ d = 3 \Rightarrow \left[\dfrac{2\cdot3}{3} \right] = [2] =2 \\ \\ \\ d=4 \Rightarrow \left[\dfrac{2\cdot4}{3} \right] = \left[\dfrac{8}{3} \right] = \left[2\dfrac{2}{3} \right] =2[/tex]

[tex]\it d=5 \Rightarrow \left[\dfrac{2\cdot5}{3} \right] = \left[\dfrac{10}{3} \right] = \left[3\dfrac{1}{3} \right] =3\ \ (nu\ \ convine) \\ \\ \\ Deci,\ \ d\in \{3,\ \ 4\}[/tex]