Question

Va rog sa ma ajutati!
Aratati ca:(poza)

Answer 1

Răspuns:

Explicație pas cu pas:

8 < 9, deci $\sqrt{8} < 3$

27 < 32, deci $3 < \sqrt[3]{32}$

Deci $2^{\frac{3}{2}} < 3 < 2^{\frac{5}{3}}$

$log_{2} 2^{\frac{3}{2}} < log_{2} 3 < log_{2} 2^{\frac{5}{3}}$

$\frac{3}{2} < log_{2} 3 < \frac{5}{3}$

$\frac{7}{2} < 2 + log_{2} 3 < \frac{11}{3}$

$\frac{5}{2} < 1 + log_{2} 3 < \frac{8}{3}$

Inmultim si obtinem   $\frac{35}{4} <(2+ log_{2} 3)(1+ log_{2} 3) < \frac{88}{9}$

cum $8 <\frac{35}{4}$   si   $\frac{88}{9} < 10$

Rezulta $8 <(2+ log_{2} 3)(1+ log_{2} 3) < 10$