Question

radical din 37 supra4,(1) + radical din 38 supra4,(2) + radical din 39supra 4,(3) + radical din 40supra4,(4)+….+ radical din 44 supra 4,(8)

Answer 1

Explicație pas cu pas:

$\sqrt{ \dfrac{37}{4.(1)} } + \sqrt{ \dfrac{38}{4.(2)} } + \sqrt{ \dfrac{39}{4.(3)} } + \sqrt{ \dfrac{40}{4.(4)} } + ... + \sqrt{ \dfrac{44}{4.(8)} } =$

$= \sqrt{ \dfrac{37}{ \dfrac{41 - 4}{9} } } + \sqrt{ \dfrac{38}{\dfrac{42 - 4}{9}} } + \sqrt{ \dfrac{39}{\dfrac{43 - 4}{9}} } + \sqrt{ \dfrac{40}{\dfrac{44 - 4}{9}} } + ... + \sqrt{ \dfrac{44}{\dfrac{48 - 4}{9}} }$

$= \sqrt{ \dfrac{37}{ \dfrac{37}{9} } } + \sqrt{ \dfrac{38}{\dfrac{38}{9}} } + \sqrt{ \dfrac{39}{\dfrac{39}{9}} } + \sqrt{ \dfrac{40}{\dfrac{40}{9}} } + ... + \sqrt{ \dfrac{44}{\dfrac{44}{9}} }$

$= \sqrt{9} + \sqrt{9} + \sqrt{9} + \sqrt{9} + ... + \sqrt{9}$

$= \underbrace{3 + 3 + 3 + ... + 3}_{8} = 8 \cdot 3 = 24$