Question

5. Fie a =
6. Fie a =
169
36
-√1,(7)+√30,25.
√1,21-√0,01
√0,04 +√0,09
Calculaţi √a.
√17²-152. Arătaţi că a este număr natural pătrat perfect. in poza ex 5 si ex 6

Answer 1

Explicație pas cu pas:

5.

$\sqrt{ \dfrac{169}{36} } + \sqrt{1.(7)} + \sqrt{30.25} = \sqrt{ \dfrac{ {13}^{2} }{ {6}^{2} } } + \sqrt{ \dfrac{17 - 1}{9}} + \sqrt{ \dfrac{3025}{100} } = \dfrac{13}{6} + \sqrt{ \dfrac{16}{9}} + \sqrt{ \dfrac{ {55}^{2} }{ {10}^{2} } } = \dfrac{^{5)} 13}{6} + \dfrac{^{10)}4}{3} + \dfrac{^{3)}55}{10} = \dfrac{5 \cdot 13 + 10 \cdot 4 + 3 \cdot 55}{30} = \dfrac{65 + 40 + 165}{30} = \dfrac{270^{(30}}{30} = 9$

$\sqrt{a} = \sqrt{9} = \sqrt{ {3}^{2} } = \bf 3$

6.

$\dfrac{ \sqrt{1.21} - \sqrt{0.01} }{ \sqrt{0.04} + \sqrt{0.09} } \cdot \sqrt{ {17}^{2} - {15}^{2} } = \dfrac{ \sqrt{ \dfrac{121}{100} } - \sqrt{\dfrac{1}{100}} }{ \sqrt{\dfrac{4}{100}} + \sqrt{\dfrac{9}{100}} } \cdot \sqrt{289 - 225} = \dfrac{ \sqrt{ \dfrac{ {11}^{2} }{ {10}^{2} } } - \sqrt{\dfrac{1}{ {10}^{2} }} }{ \sqrt{\dfrac{ {2}^{2} }{{10}^{2} }} + \sqrt{\dfrac{ {3}^{2} }{ {10}^{2} }} } \cdot \sqrt{64} = \dfrac{ \dfrac{11}{10} - \dfrac{1}{10} }{\dfrac{2}{10} + \dfrac{3}{10}} \cdot \sqrt{ {8}^{2} } = \dfrac{ \dfrac{10}{10}}{\dfrac{5}{10}} \cdot 8 = \dfrac{1}{\dfrac{1}{2}} \cdot 8 = 2 \cdot 8 = 16 = \bf {4}^{2}$