1.
a)
[tex] \frac{5}{3 \sqrt{2} } = \frac{5 \sqrt{2} }{3 \sqrt{2} \sqrt{2} } = \frac{5 \sqrt{2} }{6} [/tex]
[tex] \frac{7}{2 \sqrt{3} } = \frac{7 \sqrt{3} }{6} [/tex]
Se elimina numitorii:
[tex]5 \sqrt{2} \: \: \: si \: \: \: 7 \sqrt{ 3} [/tex]
Ridici la patrat:
[tex]50 \: \: \: si \: \: \: 147[/tex]
=> Primul este mai mic
b)
[tex] \sqrt{ {5}^{2} - {4}^{2} } \times \sqrt{ {5}^{2} \times {4}^{2} } \\ = \sqrt{(5 - 4)(5 + 4)} \times 5 \times 4 \\ \sqrt{1 \times 9} \times 5 \times 4 \\ 3 \times 5 \times 4 \\ 60[/tex]
c)
[tex] |0.4 - 1| + |0.4 + 2| = \\ | - 0.6| + |2.4| \\ 0.6 + 2.4 \\ 3[/tex]
2.
[tex]a = \sqrt{7} - \sqrt{2} [/tex]
[tex]b = \sqrt{7} + \sqrt{2} [/tex]
a)
[tex] \frac{1}{a} + \frac{1}{b} \\ \frac{1}{ \sqrt{7} - \sqrt{2} } + \frac{1}{ \sqrt{7} + \sqrt{2} } [/tex]
Rationalizezi cu conjugata in ambele:
=>
[tex] \frac{ \sqrt{7} + \sqrt{2} }{5} + \frac{ \sqrt{7} - \sqrt{2} }{5} \\ \frac{ \sqrt{7} + \sqrt{2} + \sqrt{7} - \sqrt{2} }{5} = \frac{2 \sqrt{7} }{5} [/tex]
Acum trebuie sa aratam ca rezultatul de mai sus este mai mare decat 4/5 dar mai mic ca 6/5. Faci exact cum am comparat mai sus. Ridici la patrat etc.
b)
[tex]ma = \frac{a + b}{2} \\ = \frac{ \sqrt{7} - \sqrt{2} + \sqrt{7} + \sqrt{2} }{2} \\ = \frac{2 \sqrt{7} }{2} \\ \sqrt{7} [/tex]
c)
[tex](a - b)^{2} = a^{2} - 2ab + {b}^{2} \\ = ( \sqrt{7} + \sqrt{2} )^{2} - 2(( \sqrt{7} + \sqrt{2} )( \sqrt{7} - \sqrt{2} )) + ( \sqrt{7} - \sqrt{2} )^{2} \\ = 7 + 2 \sqrt{14} + 2 - 2(7 - 2) + 7 - 2 \sqrt{14} + 2 \\ = 7 + 2- 2(7 - 2) + 7 + 2 \\ = 8[/tex]
