Question

Va rog frumos ex 5 și 6 tottt va roggg​

Answer 1

5.

a) $\it \frac{x}{2\sqrt{3} + 3 } = \frac{2\sqrt{3} - 3}{\sqrt{3} } <=> x\sqrt{3} = (2\sqrt{3} + 3)(2\sqrt{3} - 3) <=> x\sqrt{3} = (2\sqrt{3})^{2} - 3^{2} <=> x\sqrt{3} = 12 - 9 <=> x\sqrt{3} = 3 <=> x = ~^{\sqrt{3}) } \frac{3}{\sqrt{3} } <=> x = \frac{3\sqrt{3} }{3} <=> x = \sqrt{3}$

b) $\it \frac{\sqrt{13 + \sqrt{1296} } }{x} = \frac{3^{3} - 3^{2} - 3 - 3^{0} }{\sqrt{23 + \sqrt{1681} } } <=> \frac{\sqrt{13 + 36} }{x} = \frac{27 - 9 - 3 - 1}{\sqrt{23 + 41} } <=> \frac{\sqrt{49} }{x} = \frac{14}{\sqrt{64} } <=> \\ \frac{7}{x} = \frac{14}{8} <=> x = \frac{7 \cdot 8}{14} <=> x = \frac{56}{14}  <=> x = 4$

6.

a) $\it (~^{\sqrt{2}) } \frac{\sqrt{3} - 1 }{\sqrt{2} } - ~^{\sqrt{3} - 1) } \frac{\sqrt{2} }{\sqrt{3} + 1 })^{15} = (\frac{\sqrt{2}(\sqrt{3} - 1) }{2} - \frac{\sqrt{2(\sqrt{3} - 1)} }{(\sqrt{3)}^{2} - 1^{2}  })^{15}  = (\frac{\sqrt{6} - \sqrt{2}  }{2} - \frac{\sqrt{6} - \sqrt{2}  }{2} )^{15} = (\frac{\sqrt{6} - \sqrt{2} - \sqrt{6} + \sqrt{2}   }{2} )^{15} = 0^{15} = 0$

b) $\it ~^{\sqrt{7} + \sqrt{2}) } \frac{5}{\sqrt{7} - \sqrt{2}  } + ~^{3 - \sqrt{7} )} \frac{2}{3 + \sqrt{7} } = \frac{5(\sqrt{7} + \sqrt{2}) }{(\sqrt{7})^{2} - (\sqrt{2} )^{2}   } + \frac{2(3 - \sqrt{7} )}{3^{2} - (\sqrt{7})^{2}  } = \frac{5(\sqrt{7} + \sqrt{2}) }{5} + \frac{2(3 - \sqrt{7} )}{2} = \sqrt{7} + \sqrt{2} + 3 - \sqrt{7} = \sqrt{2} + 3 = 3 + \sqrt{2}$

c) $\it \sqrt{\sqrt{5} (\sqrt{5} - \sqrt{7}) - \sqrt{7}(\sqrt{5} - \sqrt{7})  } = \sqrt{5 - \sqrt{35} - \sqrt{35} + 7  } = \sqrt{12 - 2\sqrt{35} } = \sqrt{(\sqrt{5} - \sqrt{7})^{2}  } =  |\sqrt{5} - \sqrt{7} | = -(\sqrt{5} - \sqrt{7}) = -\sqrt{5} + \sqrt{7} = \sqrt{7}  - \sqrt{5}$