Question

Ex 1,2 va rog daka puteti facets si restu

Answer 1

   
[tex]\displaystyle \\ 1) \\ 2401 = 7^4 = \Big( \frac{1}{7} \Big)^{-4}=49^2=\Big( \frac{1}{49} \Big)^{-2} \\ \\ 2) \\ \left( \frac{5x^3y^{-6}}{20x^{-3}y^{-13}} \right)^{-1}= \left(\frac{x^{3-(-3)}y^{-6-(-13)}}{4} \right)^{-1}= \\ \\ =\left(\frac{x^{6}y^{7}}{4} \right)^{-1}=\frac{x^{6\cdot(-1)}y^{7\cdot(-1)}}{4^{-1}}= \frac{x^{-6}y^{-7}}{4^{-1}}= \boxed{\frac{y^{-7}}{4^{-1}x^{6}}} \Longrightarrow~~\text{Adevarat} [/tex]


[tex]\displaystyle \\ 3) \\ 6-5y \geq 0 \\ 6 \geq 5y \\ 5y \leq 6 \\ \\ y \leq \frac{6}{5} \\ \\ y \in \left(-\infty,~~ \frac{6}{5} \right] \\ \\ \\ 4) \\ \sqrt{117^2-108^2}= \sqrt{(9\times 13)^2-(9\times 12)^2}= \\ = \sqrt{9^2\times 13^2-9^2\times 12^2}= \sqrt{9^2(13^2-12^2)}= \\ = 9 \sqrt{13^2-12^2}=9 \sqrt{169-144}=9 \sqrt{25}=9\times 5 = \boxed{45} \\ \\ \\ [/tex]


[tex]\displaystyle \\ 5a) \\ \sqrt{72} - \sqrt{ \frac{12}{25} } + \sqrt{1 \frac{7}{25} } = \sqrt{36\times 2} - \sqrt{ \frac{4\times3}{25} } + \sqrt{\frac{32}{25} } = \\ \\ = \sqrt{36\times 2} - \sqrt{ \frac{4\times3}{25} } + \sqrt{\frac{16\times 2}{25} } = \boxed{6 \sqrt{2} - \frac{2}{5} \sqrt{3}+ \frac{4}{5} \sqrt{2}} [/tex]


[tex]5b)\\\left(\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\right)^2=\\\\ =\left(\sqrt{4+5+4\sqrt{5}}-\sqrt{4+5-4\sqrt{5}}\right)^2=\\\\ =\left(\sqrt{4+4\sqrt{5}+5}-\sqrt{4-4\sqrt{5}+5}\right)^2=\\\\ =\left(\sqrt{2^2+2\times2\times\sqrt{5}+(\sqrt{5})^2}-\sqrt{2^2+ 2\times2\times(-\sqrt{5})+(-\sqrt{5})^2}\right)^2=\\\\ \left(\sqrt{(2+\sqrt{5})^2}-\sqrt{(2-\sqrt{5})^2}\right)^2=\left(2+\sqrt{5}-(2-\sqrt{5})\right)^2=\\\\ ==\left(2+\sqrt{5}-2+\sqrt{5})\right)^2=(2\sqrt{5})^2 = 4\times 5=\boxed{20} [/tex]