Question

Calculeaza aplicand metoda cea mai rationala

Answer 1

a)$\frac{7,01^{2}-9,99^{2}}{34} =\frac{(7,01-9,99)(7,01+9,99)}{34}=\frac{-2,98*17}{34}=\frac{-2,98}{2}=-1,49$

b)$\sqrt{7-4\sqrt{3}} \sqrt{7+4\sqrt{3}}=\sqrt{(7-4\sqrt{3})(7+4\sqrt{3})}=\sqrt{7^{2}-(4\sqrt{x})^{2}} =\sqrt{49-48} =\sqrt{1}=1$

c)$\frac{22}{3\sqrt{3}-4}+3-6\sqrt{3}$ rationalizam numitorul ⇒

$\frac{(3\sqrt{3}+4)22}{(3\sqrt{3}+4)(3\sqrt{3}-4)}+3-6\sqrt{3}=\frac{22(3\sqrt{3}+4)}{27-4}+3-6\sqrt{3}=\frac{22(3\sqrt{3}+4)}{23}+3-6\sqrt{3}$ aducem la acelasi numitor⇒$\frac{22(3\sqrt{3}+4)+23*3-23*6\sqrt{3}}{23}=\frac{66\sqrt{3}+88+69-138\sqrt{3}}{23}=\frac{-72\sqrt{3}+157}{23}$