Question

ma puteti ajuta? dau coroana

Answer 1

Răspuns:

$\sqrt{\frac{1}{16} } :(\sqrt{\frac{11}{9} } +\sqrt{\frac{49}{36}} - \sqrt{\frac{9}{144}}= \frac{1}{4} :(\frac{\sqrt{11}}{3} +\frac{7}{6} -\frac{3}{12})= \frac{1}{4} :(\frac{4)\sqrt{11}}{3} +\frac{2)7}{6} -\frac{3}{12})=$

$=\frac{1}{4}:\frac{4\sqrt{11}+14-3}{12} =\frac{1}{4} * \frac{12}{4\sqrt{11} +11}$

se reduc 4 cu 12 (ramane 3) si se baga 4 sub radical si se rationalizeaza numitorul

$=\frac{\sqrt{176} -11)3}{\sqrt{176} +11} =\frac{3(\sqrt{176} -11)}{176 -121}=\frac{3(4\sqrt{11} -11)}{55} =\frac{3\sqrt{11}(4-\sqrt{11}) }{55}$

Explicație pas cu pas:

Cred ca ai gresit cand ai copiat exercitiul. Dupa cum vezi, rezultatul este putin ciudat.

Daca in loc de $\frac{11}{9}$ ar fi $\frac{1}{9}$, problema ar arata mult mai frumos:

$\sqrt{\frac{1}{16} } :(\sqrt{\frac{1}{9} } +\sqrt{\frac{49}{36}} - \sqrt{\frac{9}{144}}= \frac{1}{4} :(\frac{1}{3} +\frac{7}{6} -\frac{3}{12})= \frac{1}{4} :(\frac{4)1}{3} +\frac{2)7}{6} -\frac{3}{12})=\frac{1}{4} :\frac{4+14-3}{12}=\frac{1}{4} :\frac{15}{12} =\frac{1}{4} *\frac{4}{5} =\frac{1}{5}$