matematica di seconda media

Question

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gramor · Accepted Answer

[attach]51892[/attach] ============================================================  sqrt { dfrac { big { frac {9}{20}+ big [ frac {9}{5}- frac {3}{5}· big ( frac {6-1}{3} big )^2 big ]· frac {3}{2} big } :  big ( frac {3+10}{5} big )}{ frac {2}{3}· big [ frac {1}{6}+ frac {2}{13}· big ( frac {1+12}{4} big ) big ]}} = =  sqrt { dfrac { big { frac {9}{20}+ big [ frac {9}{5}- frac {3}{5}· big ( frac {5}{3} big )^2 big ]· frac {3}{2} big } :  frac {13}{5}}{ frac {2}{3}· big [ frac {1}{6}+ frac {2}{13}· frac {13}{4} big ]}} = =  sqrt { dfrac { big { frac {9}{20}+ big [ frac {9}{5}- frac {3}{5}· frac {25}{9} big ]· frac {3}{2} big }· frac {5}{13}}{ frac {2}{3}· big [ frac {1}{6}+ frac {1}{1}· frac {1}{2} big ]}} = =  sqrt { dfrac { big { frac {9}{20}+ big [ frac {9}{5}- frac {1}{1}· frac {5}{3} big ]· frac {3}{2} big }· frac {5}{13}}{ frac {2}{3}· big [ frac {1}{6}+ frac {1}{2} big ]}} = =  sqrt { dfrac { big { frac {9}{20}+ big [ frac {9}{5}- frac {5}{3} big ]· frac {3}{2} big }· frac {5}{13}}{ frac {2}{3}· big [ frac {1+3}{6} big ]}} = =  sqrt { dfrac { big { frac {9}{20}+ big [ frac {27-25}{15} big ]· frac {3}{2} big }· frac {5}{13}}{ frac {2}{3}· frac {4}{6}}} = =  sqrt { dfrac { big { frac {9}{20}+ frac {2}{15}· frac {3}{2} big }· frac {5}{13}}{ frac {2}{3}· frac {2}{3}}} = =  sqrt { dfrac { big { frac {9}{20}+ frac {1}{5}· frac {1}{1} big }· frac {5}{13}}{ frac {4}{9}}} = =  sqrt { big { frac {9}{20}+ frac {1}{5} big }· frac {5}{13}· frac {9}{4}} = =  sqrt { big { frac {9+4}{20} big }· frac {45}{52}} = =  sqrt { frac {13}{20}· frac {45}{52}} = =  sqrt { frac {1}{4}· frac {9}{4}} = =  sqrt { frac {9}{16}} =  frac {3}{4}  ====================================

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