Question

Ajutor! Cu cât este egala inmultirea? ​

Answer 1

Varianta I:

$(\frac{3}{2}+\frac{4}{3}+\frac{5}{4})^{2}$×$(\frac{2}{4}-\frac{1}{4}-\frac{1}{6}-\frac{1}{12})$

=$(\frac{3^{2}}{2^{2} } +\frac{4^{2} }{3^{2}}+ \frac{5^{2}}{4^{2} }+2*\frac{3}{2} *\frac{4}{3}+2*\frac{3}{2} *\frac{5}{4}+2*\frac{4}{3}*\frac{5}{4} } )*(\frac{2}{4} -\frac{1}{4}-\frac{1}{6} -\frac{1}{12} )$

$=(\frac{9}{4} +\frac{16}{9} +\frac{5}{16} +4 +\frac{15}{4} +\frac{10}{3} )*(\frac{1}{4} -\frac{1}{6} -\frac{1}{12} )\\\\=(\frac{9}{4} +\frac{16}{9} +\frac{5}{16} +4 +\frac{15}{4} +\frac{10}{3} )*(\frac{3}{12} -\frac{2}{12} -\frac{1}{12} )\\=(\frac{9}{4} +\frac{16}{9} +\frac{5}{16} +4 +\frac{15}{4} +\frac{10}{3} )*0\\=0$

Varianta II

$(\frac{3}{2}+\frac{4}{3}+\frac{5}{4})^{2}*(\frac{2}{4}-\frac{1}{4}-\frac{1}{6}-\frac{1}{12})$

$=(\frac{3}{2}+\frac{4}{3}+\frac{5}{4})*(\frac{3}{2}+\frac{4}{3}+\frac{5}{4})*(\frac{3}{12} -\frac{2}{12} -\frac{1}{12} )=0$

Explicatie :

In Varianta I nu mi-am dat seama ca a doua paranteza ne da 0 si am aplicat formula de calcul prescurtat ,dar nu era necesar,iar in a doua varianta ,am sintetizat mai pe scurt,fara sa mai aplic a doua formula!