Ag iomadachadh bhloighean
Nuair a bhios tu ag iomadachadh bhloighean, iomadaich na h-脿ireamhan air a鈥 bh脿rr (脿ireamhaichean) c貌mhla, agus iomadaich na h-脿ireamhan air a鈥 bhonn (se貌rsaichean) c貌mhla. An uair sin s矛mplich no tha e uaireannan nas fhasa s矛mpleachadh an toiseach.
Eisimpleir
Obraich a-mach: \(\frac{4}{5} \times \frac{5}{6}\)
D貌igh 1
Iomadaich agus an uair sin dubh 脿s:
\(= \frac{{4 \times 5}}{{5 \times 6}}\)
\(= \frac{{20}}{{30}} = \frac{2}{3}\)
D貌igh 2
S矛mplich an toiseach, an uair sin iomadaich:
\(= \frac{{4 \times 5}}{{5 \times 6}}\)
(Dubh 脿s an d脿 5 le 5, dubh 脿s 4 agus 6 le 2)
\(= \frac{{2 \times 1}}{{1 \times 3}} = \frac{2}{3}\)
Feuch a-nis na ceistean gu-h-矛osal.
Question
Obraich a-mach: \(\frac{5}{6} \times \frac{3}{8}\)
\(= \frac{{5 \times 3}}{{6 \times 8}}\)
\(= \frac{{5 \times 1}}{{2 \times 8}}\)
\(= \frac{5}{{16}}\)
Question
Obraich a-mach: \(1\frac{1}{2} \times 2\frac{1}{6}\)
\(= \frac{3}{2} \times \frac{{13}}{6}\)
Mus iomadaich thu 脿ireamhan measgte, atharraich iad gu bloighean anabharr (am b脿rr nas motha) an toiseach:
\(= \frac{{1 \times 13}}{{2 \times 2}}\)
S矛mplich am bloigh le bhith a鈥 dubhadh 脿s gu trastach. An seo th猫id agad air 3 agus 6 a roinn le 3.
\(= \frac{{13}}{4}\)
Faodaidh tu a-nis do fhreagairt atharrachadh air ais bho bhloigh anabharr gu 脿ireamh mheasgte.
\(= 3\frac{1}{4}\)