Scipeáil chuig an bpríomhábhar
Luacháil
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Fairsingigh
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Fadhbanna den chineál céanna ó Chuardach Gréasáin

Roinn

\frac{\frac{\left(x+4\right)\left(x+4\right)}{\left(x+3\right)\left(x+4\right)}-\frac{\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x+4\right)}}{\frac{14}{x^{2}+7x+12}}
Chun cothromóidí a shuimiú nó a dhealú, fairsingigh iad chun a n-ainmneoirí a mheaitseáil. Is é an t-iolrach is lú coitianta de x+3 agus x+4 ná \left(x+3\right)\left(x+4\right). Méadaigh \frac{x+4}{x+3} faoi \frac{x+4}{x+4}. Méadaigh \frac{x-3}{x+4} faoi \frac{x+3}{x+3}.
\frac{\frac{\left(x+4\right)\left(x+4\right)-\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x+4\right)}}{\frac{14}{x^{2}+7x+12}}
Tá an t-ainmneoir céanna ag \frac{\left(x+4\right)\left(x+4\right)}{\left(x+3\right)\left(x+4\right)} agus \frac{\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x+4\right)} agus, mar sin, is féidir iad a dhealú trína n-uimhreoirí a dhealú.
\frac{\frac{x^{2}+4x+4x+16-x^{2}-3x+3x+9}{\left(x+3\right)\left(x+4\right)}}{\frac{14}{x^{2}+7x+12}}
Déan iolrúcháin in \left(x+4\right)\left(x+4\right)-\left(x-3\right)\left(x+3\right).
\frac{\frac{8x+25}{\left(x+3\right)\left(x+4\right)}}{\frac{14}{x^{2}+7x+12}}
Cumaisc téarmaí comhchosúla in: x^{2}+4x+4x+16-x^{2}-3x+3x+9.
\frac{\left(8x+25\right)\left(x^{2}+7x+12\right)}{\left(x+3\right)\left(x+4\right)\times 14}
Roinn \frac{8x+25}{\left(x+3\right)\left(x+4\right)} faoi \frac{14}{x^{2}+7x+12} trí \frac{8x+25}{\left(x+3\right)\left(x+4\right)} a mhéadú faoi dheilín \frac{14}{x^{2}+7x+12}.
\frac{\left(x+3\right)\left(x+4\right)\left(8x+25\right)}{14\left(x+3\right)\left(x+4\right)}
Fachtóirigh na sloinn nach bhfuil fachtóirithe cheana.
\frac{8x+25}{14}
Cealaigh \left(x+3\right)\left(x+4\right) mar uimhreoir agus ainmneoir.
\frac{\frac{\left(x+4\right)\left(x+4\right)}{\left(x+3\right)\left(x+4\right)}-\frac{\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x+4\right)}}{\frac{14}{x^{2}+7x+12}}
Chun cothromóidí a shuimiú nó a dhealú, fairsingigh iad chun a n-ainmneoirí a mheaitseáil. Is é an t-iolrach is lú coitianta de x+3 agus x+4 ná \left(x+3\right)\left(x+4\right). Méadaigh \frac{x+4}{x+3} faoi \frac{x+4}{x+4}. Méadaigh \frac{x-3}{x+4} faoi \frac{x+3}{x+3}.
\frac{\frac{\left(x+4\right)\left(x+4\right)-\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x+4\right)}}{\frac{14}{x^{2}+7x+12}}
Tá an t-ainmneoir céanna ag \frac{\left(x+4\right)\left(x+4\right)}{\left(x+3\right)\left(x+4\right)} agus \frac{\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x+4\right)} agus, mar sin, is féidir iad a dhealú trína n-uimhreoirí a dhealú.
\frac{\frac{x^{2}+4x+4x+16-x^{2}-3x+3x+9}{\left(x+3\right)\left(x+4\right)}}{\frac{14}{x^{2}+7x+12}}
Déan iolrúcháin in \left(x+4\right)\left(x+4\right)-\left(x-3\right)\left(x+3\right).
\frac{\frac{8x+25}{\left(x+3\right)\left(x+4\right)}}{\frac{14}{x^{2}+7x+12}}
Cumaisc téarmaí comhchosúla in: x^{2}+4x+4x+16-x^{2}-3x+3x+9.
\frac{\left(8x+25\right)\left(x^{2}+7x+12\right)}{\left(x+3\right)\left(x+4\right)\times 14}
Roinn \frac{8x+25}{\left(x+3\right)\left(x+4\right)} faoi \frac{14}{x^{2}+7x+12} trí \frac{8x+25}{\left(x+3\right)\left(x+4\right)} a mhéadú faoi dheilín \frac{14}{x^{2}+7x+12}.
\frac{\left(x+3\right)\left(x+4\right)\left(8x+25\right)}{14\left(x+3\right)\left(x+4\right)}
Fachtóirigh na sloinn nach bhfuil fachtóirithe cheana.
\frac{8x+25}{14}
Cealaigh \left(x+3\right)\left(x+4\right) mar uimhreoir agus ainmneoir.