Scipeáil chuig an bpríomhábhar
Luacháil
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Difreálaigh w.r.t. r
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Fadhbanna den chineál céanna ó Chuardach Gréasáin

Roinn

\frac{4\left(5r-2\right)}{\left(5r-2\right)\left(2r+5\right)}+\frac{3\left(2r+5\right)}{\left(5r-2\right)\left(2r+5\right)}
Chun cothromóidí a shuimiú nó a dhealú, fairsingigh iad chun a n-ainmneoirí a mheaitseáil. Is é an t-iolrach is lú coitianta de 2r+5 agus 5r-2 ná \left(5r-2\right)\left(2r+5\right). Méadaigh \frac{4}{2r+5} faoi \frac{5r-2}{5r-2}. Méadaigh \frac{3}{5r-2} faoi \frac{2r+5}{2r+5}.
\frac{4\left(5r-2\right)+3\left(2r+5\right)}{\left(5r-2\right)\left(2r+5\right)}
Tá an t-ainmneoir céanna ag \frac{4\left(5r-2\right)}{\left(5r-2\right)\left(2r+5\right)} agus \frac{3\left(2r+5\right)}{\left(5r-2\right)\left(2r+5\right)} agus, mar sin, is féidir iad a shuimiú trína n-uimhreoirí a shuimiú.
\frac{20r-8+6r+15}{\left(5r-2\right)\left(2r+5\right)}
Déan iolrúcháin in 4\left(5r-2\right)+3\left(2r+5\right).
\frac{26r+7}{\left(5r-2\right)\left(2r+5\right)}
Cumaisc téarmaí comhchosúla in: 20r-8+6r+15.
\frac{26r+7}{10r^{2}+21r-10}
Fairsingigh \left(5r-2\right)\left(2r+5\right)
\frac{\mathrm{d}}{\mathrm{d}r}(\frac{4\left(5r-2\right)}{\left(5r-2\right)\left(2r+5\right)}+\frac{3\left(2r+5\right)}{\left(5r-2\right)\left(2r+5\right)})
Chun cothromóidí a shuimiú nó a dhealú, fairsingigh iad chun a n-ainmneoirí a mheaitseáil. Is é an t-iolrach is lú coitianta de 2r+5 agus 5r-2 ná \left(5r-2\right)\left(2r+5\right). Méadaigh \frac{4}{2r+5} faoi \frac{5r-2}{5r-2}. Méadaigh \frac{3}{5r-2} faoi \frac{2r+5}{2r+5}.
\frac{\mathrm{d}}{\mathrm{d}r}(\frac{4\left(5r-2\right)+3\left(2r+5\right)}{\left(5r-2\right)\left(2r+5\right)})
Tá an t-ainmneoir céanna ag \frac{4\left(5r-2\right)}{\left(5r-2\right)\left(2r+5\right)} agus \frac{3\left(2r+5\right)}{\left(5r-2\right)\left(2r+5\right)} agus, mar sin, is féidir iad a shuimiú trína n-uimhreoirí a shuimiú.
\frac{\mathrm{d}}{\mathrm{d}r}(\frac{20r-8+6r+15}{\left(5r-2\right)\left(2r+5\right)})
Déan iolrúcháin in 4\left(5r-2\right)+3\left(2r+5\right).
\frac{\mathrm{d}}{\mathrm{d}r}(\frac{26r+7}{\left(5r-2\right)\left(2r+5\right)})
Cumaisc téarmaí comhchosúla in: 20r-8+6r+15.
\frac{\mathrm{d}}{\mathrm{d}r}(\frac{26r+7}{10r^{2}+25r-4r-10})
Cuir an t-airí dáileacháin i bhfeidhm trí gach téarma de 5r-2 a iolrú faoi gach téarma de 2r+5.
\frac{\mathrm{d}}{\mathrm{d}r}(\frac{26r+7}{10r^{2}+21r-10})
Comhcheangail 25r agus -4r chun 21r a fháil.
\frac{\left(10r^{2}+21r^{1}-10\right)\frac{\mathrm{d}}{\mathrm{d}r}(26r^{1}+7)-\left(26r^{1}+7\right)\frac{\mathrm{d}}{\mathrm{d}r}(10r^{2}+21r^{1}-10)}{\left(10r^{2}+21r^{1}-10\right)^{2}}
Do dhá fheidhm indifreáilte ar bith, is ionann díorthach líon an dá fheidhme agus an t-ainmneoir méadaithe faoi dhíorthach an uimhreora lúide an t-uimhreoir méadaithe faoi dhíorthach an ainmneora, agus iad ar fad roinnte faoin ainmneoir cearnaithe.
\frac{\left(10r^{2}+21r^{1}-10\right)\times 26r^{1-1}-\left(26r^{1}+7\right)\left(2\times 10r^{2-1}+21r^{1-1}\right)}{\left(10r^{2}+21r^{1}-10\right)^{2}}
Is ionann díorthach iltéarmaigh agus suim dhíorthaigh a théarmaí. Is ionann díorthach téarma thairisigh agus 0. Is ionann díorthach ax^{n} agus nax^{n-1}.
\frac{\left(10r^{2}+21r^{1}-10\right)\times 26r^{0}-\left(26r^{1}+7\right)\left(20r^{1}+21r^{0}\right)}{\left(10r^{2}+21r^{1}-10\right)^{2}}
Simpligh.
\frac{10r^{2}\times 26r^{0}+21r^{1}\times 26r^{0}-10\times 26r^{0}-\left(26r^{1}+7\right)\left(20r^{1}+21r^{0}\right)}{\left(10r^{2}+21r^{1}-10\right)^{2}}
Méadaigh 10r^{2}+21r^{1}-10 faoi 26r^{0}.
\frac{10r^{2}\times 26r^{0}+21r^{1}\times 26r^{0}-10\times 26r^{0}-\left(26r^{1}\times 20r^{1}+26r^{1}\times 21r^{0}+7\times 20r^{1}+7\times 21r^{0}\right)}{\left(10r^{2}+21r^{1}-10\right)^{2}}
Méadaigh 26r^{1}+7 faoi 20r^{1}+21r^{0}.
\frac{10\times 26r^{2}+21\times 26r^{1}-10\times 26r^{0}-\left(26\times 20r^{1+1}+26\times 21r^{1}+7\times 20r^{1}+7\times 21r^{0}\right)}{\left(10r^{2}+21r^{1}-10\right)^{2}}
Chun cumhachtaí an bhoinn chéanna a mhéadú, suimigh a n-easpónaint.
\frac{260r^{2}+546r^{1}-260r^{0}-\left(520r^{2}+546r^{1}+140r^{1}+147r^{0}\right)}{\left(10r^{2}+21r^{1}-10\right)^{2}}
Simpligh.
\frac{-260r^{2}-140r^{1}-407r^{0}}{\left(10r^{2}+21r^{1}-10\right)^{2}}
Cuir téarmaí cosúla le chéile.
\frac{-260r^{2}-140r-407r^{0}}{\left(10r^{2}+21r-10\right)^{2}}
Do théarma ar bith t, t^{1}=t.
\frac{-260r^{2}-140r-407}{\left(10r^{2}+21r-10\right)^{2}}
Do théarma ar bith t ach amháin 0, t^{0}=1.