Réitigh frac{2sqrt{5}+4sqrt{3}}{sqrt{2}}-frac{2sqrt{18}-sqrt{27}}{sqrt{3}}

$$\frac { 2 \sqrt { 5 } + 4 \sqrt { 3 } } { \sqrt { 2 } } - \frac { 2 \sqrt { 18 } - \sqrt { 27 } } { \sqrt { 3 } }$$

Luacháil

$$\sqrt{10}+3\approx 6.16227766$$

Céimeanna gearra réitigh

Iolraigh an t-uimhreoir agus an t-ainmneoir faoi \sqrt{2} chun ainmneoir \frac{2\sqrt{5}+4\sqrt{3}}{\sqrt{2}} a thiontú in uimhir chóimheasta.

$$\frac{\left(2\sqrt{5}+4\sqrt{3}\right)\sqrt{2}}{\left(\sqrt{2}\right)^{2}}-\frac{2\sqrt{18}-\sqrt{27}}{\sqrt{3}}$$

Is é 2 uimhir chearnach \sqrt{2}.

$$\frac{\left(2\sqrt{5}+4\sqrt{3}\right)\sqrt{2}}{2}-\frac{2\sqrt{18}-\sqrt{27}}{\sqrt{3}}$$

Fachtóirigh 18=3^{2}\times 2. Athscríobh fréamh cearnach an toraidh \sqrt{3^{2}\times 2} mar thoradh na bhfréamhacha cearnacha \sqrt{3^{2}}\sqrt{2}. Tóg fréamh chearnach 3^{2}.

$$\frac{\left(2\sqrt{5}+4\sqrt{3}\right)\sqrt{2}}{2}-\frac{2\times 3\sqrt{2}-\sqrt{27}}{\sqrt{3}}$$

Méadaigh 2 agus 3 chun 6 a fháil.

$$\frac{\left(2\sqrt{5}+4\sqrt{3}\right)\sqrt{2}}{2}-\frac{6\sqrt{2}-\sqrt{27}}{\sqrt{3}}$$

Fachtóirigh 27=3^{2}\times 3. Athscríobh fréamh cearnach an toraidh \sqrt{3^{2}\times 3} mar thoradh na bhfréamhacha cearnacha \sqrt{3^{2}}\sqrt{3}. Tóg fréamh chearnach 3^{2}.

$$\frac{\left(2\sqrt{5}+4\sqrt{3}\right)\sqrt{2}}{2}-\frac{6\sqrt{2}-3\sqrt{3}}{\sqrt{3}}$$

Iolraigh an t-uimhreoir agus an t-ainmneoir faoi \sqrt{3} chun ainmneoir \frac{6\sqrt{2}-3\sqrt{3}}{\sqrt{3}} a thiontú in uimhir chóimheasta.

$$\frac{\left(2\sqrt{5}+4\sqrt{3}\right)\sqrt{2}}{2}-\frac{\left(6\sqrt{2}-3\sqrt{3}\right)\sqrt{3}}{\left(\sqrt{3}\right)^{2}}$$

Is é 3 uimhir chearnach \sqrt{3}.

$$\frac{\left(2\sqrt{5}+4\sqrt{3}\right)\sqrt{2}}{2}-\frac{\left(6\sqrt{2}-3\sqrt{3}\right)\sqrt{3}}{3}$$

Chun cothromóidí a shuimiú nó a dhealú, fairsingigh iad chun a n-ainmneoirí a mheaitseáil. Is é an t-iolrach is lú coitianta de 2 agus 3 ná 6. Méadaigh \frac{\left(2\sqrt{5}+4\sqrt{3}\right)\sqrt{2}}{2} faoi \frac{3}{3}. Méadaigh \frac{\left(6\sqrt{2}-3\sqrt{3}\right)\sqrt{3}}{3} faoi \frac{2}{2}.

$$\frac{3\left(2\sqrt{5}+4\sqrt{3}\right)\sqrt{2}}{6}-\frac{2\left(6\sqrt{2}-3\sqrt{3}\right)\sqrt{3}}{6}$$

Tá an t-ainmneoir céanna ag \frac{3\left(2\sqrt{5}+4\sqrt{3}\right)\sqrt{2}}{6} agus \frac{2\left(6\sqrt{2}-3\sqrt{3}\right)\sqrt{3}}{6} agus, mar sin, is féidir iad a dhealú trína n-uimhreoirí a dhealú.

$$\frac{3\left(2\sqrt{5}+4\sqrt{3}\right)\sqrt{2}-2\left(6\sqrt{2}-3\sqrt{3}\right)\sqrt{3}}{6}$$

Déan iolrúcháin in 3\left(2\sqrt{5}+4\sqrt{3}\right)\sqrt{2}-2\left(6\sqrt{2}-3\sqrt{3}\right)\sqrt{3}.

$$\frac{6\sqrt{10}+12\sqrt{6}-12\sqrt{6}+18}{6}$$

Déan áirimh in 6\sqrt{10}+12\sqrt{6}-12\sqrt{6}+18.

$$\frac{6\sqrt{10}+18}{6}$$

Roinn 6\sqrt{10}+18 faoi 6 chun \sqrt{10}+3 a fháil.

$$\sqrt{10}+3$$