Resol -frac{(sqrt{2}-1)^2}{4sqrt{2}}+frac{(sqrt{5}+sqrt{3})^2}{sqrt{15}}+frac{(sqrt{2}+1)^2}{4sqrt{2}}-frac{(sqrt{5}-sqrt{3})^2}{sqrt{15}}

$$- \frac { ( \sqrt { 2 } - 1 ) ^ { 2 } } { 4 \sqrt { 2 } } + \frac { ( \sqrt { 5 } + \sqrt { 3 } ) ^ { 2 } } { \sqrt { 15 } } + \frac { ( \sqrt { 2 } + 1 ) ^ { 2 } } { 4 \sqrt { 2 } } - \frac { ( \sqrt { 5 } - \sqrt { 3 } ) ^ { 2 } } { \sqrt { 15 } }$$

Calcula

$$5$$

Passos de solució senzills

Utilitzeu el teorema del binomi \left(a-b\right)^{2}=a^{2}-2ab+b^{2} per desenvolupar \left(\sqrt{2}-1\right)^{2}.

$$-\frac{\left(\sqrt{2}\right)^{2}-2\sqrt{2}+1}{4\sqrt{2}}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^{2}}{\sqrt{15}}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

L'arrel quadrada de \sqrt{2} és 2.

$$-\frac{2-2\sqrt{2}+1}{4\sqrt{2}}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^{2}}{\sqrt{15}}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

Sumeu 2 més 1 per obtenir 3.

$$-\frac{3-2\sqrt{2}}{4\sqrt{2}}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^{2}}{\sqrt{15}}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

Racionalitzeu el denominador de \frac{3-2\sqrt{2}}{4\sqrt{2}} multiplicant el numerador i el denominador per \sqrt{2}.

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{4\left(\sqrt{2}\right)^{2}}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^{2}}{\sqrt{15}}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

L'arrel quadrada de \sqrt{2} és 2.

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{4\times 2}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^{2}}{\sqrt{15}}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

Multipliqueu 4 per 2 per obtenir 8.

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^{2}}{\sqrt{15}}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

Utilitzeu el teorema del binomi \left(a+b\right)^{2}=a^{2}+2ab+b^{2} per desenvolupar \left(\sqrt{5}+\sqrt{3}\right)^{2}.

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(\sqrt{5}\right)^{2}+2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{\sqrt{15}}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

L'arrel quadrada de \sqrt{5} és 5.

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{5+2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{\sqrt{15}}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

Per multiplicar \sqrt{5} i \sqrt{3}, Multipliqueu els números sota l'arrel quadrada.

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{5+2\sqrt{15}+\left(\sqrt{3}\right)^{2}}{\sqrt{15}}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

L'arrel quadrada de \sqrt{3} és 3.

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{5+2\sqrt{15}+3}{\sqrt{15}}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

Sumeu 5 més 3 per obtenir 8.

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{8+2\sqrt{15}}{\sqrt{15}}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

Racionalitzeu el denominador de \frac{8+2\sqrt{15}}{\sqrt{15}} multiplicant el numerador i el denominador per \sqrt{15}.

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{\left(\sqrt{15}\right)^{2}}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

L'arrel quadrada de \sqrt{15} és 15.

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

Utilitzeu el teorema del binomi \left(a+b\right)^{2}=a^{2}+2ab+b^{2} per desenvolupar \left(\sqrt{2}+1\right)^{2}.

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{\left(\sqrt{2}\right)^{2}+2\sqrt{2}+1}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

L'arrel quadrada de \sqrt{2} és 2.

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{2+2\sqrt{2}+1}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

Sumeu 2 més 1 per obtenir 3.

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{3+2\sqrt{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

Racionalitzeu el denominador de \frac{3+2\sqrt{2}}{4\sqrt{2}} multiplicant el numerador i el denominador per \sqrt{2}.

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{4\left(\sqrt{2}\right)^{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

L'arrel quadrada de \sqrt{2} és 2.

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{4\times 2}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

Multipliqueu 4 per 2 per obtenir 8.

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

Utilitzeu el teorema del binomi \left(a-b\right)^{2}=a^{2}-2ab+b^{2} per desenvolupar \left(\sqrt{5}-\sqrt{3}\right)^{2}.

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8}-\frac{\left(\sqrt{5}\right)^{2}-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{\sqrt{15}}$$

L'arrel quadrada de \sqrt{5} és 5.

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8}-\frac{5-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{\sqrt{15}}$$

Per multiplicar \sqrt{5} i \sqrt{3}, Multipliqueu els números sota l'arrel quadrada.

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8}-\frac{5-2\sqrt{15}+\left(\sqrt{3}\right)^{2}}{\sqrt{15}}$$

L'arrel quadrada de \sqrt{3} és 3.

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8}-\frac{5-2\sqrt{15}+3}{\sqrt{15}}$$

Sumeu 5 més 3 per obtenir 8.

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8}-\frac{8-2\sqrt{15}}{\sqrt{15}}$$

Racionalitzeu el denominador de \frac{8-2\sqrt{15}}{\sqrt{15}} multiplicant el numerador i el denominador per \sqrt{15}.

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8}-\frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{\left(\sqrt{15}\right)^{2}}$$

L'arrel quadrada de \sqrt{15} és 15.

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8}-\frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{15}$$

Per afegir o restar les expressions, amplieu-les perquè els denominadors coincideixin. El mínim comú múltiple de 8 i 15 és 120. Multipliqueu -\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8} per \frac{15}{15}. Multipliqueu \frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15} per \frac{8}{8}.

$$-\frac{15\left(3-2\sqrt{2}\right)\sqrt{2}}{120}+\frac{8\left(8+2\sqrt{15}\right)\sqrt{15}}{120}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8}-\frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{15}$$

Com que -\frac{15\left(3-2\sqrt{2}\right)\sqrt{2}}{120} i \frac{8\left(8+2\sqrt{15}\right)\sqrt{15}}{120} tenen el mateix denominador, afegiu-los mitjançant l'addició dels seus numeradors.

$$\frac{-15\left(3-2\sqrt{2}\right)\sqrt{2}+8\left(8+2\sqrt{15}\right)\sqrt{15}}{120}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8}-\frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{15}$$

Feu les multiplicacions a -15\left(3-2\sqrt{2}\right)\sqrt{2}+8\left(8+2\sqrt{15}\right)\sqrt{15}.

$$\frac{-45\sqrt{2}+60+64\sqrt{15}+240}{120}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8}-\frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{15}$$

Feu el càlcul -45\sqrt{2}+60+64\sqrt{15}+240.

$$\frac{-45\sqrt{2}+300+64\sqrt{15}}{120}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8}-\frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{15}$$

Per afegir o restar les expressions, amplieu-les perquè els denominadors coincideixin. El mínim comú múltiple de 120 i 8 és 120. Multipliqueu \frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8} per \frac{15}{15}.

$$\frac{-45\sqrt{2}+300+64\sqrt{15}}{120}+\frac{15\left(3+2\sqrt{2}\right)\sqrt{2}}{120}-\frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{15}$$

Com que \frac{-45\sqrt{2}+300+64\sqrt{15}}{120} i \frac{15\left(3+2\sqrt{2}\right)\sqrt{2}}{120} tenen el mateix denominador, afegiu-los mitjançant l'addició dels seus numeradors.

$$\frac{-45\sqrt{2}+300+64\sqrt{15}+15\left(3+2\sqrt{2}\right)\sqrt{2}}{120}-\frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{15}$$

Feu les multiplicacions a -45\sqrt{2}+300+64\sqrt{15}+15\left(3+2\sqrt{2}\right)\sqrt{2}.

$$\frac{-45\sqrt{2}+300+64\sqrt{15}+45\sqrt{2}+60}{120}-\frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{15}$$

Feu el càlcul -45\sqrt{2}+300+64\sqrt{15}+45\sqrt{2}+60.

$$\frac{360+64\sqrt{15}}{120}-\frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{15}$$

Per afegir o restar les expressions, amplieu-les perquè els denominadors coincideixin. El mínim comú múltiple de 120 i 15 és 120. Multipliqueu \frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{15} per \frac{8}{8}.

$$\frac{360+64\sqrt{15}}{120}-\frac{8\left(8-2\sqrt{15}\right)\sqrt{15}}{120}$$

Com que \frac{360+64\sqrt{15}}{120} i \frac{8\left(8-2\sqrt{15}\right)\sqrt{15}}{120} tenen el mateix denominador, resteu-los mitjançant la subtracció dels seus numeradors.

$$\frac{360+64\sqrt{15}-8\left(8-2\sqrt{15}\right)\sqrt{15}}{120}$$

Feu les multiplicacions a 360+64\sqrt{15}-8\left(8-2\sqrt{15}\right)\sqrt{15}.

$$\frac{360+64\sqrt{15}-64\sqrt{15}+240}{120}$$

Feu el càlcul 360+64\sqrt{15}-64\sqrt{15}+240.

$$\frac{600}{120}$$

Dividiu 600 entre 120 per obtenir 5.

$$5$$

Factoritzar

$$5$$