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-\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}}
\left(\sqrt{2}-1\right)^{2} zabaltzeko, erabili Newton-en binomioa \left(a-b\right)^{2}=a^{2}-2ab+b^{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}}
\sqrt{2} zenbakiaren karratua 2 da.
-\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}}
3 lortzeko, gehitu 2 eta 1.
-\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}}
Adierazi \frac{3-2\sqrt{2}}{4\sqrt{2}} balioaren izendatzailea zenbaki arrazional gisa. Horretarako, egin zenbakitzailea eta izendatzailea bider \sqrt{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}}
\sqrt{2} zenbakiaren karratua 2 da.
-\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}}
8 lortzeko, biderkatu 4 eta 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}}
\left(\sqrt{5}+\sqrt{3}\right)^{2} zabaltzeko, erabili Newton-en binomioa \left(a+b\right)^{2}=a^{2}+2ab+b^{2}.
-\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}}
\sqrt{5} zenbakiaren karratua 5 da.
-\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}}
\sqrt{5} eta \sqrt{3} biderkatzeko, biderkatu erro karratuaren azpiko zenbakiak.
-\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}}
\sqrt{3} zenbakiaren karratua 3 da.
-\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}}
8 lortzeko, gehitu 5 eta 3.
-\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}}
Adierazi \frac{8+2\sqrt{15}}{\sqrt{15}} balioaren izendatzailea zenbaki arrazional gisa. Horretarako, egin zenbakitzailea eta izendatzailea bider \sqrt{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}}
\sqrt{15} zenbakiaren karratua 15 da.
-\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}}
\left(\sqrt{2}+1\right)^{2} zabaltzeko, erabili Newton-en binomioa \left(a+b\right)^{2}=a^{2}+2ab+b^{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}}
\sqrt{2} zenbakiaren karratua 2 da.
-\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}}
3 lortzeko, gehitu 2 eta 1.
-\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}}
Adierazi \frac{3+2\sqrt{2}}{4\sqrt{2}} balioaren izendatzailea zenbaki arrazional gisa. Horretarako, egin zenbakitzailea eta izendatzailea bider \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\times 2}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}
\sqrt{2} zenbakiaren karratua 2 da.
-\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}}
8 lortzeko, biderkatu 4 eta 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}}
\left(\sqrt{5}-\sqrt{3}\right)^{2} zabaltzeko, erabili Newton-en binomioa \left(a-b\right)^{2}=a^{2}-2ab+b^{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{5-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{\sqrt{15}}
\sqrt{5} zenbakiaren karratua 5 da.
-\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}}
\sqrt{5} eta \sqrt{3} biderkatzeko, biderkatu erro karratuaren azpiko zenbakiak.
-\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}}
\sqrt{3} zenbakiaren karratua 3 da.
-\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}}
8 lortzeko, gehitu 5 eta 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{\left(8-2\sqrt{15}\right)\sqrt{15}}{\left(\sqrt{15}\right)^{2}}
Adierazi \frac{8-2\sqrt{15}}{\sqrt{15}} balioaren izendatzailea zenbaki arrazional gisa. Horretarako, egin zenbakitzailea eta izendatzailea bider \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}}{15}
\sqrt{15} zenbakiaren karratua 15 da.
-\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}
Adierazpenak gehitzeko edo kentzeko, zabal itzazu izendatzaileak berdintzeko. 8 eta 15 ekuazioen multiplo komun txikiena 120 da. Egin -\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8} bider \frac{15}{15}. Egin \frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15} bider \frac{8}{8}.
\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}
-\frac{15\left(3-2\sqrt{2}\right)\sqrt{2}}{120} eta \frac{8\left(8+2\sqrt{15}\right)\sqrt{15}}{120} balioek izendatzaile bera dutenez, zenbakitzaileak batu behar dituzu zatikien batura kalkulatzeko.
\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}
Egin biderketak -15\left(3-2\sqrt{2}\right)\sqrt{2}+8\left(8+2\sqrt{15}\right)\sqrt{15} zatikian.
\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}
Egin kalkuluak hemen: -45\sqrt{2}+60+64\sqrt{15}+240.
\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}
Adierazpenak gehitzeko edo kentzeko, zabal itzazu izendatzaileak berdintzeko. 120 eta 8 ekuazioen multiplo komun txikiena 120 da. Egin \frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8} bider \frac{15}{15}.
\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}
\frac{-45\sqrt{2}+300+64\sqrt{15}}{120} eta \frac{15\left(3+2\sqrt{2}\right)\sqrt{2}}{120} balioek izendatzaile bera dutenez, zenbakitzaileak batu behar dituzu zatikien batura kalkulatzeko.
\frac{-45\sqrt{2}+300+64\sqrt{15}+45\sqrt{2}+60}{120}-\frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{15}
Egin biderketak -45\sqrt{2}+300+64\sqrt{15}+15\left(3+2\sqrt{2}\right)\sqrt{2} zatikian.
\frac{360+64\sqrt{15}}{120}-\frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{15}
Egin kalkuluak hemen: -45\sqrt{2}+300+64\sqrt{15}+45\sqrt{2}+60.
\frac{360+64\sqrt{15}}{120}-\frac{8\left(8-2\sqrt{15}\right)\sqrt{15}}{120}
Adierazpenak gehitzeko edo kentzeko, zabal itzazu izendatzaileak berdintzeko. 120 eta 15 ekuazioen multiplo komun txikiena 120 da. Egin \frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{15} bider \frac{8}{8}.
\frac{360+64\sqrt{15}-8\left(8-2\sqrt{15}\right)\sqrt{15}}{120}
\frac{360+64\sqrt{15}}{120} eta \frac{8\left(8-2\sqrt{15}\right)\sqrt{15}}{120} balioek izendatzaile bera dutenez, zenbakitzaileak kendu behar dituzu zatikien kendura kalkulatzeko.
\frac{360+64\sqrt{15}-64\sqrt{15}+240}{120}
Egin biderketak 360+64\sqrt{15}-8\left(8-2\sqrt{15}\right)\sqrt{15} zatikian.
\frac{600}{120}
Egin kalkuluak hemen: 360+64\sqrt{15}-64\sqrt{15}+240.
5
5 lortzeko, zatitu 600 120 balioarekin.