Question

resolver los siguientes productos​

Answer 1

Respuesta:

7)

$(\sqrt{2}+ \sqrt{3}) (\sqrt{6} -1)=\sqrt{2}\sqrt{6}-\sqrt{2}+\sqrt{3}\sqrt{6}-\sqrt{3}\\(\sqrt{2}+ \sqrt{3}) (\sqrt{6} -1)=\sqrt{2}\sqrt{2}\sqrt{3}- \sqrt{2}+\sqrt{3}\sqrt{2}\sqrt{3}-\sqrt{3}\\(\sqrt{2}+ \sqrt{3}) (\sqrt{6} -1)=2\sqrt{3}- \sqrt{2}+3\sqrt{2}-\sqrt{3}\\(\sqrt{2}+ \sqrt{3}) (\sqrt{6} -1)=\sqrt{3}+2\sqrt{2}$

8)

$(\sqrt{5} +\sqrt{2}) (\sqrt{10} )=\sqrt{5}\sqrt{10} +\sqrt{2}\sqrt{10} \\(\sqrt{5} +\sqrt{2}) (\sqrt{10} )=\sqrt{5}\sqrt{5}\sqrt{2} +\sqrt{2}\sqrt{5}\sqrt{2}\\(\sqrt{5} +\sqrt{2}) (\sqrt{10} )=5\sqrt{2} +2\sqrt{5}$

9)

$(\sqrt{2}+1 )^{2}=(\sqrt{2} )^{2}+2\sqrt{2}(1) +(1)^{2}\\(\sqrt{2}+1 )^{2}=2+2\sqrt{2} +1\\(\sqrt{2}+1 )^{2}=3+2\sqrt{2}$

10)

$(\sqrt{2}+\sqrt{3} )\times (\sqrt{2}-\sqrt{3} )=(\sqrt{2} )^{2}-(\sqrt{3} )^{2}\\(\sqrt{2}+\sqrt{3} )\times (\sqrt{2}-\sqrt{3} )=2-3\\(\sqrt{2}+\sqrt{3} )\times (\sqrt{2}-\sqrt{3} )=-1$

*Aplicar productos notables de un binomio conjugado

Espero se entienda, SUERTE!!!

Answer 2

Respuesta:

Explicación paso a paso:

7)

8)

9)

10)