Question

$7\sqrt{2-3\sqrt{2}+2\sqrt{2}$

Answer 1

Resposta:

)

\sf (\sqrt{5}+\sqrt{3})^2=(\sqrt{5})^2+2\cdot\sqrt{5}\cdot\sqrt{3}+(\sqrt{3})^2(5+3)2=(5)2+2⋅5⋅3+(3)2

\sf (\sqrt{5}+\sqrt{3})^2=5+2\cdot\sqrt{15}+3(5+3)2=5+2⋅15+3

\sf (\sqrt{5}+\sqrt{3})^2=5+3+2\sqrt{15}(5+3)2=5+3+215

\sf \red{(\sqrt{5}+\sqrt{3})^2=8+2\sqrt{15}}(5+3)2=8+215

b)

\sf (2-\sqrt{2})^2=2^2-2\cdot2\cdot\sqrt{2}+(\sqrt{2})^2(2−2)2=22−2⋅2⋅2+(2)2

\sf (2-\sqrt{2})^2=4-4\sqrt{2}+2(2−2)2=4−42+2

\sf (2-\sqrt{2})^2=4+2-4\sqrt{2}(2−2)2=4+2−42

\sf \red{(2-\sqrt{2})^2=6-4\sqrt{2}}(2−2)2=6−42

c)

\sf (\sqrt{7}+\sqrt{5})\cdot(\sqrt{7}-\sqrt{5})(7+5)⋅(7−5)

\sf =(\sqrt{7})^2-(\sqrt{5})^2=(7)2−(5)2

\sf =7-5=7−5

\sf =\red{2}=2

d)

\sf (a+\sqrt{b})^2=a^2+2\cdot a\cdot\sqrt{b}+(\sqrt{b})^2(a+b)2=a2+2⋅a⋅b+(b)2

\sf (a+\sqrt{b})^2=a^2+2a\sqrt{b}+b(a+b)2=a2+2ab+b

\sf \red{(a+\sqrt{b})^2=a^2+b+2a\sqrt{b}}(a+b)2=a2+b+2ab

e)

\sf (a+\sqrt{b})^2=a^2+2\cdot a\cdot\sqrt{b}+(\sqrt{b})^2(a+b)2=a2+2⋅a⋅b