Question

Quem podé me ajudar com as questões 2, 3 e 4

Answer 1

Lembrando:

$(a+b)^{2}=a^{2}+2ab+b^{2}\\(a+b)(a-b)\rightleftharpoons a^{2}-b^{2}$

$\sqrt{a}*\sqrt{b}=\sqrt{ab}$
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2.

$(\sqrt{2+\sqrt{3}} + \sqrt{2-\sqrt{3}})^{2}$
$(\sqrt{2+\sqrt{3}})^{2}+2*\sqrt{2+\sqrt{3}}*\sqrt{2-\sqrt{3}}+(\sqrt{2-\sqrt{3}})^{2}\\ (2+\sqrt{3})+2\sqrt{(2+\sqrt{3})(2-\sqrt{3})}+(2-\sqrt{3})\\2+\sqrt{3}+2\sqrt{2^{2}-(\sqrt{3})^{2}}+2-\sqrt{3}\\(2+\sqrt{3}+2-\sqrt{3})+2\sqrt{4-3}\\4+2\sqrt{1}\\4+2\\6$

Letra C
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3.

$\sqrt{27}=\sqrt{3*9}=\sqrt{3}*\sqrt{9}=\sqrt{3}*3=3\sqrt{3}$


$(\sqrt{27}+\sqrt{3}+1)^{2}=(3\sqrt{3}+\sqrt{3}+1)^{2}\\(\sqrt{27}+\sqrt{3}+1)^{2}=(3\sqrt{3}+\sqrt{3}+1)^{2}\\(\sqrt{27}+\sqrt{3}+1)^{2}=(4\sqrt{3}+1)^{2}\\(\sqrt{27}+\sqrt{3}+1)^{2}=(4\sqrt{3})^{2}+2*4\sqrt{3}*1+1^{2}\\(\sqrt{27}+\sqrt{3}+1)^{2}=4^{2}*3+8\sqrt{3}+1\\(\sqrt{27}+\sqrt{3}+1)^{2}=48+8\sqrt{3}+1\\(\sqrt{27}+\sqrt{3}+1)^{2}=49+8\sqrt{3}\\(\sqrt{27}+\sqrt{3}+1)^{2}=8\sqrt{3}+49$

$a\sqrt{3}+b=8\sqrt{3}+49$

$a=8\\b=49\\\\a+b=8+49=57$

Letra C
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4.

$\frac{x^{3}-xy^{2}}{x^{2}+xy}=\frac{x*(x^{2}-y^{2})}{x*(x+y)}$

Cortando x com x:

$\frac{x^{3}-xy^{2}}{x^{2}+xy}=\frac{x^{2}-y^{2}}{x+y}$

x² - y² é o resultado de um produto notável:

$x^{2}-y^{2}\rightleftharpoons (x+y)(x-y)$

$\frac{x^{3}-xy^{2}}{x^{2}+xy}=\frac{(x+y)(x-y)}{x+y}$

Cortando (x + y) com (x + y):

$\boxed{\boxed{\frac{x^{3}-xy^{2}}{x^{2}+xy}=x - y}}$

Como x = 53 e y = 52:

$\frac{x^{3}-xy^{2}}{x^{2}+xy}=53-52\\\\\frac{x^{3}-xy^{2}}{x^{2}+xy}=1$