Question

calcule a diferença entre √2 /2√2-3 e 2√2/2√2+3

Answer 1

$\frac{ \sqrt{2} }{2 \sqrt{2} -3} - \frac{2 \sqrt{2} }{2 \sqrt{2}+3 } = \\ \\ mmc=(2 \sqrt{2} -3)(2 \sqrt{2} +3) \\ \\ \frac{ \sqrt{2}(2 \sqrt{2} +3)-2 \sqrt{2} (2 \sqrt{2} -3) }{(2 \sqrt{2} -3)(2 \sqrt{2}+3) } = \\ \\ \frac{2 \sqrt{4}+3 \sqrt{2} -4 \sqrt{4} +6 \sqrt{2} }{4 \sqrt{4}+6 \sqrt{2} -6 \sqrt{2} -9 } = \\ \\ \frac{2.(2)+3 \sqrt{2}-4(2)+6 \sqrt{2} }{4.(2)-9} = \\ \\ \frac{3 \sqrt{2}+6 \sqrt{2}+4-8 }{8-9} = \\ \\ \frac{9 \sqrt{2}-4 }{-1} = \\ \\ -1(9 \sqrt{2} -4)=4-9 \sqrt{2}$

Answer 2

$\dfrac{ \sqrt{2} }{2 \sqrt{2}-3 } - \dfrac{2 \sqrt{2} }{2 \sqrt{2} +3} \\ \\ \dfrac{ \sqrt{2} \cdot (2 \sqrt{2} +3)-2 \sqrt{2} \cdot (2 \sqrt{2} -3) }{(2 \sqrt{2}-3) \cdot (2 \sqrt{2} +3) } \\ \\ \dfrac{2 \sqrt{2 \cdot 2}+3 \sqrt{2}-2 \cdot 2 \sqrt{2 \cdot 2}+2 \cdot 3 \sqrt{2} }{2 \cdot 2 \sqrt{2 \cdot 2}+2 \cdot 3 \sqrt{2} -3 \cdot 2 \sqrt{2} - 3 \cdot 3} \\ \\ \dfrac{2 \sqrt{4}+3 \sqrt{2}-4 \sqrt{4}+6 \sqrt{2} }{4 \sqrt{4} +6 \sqrt{2}-6 \sqrt{2}-9 }$
$\dfrac{2 \sqrt{4}+3 \sqrt{2}+6 \sqrt{2} -4 \sqrt{4} }{4 \sqrt{4} +6 \sqrt{2}-6 \sqrt{2}-9 } \\ \\ \dfrac{2 \cdot 2+ \sqrt{2} \cdot (3+6)-4 \cdot 2 }{4 \cdot 2-9} \\ \\ \dfrac{4+ \sqrt{2} \cdot 9 -8 }{8-9} \\ \\ \dfrac{-4+ \sqrt{2} \cdot 9}{-1} \\ \\ -(-4+ \sqrt{2} \cdot 9) \\ \\ 4- \sqrt{2} \cdot 9 \\ \\\boxed{ 4-9 \sqrt{2}}$

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$\dfrac{ \sqrt{2} }{2 \sqrt{2}}-3 - \dfrac{2 \sqrt{2} }{2 \sqrt{2} }+3 \\ \\ \dfrac{ \sqrt{2} }{2 \sqrt{2}} - \dfrac{2 \sqrt{2} }{2 \sqrt{2} } \\ \\ \dfrac{ \sqrt{2} \cdot (1-2) }{2 \sqrt{2} } \\ \\ \dfrac{ \sqrt{2} \cdot (-1) }{2 \sqrt{2} } \\ \\ \dfrac{-1 \sqrt{2} }{2 \sqrt{2} } \\ \\ \boxed{ -\dfrac{1}{2}}$



Bons estudos!