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vova2212 [387]
3 years ago
9

Help please! Thank you

Mathematics
1 answer:
RoseWind [281]3 years ago
3 0
The correct answer is A.
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SOMEONE HELP ME IM FREAKING OUT I LITERALLY CANT WITH THIS QUESTION IM PRAYING PLEASE HELP ME IM SO SERIOUS IM GONNA END IT PLS
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Answer:

\sf -11+7\sqrt{2}

Step-by-step explanation:

Given expression:

\sf \dfrac{3-\sqrt{32}}{1+\sqrt{2} }

Rewrite 32 as 16 · 2:

\sf \implies \dfrac{3-\sqrt{16 \cdot 2}}{1+\sqrt{2} }

Apply radical rule \sf \sqrt{a \cdot b}=\sqrt{a}\sqrt{b}

\sf \implies \dfrac{3-\sqrt{16}\sqrt{2}}{1+\sqrt{2} }

As \sf \sqrt{16}=4:

\sf \implies \dfrac{3-4\sqrt{2}}{1+\sqrt{2} }

Multiply by the conjugate:

\sf \implies \dfrac{3-4\sqrt{2}}{1+\sqrt{2} } \times \dfrac{1-\sqrt{2} }{1-\sqrt{2} }

\sf \implies \dfrac{(3-4\sqrt{2})(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})}

\sf \implies \dfrac{3-3\sqrt{2}-4\sqrt{2}+4\sqrt{2}\sqrt{2}}{1-\sqrt{2}+\sqrt{2}-\sqrt{2}\sqrt{2}}

As \sf \sqrt{2}\sqrt{2}=\sqrt{4}=2:

\sf \implies \dfrac{3-3\sqrt{2}-4\sqrt{2}+4 \cdot 2}{1-\sqrt{2}+\sqrt{2}-2}

\sf \implies \dfrac{3-7\sqrt{2}+8}{1-2}

\sf \implies \dfrac{11-7\sqrt{2}}{-1}

\sf \implies -11+7\sqrt{2}

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Answer:

I think so, but don't lose sight

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