3 years ago

Have to find x,y and z.

Mathematics

1 answer:

ELEN [110]3 years ago

6 0

Answer:

z= 2√2

x=2

y=2√2

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Step-by-step explanation:

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$~~~\dfrac{2\sqrt 2 + \sqrt 7}{2\sqrt 2 -\sqrt 7} + \dfrac{2\sqrt 2-\sqrt 7 }{2\sqrt 2+\sqrt 7}\\\\\\=\dfrac{\sqrt 8 + \sqrt 7}{\sqrt 8 -\sqrt 7} + \dfrac{\sqrt 8-\sqrt 7 }{\sqrt 8+\sqrt 7}\\\\\\=\dfrac{\left(\sqrt 8 +\sqrt 7\right)^2+\left(\sqrt 8 -\sqrt 7\right)^2 }{\left(\sqrt 8 -\sqrt 7\right)\left(\sqrt 8 +\sqrt 7\right)}\\\\\\$

$=\dfrac{\left(\sqrt 8+ \sqrt 7 + \sqrt 8 - \sqrt 7 \right)^2 - 2 \left(\sqrt 8 +\sqrt 7\right)\left(\sqrt 8 -\sqrt 7\right)}{\left(\sqrt 8 \right)^2 - \left(\sqrt 7 \right)^2}\\\\\\=\dfrac{\left(2\sqrt 8 \right)^2 - 2(8-7)}{8-7}\\\\\\=\dfrac{4 \cdot 8 -2\cdot 1}{1}\\\\\\=32 -2 \\\\\\=30$