+0  
 
+5
660
2
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Need full solution, help please :)

 Jan 11, 2015

Best Answer 

 #2
avatar+26400 
+5

Need full solution

$$\\ \small{\text{
$
\dfrac{ \left( \sqrt{3}-2\sqrt{5}\right ) \left( 2\sqrt{5}+\sqrt{3} \right) }{17\sqrt{20}}
$
}}$\\\\$
\small{\text{
$
=\dfrac{ \left( \sqrt{3}-2\sqrt{5}\right ) \left( \sqrt{3}+ 2\sqrt{5} \right) }{17\sqrt{20}} \quad |\quad (a-b)(a+b) = a^2-b^2 \quad \left( \sqrt{3}-2\sqrt{5}\right ) \left( \sqrt{3}+ 2\sqrt{5} \right) = 3-4*5
$
}}$\\\\$
\small{\text{
$
=\dfrac{ 3-4*5 }{17\sqrt{20}} $
}}$\\\\$
\small{\text{
$
=\dfrac{ -17 }{17\sqrt{20}} $
}}$\\\\$
\small{\text{
$
=-\dfrac{ 1 }{\sqrt{20}} \quad | \quad * \frac{\sqrt{20}} {\sqrt{20} } $
}}$\\\\$
\small{\text{
$
=-\dfrac{\sqrt{20} }{20} $
}}$\\\\$
\small{\text{
$
=-\dfrac{\sqrt{4*5} }{20} $
}}$\\\\$
\small{\text{
$
=-\dfrac{\sqrt{4}*\sqrt{5} } {20} $
}}$\\\\$
\small{\text{
$
=-\dfrac{2\sqrt{5} } {20} $
}}$\\\\$
\small{\text{
$
=-\dfrac{\sqrt{5} } {10} $
}}$$

 Jan 12, 2015
 #1
avatar+33665 
+5

$$(\sqrt3-2\sqrt5)(2\sqrt5 +\sqrt3)=3-4\times 5=-17$$

 

$$\frac{1}{\sqrt20}=\frac{\sqrt20}{20}=\frac{2\sqrt5}{20}=\frac{\sqrt5}{10}$$

 

So:

 

$$\frac{(\sqrt3 - 2\sqrt5)(2\sqrt5 + \sqrt3)}{17\sqrt20}=\frac{-17\sqrt5}{17\times10}=-\frac{\sqrt5}{10}$$

.

.
 Jan 12, 2015
 #2
avatar+26400 
+5
Best Answer

Need full solution

$$\\ \small{\text{
$
\dfrac{ \left( \sqrt{3}-2\sqrt{5}\right ) \left( 2\sqrt{5}+\sqrt{3} \right) }{17\sqrt{20}}
$
}}$\\\\$
\small{\text{
$
=\dfrac{ \left( \sqrt{3}-2\sqrt{5}\right ) \left( \sqrt{3}+ 2\sqrt{5} \right) }{17\sqrt{20}} \quad |\quad (a-b)(a+b) = a^2-b^2 \quad \left( \sqrt{3}-2\sqrt{5}\right ) \left( \sqrt{3}+ 2\sqrt{5} \right) = 3-4*5
$
}}$\\\\$
\small{\text{
$
=\dfrac{ 3-4*5 }{17\sqrt{20}} $
}}$\\\\$
\small{\text{
$
=\dfrac{ -17 }{17\sqrt{20}} $
}}$\\\\$
\small{\text{
$
=-\dfrac{ 1 }{\sqrt{20}} \quad | \quad * \frac{\sqrt{20}} {\sqrt{20} } $
}}$\\\\$
\small{\text{
$
=-\dfrac{\sqrt{20} }{20} $
}}$\\\\$
\small{\text{
$
=-\dfrac{\sqrt{4*5} }{20} $
}}$\\\\$
\small{\text{
$
=-\dfrac{\sqrt{4}*\sqrt{5} } {20} $
}}$\\\\$
\small{\text{
$
=-\dfrac{2\sqrt{5} } {20} $
}}$\\\\$
\small{\text{
$
=-\dfrac{\sqrt{5} } {10} $
}}$$

heureka Jan 12, 2015

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