Question

Simplify the given question (in attachment) ..as fast as possible..​

Answer 1

Answer:

A + B = 20 + 2root(10)

Step-by-step explanation:

( root(5) + root(2) )^2 = 5 + 2 + 2root(10) = 7 + 2root(10) = A

( root(8) - root(5) )^2 = 8 + 5 - 4root(10) = 13 -4root(10) = B

A + B = 20 + 2root(10)

hope this helps

Answer 2

Question

• $\small \sf{ (\sqrt{5} + \sqrt{2} )}^{2} + {( \sqrt{8} - \sqrt{5} ) }^{2}$

Step by step explanation

$\sf{ ~~~~~:\implies~~(\sqrt{5} + \sqrt{2} )}^{2} + {( \sqrt{8} - \sqrt{5} ) }^{2}$

• $\sf{Using \: Identity} \: \footnotesize(a+b)²=a²+b²+2ab \\ \sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \footnotesize(a-b)²=a²+b²-2ab}$

$\sf{ ~~~~~:\implies~~(\sqrt{5} )^{2} + (\sqrt{2} )^{2} + 2 \times \sqrt{5} \times \sqrt{2} + ( \sqrt{8} )^{2} + ( \sqrt{5} ) ^{2} - 2 \times \sqrt{8} \times \sqrt{5} }$

$\sf{ ~~~~~:\implies~~5 + 2 + 2 \sqrt{10} + 8 + 5 - 4 \sqrt{10} }$

$\sf{ ~~~~~:\implies~~7 + 13 - 2 \sqrt{10} }$

$\sf{ ~~~~~:\implies~~20 - 2 \sqrt{10} }$

${ \therefore \small \sf{ (\sqrt{5} + \sqrt{2} )}^{2} + {( \sqrt{8} - \sqrt{5} ) }^{2} = \bf20 - 2 \sqrt{10} }$

$\tiny\begin{gathered}\small{\small{\small{\small{\small{\small{\small{\small{\small{\small{\begin{gathered}\begin{gathered}\begin{gathered}\\\\\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \bigstar \: \underline{\bf{\blue{More \: Useful \: Formula}}}\\ {\boxed{\begin{array}{cc}\dashrightarrow \sf(a + b)^{2} = {a}^{2} + {b}^{2} + 2ab \\\\\dashrightarrow \sf(a - b)^{2} = {a}^{2} + {b}^{2} - 2ab \\\\\dashrightarrow \sf(a + b)(a - b) = {a}^{2} - {b}^{2} \\\\\dashrightarrow \sf(a + b) ^{3} = {a}^{3} + b^{3} + 3ab(a + b) \\\\ \dashrightarrow\sf(a - b) ^{3} = {a}^{3} - b^{3} - 3ab(a - b) \\ \\\dashrightarrow\sf a ^{3} + {b}^{3} = (a + b)(a ^{2} + {b}^{2} - ab) \\\\\dashrightarrow \sf a ^{3} - {b}^{3} = (a - b)(a ^{2} + {b}^{2} + ab \\\\\dashrightarrow \sf{a²+b²=(a+b)²-2ab}\\ \end{array}}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}}}}}}}}}}}\end{gathered}$