Question

The greatest among ( sqrt(7 )- sqrt(5)),(sqrt(5) - sqrt(3)),( sqrt(9)- sqrt(7) )and (sqrt(11) - sqrt(9)) is : *​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

$(\sqrt{5} - \sqrt{3} )$

Step-by-step explanation:

Given Numbers are:

$(\sqrt{7} -\sqrt5}),(\sqrt{5}-\sqrt{3}),(\sqrt{9}-\sqrt{7}),(\sqrt{11}-\sqrt{9})$

Let us unrationalize the following terms:

⇒  $(\sqrt{7} -\sqrt{5}) * \frac{(\sqrt{7} +\sqrt{5})}{(\sqrt{7} +\sqrt{5})}$      

We know that $(a+b)(a-b) = a^{2}-b^{2}$

⇒ $\frac{(\sqrt{7})^{2} - (\sqrt{5})^{2}}{\sqrt{7}+\sqrt{5}}$

⇒ $\frac{2}{\sqrt{7}+\sqrt{5}}$

Similarly,

$(\sqrt{5} - \sqrt{3}) = \frac{2}{\sqrt{5} + \sqrt{3}}$

$(\sqrt{9} - \sqrt{7}) = \frac{2}{\sqrt{9} + \sqrt{7}}$

$(\sqrt{11}- \sqrt{9}) = \frac{2}{\sqrt{11}+ \sqrt{9}}$

We know that $3<5<7<9<11$

Square rooting all sides

⇒ $\sqrt{3}<\sqrt{5}<\sqrt{7} <\sqrt{9}<\sqrt{11}$

This would also mean that,

$(\sqrt{5}+\sqrt{3}) < (\sqrt{7} + \sqrt{5})<(\sqrt{9} + \sqrt{7}) < (\sqrt{11} + \sqrt{9})$

Taking reciprocal on all sides.

⇒ $\frac{1}{(\sqrt{5}+\sqrt{3})} > \frac{1}{(\sqrt{7} + \sqrt{5})} >\frac{1}{(\sqrt{9} + \sqrt{7})} > \frac{1}{(\sqrt{11} + \sqrt{9})}$

Multiplying by $2$ on all sides.

⇒ $\frac{2}{(\sqrt{5}+\sqrt{3})} > \frac{2}{(\sqrt{7} + \sqrt{5})} >\frac{2}{(\sqrt{9} + \sqrt{7})} > \frac{2}{(\sqrt{11} + \sqrt{9})}$

⇒ $(\sqrt{5}-\sqrt{3})>(\sqrt{7} -\sqrt5})>(\sqrt{9}-\sqrt{7})>(\sqrt{11}-\sqrt{9})$

∴ The greatest among $(\sqrt{7} -\sqrt5}),(\sqrt{5}-\sqrt{3}),(\sqrt{9}-\sqrt{7}),(\sqrt{11}-\sqrt{9})$ is $(\sqrt{5} - \sqrt{3} )$