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How to compare the reciprocal sum of the arithmetic square root of 7 with -3?
Hello!

The reciprocal of the arithmetic square root of 7 =-√7

√7 & lt; √9=3

-√7 & gt; -3

So: the reciprocal of the arithmetic square root of 7 >; -3

How to compare the radical sign 1 1 plus 1 divided by 2 and 2: apply the idea of backward deduction;

1 1 & gt; nine

√ 1 1 & gt; √9=3

√ 1 1+ 1 & gt; 3+ 1=4

(√ 1 1+ 1)/2 & gt; 4/2=2

So: root number 1 1 plus 1 divided by 2 >; 2

I hope it helps you!