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Simplify the following:-(1) (✓11+✓7) (✓11-✓7)
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hope this helps u
we know that
[a-b][a+b] = a²-b²
(✓11+✓7) (✓11-✓7) which is in the format of [a+b][a-b] = a²-b²
so (✓11+✓7) (✓11-✓7)=(√11)²-(√7)² here root and square get cancelled
= 11-7
(✓11+✓7) (✓11-✓7) = 6
sub the value of (✓11+✓7) (✓11-✓7) =6
[1] [(✓11+✓7) (✓11-✓7)]
1*6
=6 is the answer
plz mark as brainlist answer
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