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anuritha:
which question friend
the sums are easy but what question
find the value a and b
first question
and i am answering for it
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2
if that is first question answer is here
√2+1/√2-1-√2-1/√2+1
let we take the LCM therefore we get ,
[√2+1]^2-[√2-1]^2/[√2-1][√2+1]
we know that
[a+b][a-b]=a^2-b^2 and[ a+b]^2=a^2+b^2+2ab
so
[(√2)^2+1+2√2]-[(√2)+1-2√2]/(√2)^2-1
[3+2√2]-[3-2√2]/2-1
3+2√2-3+2√2/1
+4√2 is the answer
therefore a=0 and b =4
√2+1/√2-1-√2-1/√2+1
let we take the LCM therefore we get ,
[√2+1]^2-[√2-1]^2/[√2-1][√2+1]
we know that
[a+b][a-b]=a^2-b^2 and[ a+b]^2=a^2+b^2+2ab
so
[(√2)^2+1+2√2]-[(√2)+1-2√2]/(√2)^2-1
[3+2√2]-[3-2√2]/2-1
3+2√2-3+2√2/1
+4√2 is the answer
therefore a=0 and b =4
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great job
thanks plz mark it as brainliest if u understand
Thank u friend rittwotson
ur welcome dear its mah pleasure.
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