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If a = (√3 + √2)-3 and b = (√3 - √2)-3, find (a + 1)-1 + (b + 1)-1 . |
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Answer» a = (\(\sqrt3 + \sqrt2\))-3 and b = (\(\sqrt3 - \sqrt2\))-3 Now (a + 1)-1 + (b + 1)-1 = \(\frac1{a+1}+\frac1{b+1}=\frac{a+b+2}{(a+1)(b+1)}\) \(\therefore\) (a + 1)-1 + (b + 1)-1 = \(\frac{a+b+2}{ab+a+b+1}\)-----(1) Now, ab = (\(\sqrt3 + \sqrt2\))-3 (\(\sqrt3 - \sqrt2\))-3 = ((\(\sqrt3 + \sqrt2\))(\(\sqrt3 - \sqrt2\)))-3 (\(\because\) ambm = (ab)m) = ((√3)2 - (√2)2)-3(\(\because\) (a + b) (a - b) = a2 - b2) = (3 - 2)-3 = 1-3 = \(\frac1{1^3}=1\) \(\therefore\) (a + 1)-1 + (b + 1)-1 = \(\frac{a+b+2}{1 + a + b+1}\) (From 1) = \(\frac{a+b+2}{a + b + 2}=1\) Hence, (a + 1)-1 + (b + 1)-1 = 1 |
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