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(x + 1)/(x - 1) - (x - 1)/(x + 1) = {(x + 1)^{2} - (x - 1)^{2}} / {(x - 1) (x + 1)}
= (x^{2} + 1 + 2x) - (x^{2} + 1 - 2x) / (x^{2} - 1)
= 4x / (x^{2} - 1) = 4(√3 + √2) / {(√3 + √2)^{2} - 1}
(x + 1)/(x - 1) - (x - 1)/(x + 1) = {(x + 1)^{2} - (x - 1)^{2}} / {(x - 1) (x + 1)}
= (x^{2} + 1 + 2x) - (x^{2} + 1 - 2x) / (x^{2} - 1)
= 4x / (x^{2} - 1) = 4(√3 + √2) / {(√3 + √2)^{2} - 1}
= 4(√3 + √2) / (3 + 2 + 2√6 - 1) = 4(√3 + √2) / (4 + 2 √6)
= 4(√3 + √2) / 2 √2 (√3 + √2)
= √2
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∛8 and ∛1000 = 8^{1/3} and 1000^{1/3}
∛8 and ∛1000 = 8^{1/3} and 1000^{1/3}
Since 1000 > 8
∴ ∛1000 > ∛8
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√81 and √1 = 81^{1/2} and 1^{1/2}
√81 and √1 = 81^{1/2} and 1^{1/2}
Since, 1 < 81
∴ √1 < √81
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Let the number be x.
According to the question,
x^{2} + (14)^{3} = 4425
Let the number be x.
According to the question,
x^{2} + (14)^{3} = 4425
⇒ x^{2} = 4425 - (14)^{3}
⇒ x^{2} = 4425 - 2744
⇒ x^{2} = 1681
∴ x = √1681 = 41
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∜10 and ∛8 = 10^{1/4} and 8^{1/3}
Since, LCM of 4 and 3 is 12.
∴ 10^{1/4} = 10^{3/12} = (10^{3})^{1/12} = (1000)^{1/12} = ^{12}√1000
and 8^{1/3} = 8^{4/12} = (8^{4})^{1/12} (4096)^{1/2} = ^{12}√4096
∴ 4096 > 1000
∜10 and ∛8 = 10^{1/4} and 8^{1/3}
Since, LCM of 4 and 3 is 12.
∴ 10^{1/4} = 10^{3/12} = (10^{3})^{1/12} = (1000)^{1/12} = ^{12}√1000
and 8^{1/3} = 8^{4/12} = (8^{4})^{1/12} (4096)^{1/2} = ^{12}√4096
∴ 4096 > 1000
Hence, 8^{1/3} >10^{1/4}
∴ ∛8 > ∜10
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