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What is the value of the following expression:
\(\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{120} + \sqrt{121}}\)
A. \(1\)
B. \(9\)
C. \(10\)
D. \(11\)
E. \(12\)
\(\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{120} + \sqrt{121}}\)
A. \(1\)
B. \(9\)
C. \(10\)
D. \(11\)
E. \(12\)
26 Nov 2021, 04:13
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Official Solution:
What is the value of the following expression:
\(\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{120} + \sqrt{121}}\)
A. \(1\)
B. \(9\)
C. \(10\)
D. \(11\)
E. \(12\)
Rationalize by multiplying the denominator and the numerator of each fraction by \(\sqrt{x}-\sqrt{x-1}\) (this algebraic manipulation is called rationalization and is performed to eliminate irrational expression in the denominator). As in the denominator we have \(\sqrt{x}+\sqrt{x-1}\) we'll get \((\sqrt{x}+\sqrt{x-1})(\sqrt{x}-\sqrt{x-1})=x-(x-1)=1\).
We'll be left with the following:
\((\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+....+(\sqrt{121}-\sqrt{120})=-\sqrt{1}+\sqrt{121}=-1+11=10\)
Answer: C
What is the value of the following expression:
\(\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{120} + \sqrt{121}}\)
A. \(1\)
B. \(9\)
C. \(10\)
D. \(11\)
E. \(12\)
Rationalize by multiplying the denominator and the numerator of each fraction by \(\sqrt{x}-\sqrt{x-1}\) (this algebraic manipulation is called rationalization and is performed to eliminate irrational expression in the denominator). As in the denominator we have \(\sqrt{x}+\sqrt{x-1}\) we'll get \((\sqrt{x}+\sqrt{x-1})(\sqrt{x}-\sqrt{x-1})=x-(x-1)=1\).
We'll be left with the following:
\((\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+....+(\sqrt{121}-\sqrt{120})=-\sqrt{1}+\sqrt{121}=-1+11=10\)
Answer: C
28 Apr 2023, 12:33
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Bunuel
Official Solution:
What is the value of the following expression:
\(\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{120} + \sqrt{121}}\)
A. \(1\)
B. \(9\)
C. \(10\)
D. \(11\)
E. \(12\)
Rationalize by multiplying the denominator and the numerator of each fraction by \(\sqrt{x}-\sqrt{x-1}\) (this algebraic manipulation is called rationalization and is performed to eliminate irrational expression in the denominator). As in the denominator we have \(\sqrt{x}+\sqrt{x-1}\) we'll get \((\sqrt{x}+\sqrt{x-1})(\sqrt{x}-\sqrt{x-1})=x-(x-1)=1\).
We'll be left with the following:
\((\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+....+(\sqrt{121}-\sqrt{120})=-\sqrt{1}+\sqrt{121}=-1+11=10\)
Answer: C
What is the value of the following expression:
\(\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{120} + \sqrt{121}}\)
A. \(1\)
B. \(9\)
C. \(10\)
D. \(11\)
E. \(12\)
Rationalize by multiplying the denominator and the numerator of each fraction by \(\sqrt{x}-\sqrt{x-1}\) (this algebraic manipulation is called rationalization and is performed to eliminate irrational expression in the denominator). As in the denominator we have \(\sqrt{x}+\sqrt{x-1}\) we'll get \((\sqrt{x}+\sqrt{x-1})(\sqrt{x}-\sqrt{x-1})=x-(x-1)=1\).
We'll be left with the following:
\((\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+....+(\sqrt{121}-\sqrt{120})=-\sqrt{1}+\sqrt{121}=-1+11=10\)
Answer: C
Why do we multiply by \(\sqrt{x}-\sqrt{x-1}\) when rationalizing? Can you explain how this step is done?
