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Difficulty:
55%
(hard)
Question Stats:
61% (01:50) correct
39%
(02:27)
wrong
based on 161
sessions
History
Date
Time
Result
Not Attempted Yet
\(\frac{1}{√9 − √7} − \frac{1}{√7 − √5 }+ \frac{1}{√5 − √3} − \frac{1}{√3 − √1}\) =
A) \(\frac{1}{2}\)
B) 1
C) 2
D) 2√2
E) \( 4 − √7 + √5 − √3\)
A) \(\frac{1}{2}\)
B) 1
C) 2
D) 2√2
E) \( 4 − √7 + √5 − √3\)
12 Apr 2021, 08:19
Kudos
Bookmarks
sjuniv32
\(\frac{1}{√9 − √7} − \frac{1}{√7 − √5 }+ \frac{1}{√5 − √3} − \frac{1}{√5 − √3}\) =
A) \(\frac{1}{2}\)
B) 1
C) 2
D) 2√2
E) \( 4 − √7 + √5 − √3\)
A) \(\frac{1}{2}\)
B) 1
C) 2
D) 2√2
E) \( 4 − √7 + √5 − √3\)
The question is faulty. it should be as below
\(\frac{1}{√9 − √7} − \frac{1}{√7 − √5 }+ \frac{1}{√5 − √3}\) -\(\frac{1}{√3 − √1} \)
Let us remove the square root from the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator.
Also, remember in all such questions: \(a^2-b^2=(a-b)(a+b)\)
\(\frac{1}{√9 − √7} − \frac{1}{√7 − √5 }+ \frac{1}{√5 − √3} − \frac{1}{√3 − √1}\) =
\(\frac{1*(√9 + √7)}{(√9 − √7)(√9 + √7)} − \frac{1*(√7 + √5)}{(√7 − √5)(√7 + √5) }+ \frac{1*(√5 + √3)}{(√5 - √3)(√5 + √3)} − \frac{(√3 + √1)}{(√3 -√1)(√3 + √1)}\)
\(\frac{(√9 + √7)}{9-7} − \frac{(√7 + √5)}{7-5 }+ \frac{(√5 + √3)}{5-3} − \frac{(√3 + √1)}{3-1}\)
\(\frac{(√9 + √7)}{2} − \frac{(√7 + √5)}{2 }+ \frac{(√5 + √3)}{2} − \frac{(√3 + √1)}{2}\)
\(\frac{(√9 + √7)-(√7 + √5)+(√5 + √3)-(√3 + √1)}{2}=\frac{√9 - √1}{2}=\frac{3-1}{2}=\frac{2}{2}=1\)
B
17 Nov 2024, 05:05
Kudos
Bookmarks
chetan2u
sjuniv32
\(\frac{1}{√9 − √7} − \frac{1}{√7 − √5 }+ \frac{1}{√5 − √3} − \frac{1}{√5 − √3}\) =
A) \(\frac{1}{2}\)
B) 1
C) 2
D) 2√2
E) \( 4 − √7 + √5 − √3\)
A) \(\frac{1}{2}\)
B) 1
C) 2
D) 2√2
E) \( 4 − √7 + √5 − √3\)
The question is faulty. it should be as below
\(\frac{1}{√9 − √7} − \frac{1}{√7 − √5 }+ \frac{1}{√5 − √3}\) -\(\frac{1}{√3 − √1} \)
Let us remove the square root from the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator.
Also, remember in all such questions: \(a^2-b^2=(a-b)(a+b)\)
\(\frac{1}{√9 − √7} − \frac{1}{√7 − √5 }+ \frac{1}{√5 − √3} − \frac{1}{√3 − √1}\) =
\(\frac{1*(√9 + √7)}{(√9 − √7)(√9 + √7)} − \frac{1*(√7 + √5)}{(√7 − √5)(√7 + √5) }+ \frac{1*(√5 + √3)}{(√5 - √3)(√5 + √3)} − \frac{(√3 + √1)}{(√3 -√1)(√3 + √1)}\)
\(\frac{(√9 + √7)}{9-7} − \frac{(√7 + √5)}{7-5 }+ \frac{(√5 + √3)}{5-3} − \frac{(√3 + √1)}{3-1}\)
\(\frac{(√9 + √7)}{2} − \frac{(√7 + √5)}{2 }+ \frac{(√5 + √3)}{2} − \frac{(√3 + √1)}{2}\)
\(\frac{(√9 + √7)-(√7 + √5)+(√5 + √3)-(√3 + √1)}{2}=\frac{√9 - √1}{2}=\frac{3-1}{2}=\frac{2}{2}=1\)
B
