If \(\left( {x + \frac{1}{x}} \right){ \text{ = 2,}}\) then \(\left( {x - \frac{1}{x}} \right)\) is equal to = ?
Quiz Recommendation System API Link - https://fresherbell-quiz-api.herokuapp.com/fresherbell_quiz_api
# | Quiz |
---|---|
1
Discuss
|
The value of \({ \text{3}}\frac{1}{2} - \left[ {2\frac{1}{4} \div \left\{ {1\frac{1}{4} - \frac{1}{2}\left( {1\frac{1}{2} - \frac{1}{3} - \frac{1}{6}} \right)} \right\}} \right]\) = ?
Solution |
2
Discuss
|
If \( \left( {x + \frac{1}{x}} \right){ \text{ = }}\sqrt {13} { \text{,}} \) then the value of \(\left( {{x^3} - \frac{1}{{{x^3}}}} \right)\) is = ?
Solution |
3
Discuss
|
If \(\frac{a}{b}{ \text{ + }}\frac{b}{a}{ \text{ = 2,}}\) then the value of (a - b) is = ?
Solution |
4
Discuss
|
\(\frac{{225}}{{836}} \times \frac{{152}}{{245}} \div 1\frac{{43}}{{77}} = ?\)
Solution |
5
Discuss
|
\(\left( {x + \frac{1}{x}} \right)\left( {x - \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} - 1} \right)\left( {{x^2} + \frac{1}{{{x^2}}} + 1} \right)\) is equal to ?
Solution |
6
Discuss
|
(755.0% of 523.0) / 777.0 = x
Solution |
7
Discuss
|
There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is:
Solution |
8
Discuss
|
\(\frac{{{{\left( {469 + 174} \right)}^2} - {{\left( {469 - 174} \right)}^2}}}{{469 \times 174}} \) = ?
Solution |
9
Discuss
|
The cost of 5 pendants and 8 chains is Rs. 145785. What would be the cost of 15 pendants and 24 chains ?
Solution |
10
Discuss
|
The simplified value of \(\frac{{\left( {1 + \frac{1}{{1 + \frac{1}{{100}}}}} \right)\left( {1 + \frac{1}{{1 + \frac{1}{{100}}}}} \right) - \left( {1 - \frac{1}{{1 + \frac{1}{{100}}}}} \right)\left( {1 - \frac{1}{{1 + \frac{1}{{100}}}}} \right)}}{{\left( {1 + \frac{1}{{1 + \frac{1}{{100}}}}} \right) + \left( {1 - \frac{1}{{1 + \frac{1}{{100}}}}} \right)}} = ?\)
Solution |
# | Quiz |
Copyright © 2020 Inovatik - All rights reserved