(1+11+1100)(1+11+1100)−(1−11+1100)(1−11+1100)(1+11+1100)+(1−11+1100)=?(1+11+1100)(1+11+1100)−(1−11+1100)(1−11+1100)(1+11+1100)+(1−11+1100)=?\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)}} = ?" /> (1+11+1100)(1+11+1100)−(1−11+1100)(1−11+1100)(1+11+1100)+(1−11+1100)=?(1+11+1100)(1+11+1100)−(1−11+1100)(1−11+1100)(1+11+1100)+(1−11+1100)=?\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)}} = ?" /> (1+11+1100)(1+11+1100)−(1−11+1100)(1−11+1100)(1+11+1100)+(1−11+1100)=?(1+11+1100)(1+11+1100)−(1−11+1100)(1−11+1100)(1+11+1100)+(1−11+1100)=?\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)}} = ?" />

India’s No.1 Educational Platform For UPSC,PSC And All Competitive Exam

Download Our App

The simplified value of $\frac{(1 + \frac{1}{1 + \frac{1}{100}}) (1 + \frac{1}{1 + \frac{1}{100}}) - (1 - \frac{1}{1 + \frac{1}{100}}) (1 - \frac{1}{1 + \frac{1}{100}})}{(1 + \frac{1}{1 + \frac{1}{100}}) + (1 - \frac{1}{1 + \frac{1}{100}})} = ?$ $\frac{{\left( {1 + \frac{1}{{1 + \frac{1}{{100}}}}} \right)\left( {1 + \frac{1}{{1 + \frac{1}{{100}}}}} \right) - \left( {1 - \frac{1}{{1 + \frac{1}{{100}}}}} \right)\left( {1 - \frac{1}{{1 + \frac{1}{{100}}}}} \right)}}{{\left( {1 + \frac{1}{{1 + \frac{1}{{100}}}}} \right) + \left( {1 - \frac{1}{{1 + \frac{1}{{100}}}}} \right)}} = ?$

1100
2 $\frac{200}{101}$ $\frac{{200}}{{101}}$
3200
4 $\frac{202}{100}$ $\frac{{202}}{{100}}$

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given expression , \\ \frac{a^{2} - b^{2}}{a + b} = a - b \\ = (1 + \frac{1}{1 + \frac{1}{100}}) - (1 - \frac{1}{1 + \frac{1}{100}}) \\ = 2 \times \frac{1}{(101 / 100)} \\ = 2 \times \frac{100}{101} \\ = \frac{200}{101} \end{aligned}

$\eqalign{ & {\text{Given expression ,}} \cr & \frac{{{a^2} - {b^2}}}{{a + b}} = a - b \cr & = \left( {1 + \frac{1}{{1 + \frac{1}{{100}}}}} \right) - \left( {1 - \frac{1}{{1 + \frac{1}{{100}}}}} \right) \cr & = 2 \times \frac{1}{{\left( {101/100} \right)}} \cr & = 2 \times \frac{{100}}{{101}} \cr & = \frac{{200}}{{101}} \cr}$

Workspace Report

Post your Comments

", "text": "The simplified value of

\frac{(1 + \frac{1}{1 + \frac{1}{100}}) (1 + \frac{1}{1 + \frac{1}{100}}) - (1 - \frac{1}{1 + \frac{1}{100}}) (1 - \frac{1}{1 + \frac{1}{100}})}{(1 + \frac{1}{1 + \frac{1}{100}}) + (1 - \frac{1}{1 + \frac{1}{100}})} = ?

$\frac{{\left( {1 + \frac{1}{{1 + \frac{1}{{100}}}}} \right)\left( {1 + \frac{1}{{1 + \frac{1}{{100}}}}} \right) - \left( {1 - \frac{1}{{1 + \frac{1}{{100}}}}} \right)\left( {1 - \frac{1}{{1 + \frac{1}{{100}}}}} \right)}}{{\left( {1 + \frac{1}{{1 + \frac{1}{{100}}}}} \right) + \left( {1 - \frac{1}{{1 + \frac{1}{{100}}}}} \right)}} = ?$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Given expression , \\ \frac{a^{2} - b^{2}}{a + b} = a - b \\ = (1 + \frac{1}{1 + \frac{1}{100}}) - (1 - \frac{1}{1 + \frac{1}{100}}) \\ = 2 \times \frac{1}{(101 / 100)} \\ = 2 \times \frac{100}{101} \\ = \frac{200}{101} \end{aligned}

$\eqalign{ & {\text{Given expression ,}} \cr & \frac{{{a^2} - {b^2}}}{{a + b}} = a - b \cr & = \left( {1 + \frac{1}{{1 + \frac{1}{{100}}}}} \right) - \left( {1 - \frac{1}{{1 + \frac{1}{{100}}}}} \right) \cr & = 2 \times \frac{1}{{\left( {101/100} \right)}} \cr & = 2 \times \frac{{100}}{{101}} \cr & = \frac{{200}}{{101}} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Given expression , \\ \frac{a^{2} - b^{2}}{a + b} = a - b \\ = (1 + \frac{1}{1 + \frac{1}{100}}) - (1 - \frac{1}{1 + \frac{1}{100}}) \\ = 2 \times \frac{1}{(101 / 100)} \\ = 2 \times \frac{100}{101} \\ = \frac{200}{101} \end{aligned}

$\eqalign{ & {\text{Given expression ,}} \cr & \frac{{{a^2} - {b^2}}}{{a + b}} = a - b \cr & = \left( {1 + \frac{1}{{1 + \frac{1}{{100}}}}} \right) - \left( {1 - \frac{1}{{1 + \frac{1}{{100}}}}} \right) \cr & = 2 \times \frac{1}{{\left( {101/100} \right)}} \cr & = 2 \times \frac{{100}}{{101}} \cr & = \frac{{200}}{{101}} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

Test

Classes

E-Book