Expand: (i) (1001)^2 (ii) $\left(99\frac{1}{3}\right)^2$ (iii)(a/x+x/a)^2

Expand: (i) (1001)^2 (ii) $\left(99\frac{1}{3}\right)^2$ (iii)(a/x+x/a)^2 || Find The Squares of Each of The Following (i) 1001 (ii) $99\frac{1}{3}$ (iii) (a/x+x/a)^2 [S.Chand ICSE Class 9 Fundamentals of Mathematics Chapter 3 Exercise 3.1 Expansion and Factorisation Solution ]

(i) (1001)2

=(1000+1)2

= (1000)2 +2.(1000).1 +(1)2

= 1000000+2000+1

=1002001 [Ans]

(ii) $\left(99\frac{1}{3}\right)^2$

= $\left(99\frac{1}{3}\right)^2$

= $\left(99 + \frac{1}{3}\right)^2$

= $\left(100 – 1 + \frac{1}{3}\right)^2$

= $\left(100 + \frac{1}{3} – 1\right)^2$

= $\left(100 – \frac{2}{3}\right)^2$

=$ \left(100\right)^2 – 2.100\ldotp \frac{2}{3} + \left(\frac{2}{3}\right)^2$

= $10000 – \frac{400}{3} + \frac{4}{9}$

= $\frac{90000 – 1200 + 4}{9}$

= $\frac{90000 – 1196}{9}$

= $\frac{88804}{9}$

= $9867\frac{1}{9}$ [Ans]

(iii) $\left(\frac{{\mathrm{a}}}{{\mathrm{x}}} + \frac{{\mathrm{x}}}{{\mathrm{a}}}\right)^2$

= $\left(\frac{{\mathrm{a}}}{{\mathrm{x}}}\right)^2 + 2.\frac{{\mathrm{a}}}{{\mathrm{x}}}\ldotp \frac{{\mathrm{x}}}{{\mathrm{a}}} + \left(\frac{{\mathrm{x}}}{{\mathrm{a}}}\right)^2$

= $\frac{{\mathrm{a}}^2}{{\mathrm{x}}^2} + \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2} + 2$ [Ans]

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