Question 79

Determine the value of $$'x'$$ if $$x=\frac{(943+864)^{2}-(943-864)^{2}}{(1886\times1728)}$$

Solution

Expression : $$x=\frac{(943+864)^{2}-(943-864)^{2}}{(1886\times1728)}$$

Using, $$a^2-b^2=(a+b)(a-b)$$, where $$a=(943+864)$$ and $$b=(943-864)$$

= $$\frac{[(943+864)+(943-864)]\times[(943+864)-(943-864)]}{(1886\times1728)}$$

= $$\frac{(943+943)\times(864+864)}{(1886\times1728)}$$

= $$\frac{(1886)\times(1728)}{(1886\times1728)}=1$$

=> Ans - (A)

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