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Question 41
If $$a = 0.25, b = - 0.05$$ and $$c = 0.5$$, the value of $$\frac{a^2 - b^2 - c^2 - 2bc}{a^2 + b^2 - 2ab - c^2}$$ is
Solution
We have $$\frac{a^2-b^2-c^2-2bc}{a^2+b^2-c^2-2ab}=\frac{a^2-\left(b+c\right)^2}{\left(a-b\right)^2-c^2}=\frac{\left(a+b+c\right)\left(a-b-c\right)}{\left(a-b+c\right)\left(a-b-c\right)}$$
Substituting the values of a,b and c we get
$$\frac{\left(a+b+c\right)\left(a-b-c\right)}{\left(a-b+c\right)\left(a-b-c\right)}=\frac{0.7\times\ \left(-0.2\right)}{\left(0.35\right)\left(-0.2\right)}=2$$
