Question

If $\Rightarrow$and $1 = A + C$then.

• A. $\therefore$
• B. $A = \frac{3}{8},B = \frac{3}{4},C = \frac{5}{8}$
• C. $\therefore$
• D. $\frac{x^{2} + 1}{(x^{2} + 4)(x - 2)} = \frac{\frac{3}{8}x + \frac{3}{4}}{x^{2} + 4} + \frac{\frac{5}{8}}{x - 2}$

Answer 1

Answer: (D)

$\frac{x^{2} + 1}{(x^{2} + 4)(x - 2)} = \frac{\frac{3}{8}x + \frac{3}{4}}{x^{2} + 4} + \frac{\frac{5}{8}}{x - 2}$

Answer 2

$x^{2} + x - 20 = 0$

= $\Rightarrow$

= $(x - 4)(x + 5) = 0$

= $\Rightarrow$

= $x = 4, - 5$

= $x = 4$

= $\log_{4}18 = \frac{1}{2}\log_{2}(3^{2}.2) = \frac{1}{2}(2\log_{2}3 + \log_{2}2)$, $= \log_{2}3 + \frac{1}{2},$ $(0.05)^{\log_{\sqrt{20}}(0.1 + 0.01 + ......)} = \left( \frac{1}{20} \right)^{2\log_{20}\left( \frac{0.1}{1 - 0.1} \right)}$.