Question

HELP -_-
Может кто знает

Answer 1

$\frac{10a+b}{b ^{2} -4a ^{2} } + \frac{4a+2b}{b ^{2} +4ab+4a^2} = \frac{10a+b}{(b-2a)(b+2a)} + \frac{4a+2b}{(b+2a)(b+2a)} = \\ \\ = \frac{(10a+b)(b+2a)+(4a+2b)(b-2a)}{(b-2a)(b+2a)(b+2a)} = \frac{10ab+20a^2+b^2+2ab+4ab-8a^2+2b^2-4ab}{(b-2a)(b+2a)(b+2a)} = \\ \\ = \frac{12ab+12a^2+3b^2}{(b-2a)(b+2a)(b+2a)} = \frac{3(4ab+4a^2+b^2)}{(b-2a)(b+2a)(b+2a)} = \frac{3(b+2a)^2}{(b-2a)(b+2a)^2} = \frac{3}{b-2a} \\ \\ a=0.25;b=4.5$

$\frac{3}{4.5-2*0.25} = \frac{3}{4.5-0.5} = \frac{3}{4} = \frac{3*25}{4*25} = \frac{75}{100} =0.75$