Question

b/a2-b2+b/a2+ab при а=1.1 b=0.9

Answer 1

$\frac{b}{ a^{2} - b^{2} } + \frac{b}{ a^{2} +ab} = \frac{b}{ (a - b)(a+b)} + \frac{b}{ a(a +b)} =$
$= \frac{ab}{ a(a - b)(a+b)} + \frac{b(a - b)}{ a(a +b)(a - b)} =$
$= \frac{ab}{ a(a - b)(a+b)} + \frac{ba - b^2}{ a(a +b)(a - b)} =$
$= \frac{ab+ab-b^2}{ a(a - b)(a+b)} = \frac{2ab-b^2}{ a(a - b)(a+b)}$

при а=1.1 b=0.9
$\frac{2*1,1*0,9-0,9^2}{ 1,1(1,1 - 0,9)(1,1+0,9)} = \frac{1,98-0,81}{ 1,1*0,2*2} =$
$= \frac{1,17}{ 0,44} = \frac{117}{44}$