Question

номер 41.5
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Answer 1

41.5
1.
$\frac{a^2 - 25}{a+3} * \frac{1}{a^2+5a} - \frac{a+5}{a^2-3a} = \frac{a^2 - 5^2}{a+3} * \frac{1}{a(a+5)} - \frac{a+5}{a(a-3)} = \\ \\ = \frac{(a-5)(a+5)}{a+3} * \frac{1}{a(a+5)} - \frac{a+5}{a(a-3)} = \frac{(a-5)(a+5)*1}{(a+3) * a(a+5)} - \frac{a+5}{a(a-3)} = \\ \\ = \frac{(a-5)*1*1}{(a+3)*a*1} - \frac{a+5}{a(a-3)} = \frac{a-5}{a(a+3)} - \frac{a+5}{a(a-3)} = \frac{(a-5)(a-3)-(a+5)(a+3)}{a(a+3)(a-3)} =$
$= \frac{a^2-3a-5a+15 - (a^2+3a+5a+15)}{a(a+3)(a-3)} = \frac{a^2-8a+15-a^2-8a-15}{a(a+3)(a-3)} = \frac{-16a}{a(a+3)(a-3)} = \\ \\ = \frac{-16}{(a+3)(a-3)} = - \frac{16}{a^2-3^2} = - \frac{16}{a^2 - 9} \\ \\ 1) a = 2 \\ \\ - \frac{16}{2^2 - 9} = - \frac{16}{4 - 9} = - \frac{16}{-5} = - ( - 3.2) = 3.2 \\ \\ 2) a = - 4 \\ \\ - \frac{16}{(-4)^2 - 9} = - \frac{16}{16 - 9} = - \frac{16}{7} = - 2 \frac{2}{7}$

2.
$\frac{1 - 2x}{2x+1} + \frac{x^2+3x}{4x^2 - 1} : \frac{3+x}{4x+2} = \frac{1 - 2x}{2x+1} + \frac{x(x+3)}{(2x)^2 - 1^2} : \frac{3+x}{2(2x+1)} = \\ \\ = \frac{1 - 2x}{2x+1} + \frac{x(x+3)}{(2x-1)(2x+1)} * \frac{2(2x+1)}{x+3} =\frac{1 - 2x}{2x+1} + \frac{x(x+3)*2(2x+1)}{(2x-1)(2x+1)(x+3)} = \\ \\ = \frac{1 - 2x}{2x+1} + \frac{x*1*2*1}{(2x-1)*1*1} = \frac{1 - 2x}{2x+1} + \frac{2x}{2x-1} = \frac{(1-2x)(2x-1) +2x(2x+1)}{(2x+1)(2x-1)} =$
$= \frac{2x-1-4x^2+2x+4x^2+2x}{(2x+1)(2x-1)} = \frac{6x-1}{(2x)^2-1^2} = \frac{6x-1}{4x^2 - 1} \\ \\ 1) x = - 1 \\ \\ \frac{6*(-1)-1}{4*(-1)^2 - 1}= \frac{- 6 - 1}{4*1 - 1} = \frac{-7}{3} = - 2 \frac{1}{3} \\ \\ 2) x = -2.5 \\ \\ \frac{6*(-2.5)-1}{4*(-2.5)^2 - 1}= \frac{- 15 - 1}{4 *6.25 - 1} = \frac{-16}{25 - 1} = \frac{-16}{24} = - \frac{2}{3}$