Question

41.2 41.3 помогите срочно пожалуйста

Answer 1

41.2
1)
$(\frac{2m+1}{2m-1} - \frac{2m-1}{2m+1} ) : \frac{4m}{10m-5} = \\ \\ =\frac{(2m+1)(2m+2)-(2m-1)(2m-1)}{(2m-1)(2m+1)} : \frac{4m}{5(2m-1)} = \\ \\ = \frac{(2m+1)^2-(2m-1)^2}{(2m-1)(2m+1)} * \frac{5(2m-1)}{4m} = \\ \\ = \frac{4m^2+2*2m*1+1^2-(4m^2-2*2m*1+1^2)}{(2m-1)(2m+1)} * \frac{5(2m-1)}{4m} = \\ \\ = \frac{4m^2+4m+1-4m^2+4m-1}{(2m-1)(2m+1)} * \frac{5(2m-1)}{4m} = \\ \\ = \frac{(2*4m)*5(2m-1)}{(2m-1)(2m+1)*4m} = \frac{2*5}{2m+1} = \frac{10}{2m+1}$

2)
$( \frac{y+3}{y^2+9} )*( \frac{y+3}{y-3} + \frac{y-3}{y+3} ) = \\ \\ = ( \frac{y+3}{y^2+9} )* \frac{(y+3)^2 + (y-3)^2}{(y-3)(y+3)} = \\ \\ = \frac{y+3}{y^2+9} * \frac{y^2+6y+9 +y^2-6y+9}{(y-3)(y+3)} = \\ \\ = \frac{y+3}{y^2+9} * \frac{2y^2+2*9}{(y-3)(y+3)} = \frac{y+3}{y^2+9} * \frac{2(y^2+9)}{(y-3)(y+3)} = \\ \\ = \frac{(y+3) * 2*(y^2+9)}{(y^2+9)*(y-3)(y+3)}= \frac{2}{y-3}$

41.3
1)
$\frac{n^2-9}{2n^2+1} * ( \frac{6n+1}{n-3} + \frac{6n-1}{n+3} )= \\ \\ = \frac{n^2-3^2}{2n^2+1} * \frac{(6n+1)(n+3) + (6n-1)(n-3)}{(n-3)(n+3)} = \\ \\ = \frac{(n-3)(n+3}{2n^2+1} * \frac{6n^2+18n+n+3 +6n^2-18n-n+3}{(n-3)(n+3)}= \\ \\ = \frac{(n-3)(n+3)}{2n^2+1} * \frac{12n^2+6}{(n-3)(n+3)}= \frac{(n-3)(n+3)}{2n^2+1} * \frac{6(2n^2+1)}{(n-3)(n+3)}= \\ \\ = \frac{6* (n-3)(n+3)*(2n^2+1)}{(n-3)(n+3)(2n^2+1)} = \frac{6}{1} =6$

2)
$(\frac{6x+y}{x-6y} + \frac{6x-y}{x+6y} ) : \frac{x^2+y^2}{x^2-36y^2} = \\ \\ = \frac{(6x+y)(x+6y) + (6x-y)(x-6y)}{(x-6y)(x+6y)} : \frac{x^2+y^2}{x^2-36y^2}= \\ \\ = = \frac{6x^2+36xy+xy+6y^2+6x^2-36xy -xy +6y^2}{x^2-(6y)^2} * \frac{x^2-36y^2}{x^2+y^2} = \\ \\ = \frac{12x^2+12y^2}{x^2-36y^2} * \frac{x^2-36y^2}{x^2+y^2}= \frac{12(x^2+y^2)(x^2-36y^2)}{(x^2+y^2)(x^2-36y^2)} = \\ \\ = \frac{12}{1} = 12$