Question

Помогите хоть кто нибудь срочноооооооооо даю 10 баллов​

Answer 1

1.

$3x^{-2}y^{3} (-5x^{-3} y^{4} )^{-3}=3x^{-2}y^{3} *125x^{9}y^{-12} =375x^{7}y^{-9}$

2.

$(\frac{x-9}{x} -\frac{y-9}{y})*\frac{xy}{x^{2}-y^{2}}=\frac{y(x-9)-x(y-9)}{xy} *\frac{xy}{(x-y)(x+y)} =\frac{xy-9y-xy+9x}{xy(x-y)(x+y)} =\\ \\ \frac{0}{xy(x-y)(x+y)} =0$

3.

$....................\\ \\ \frac{5x^{2}+8 }{(x+4)(x-4)}-\frac{2x-1}{x+4} +\frac{3x-1}{4-x} =0\\ \\ \frac{5x^{2}+8 }{(x+4)(x-4)}-\frac{2x-1}{x+4} -\frac{3x-1}{x-4} =0\\ \\ 5x^{2} +8-(2x-1)(x-4)-(3x-1)(x+4)=0\\ \\ 5x^{2} +8-2x^{2} +4x+x-4-3x^{2} -12x+x+4=0\\ \\ 8-6x=0\\ \\ 6x=8\\ \\ x=8:6\\ \\x=1\frac{1}{3}$

4.

$.........=(\frac{8b}{b+7} -\frac{15b}{(b+7)(b+7)} )*\frac{(b-7)(b+7)}{8b+41} +\frac{7(b-7)}{b+7} =\\ \\ \frac{8b(b+7)-15b}{(b+7)(b+7)} *\frac{(b-7)(b+7)}{8b+41} +\frac{7(b-7)}{b+7} =\frac{8b^{2}+56b-15b }{b+7} *\frac{b-7}{8b+41} +\frac{7(b-7)}{b+7} =\\ \\ \frac{8b^{2}+41b }{b+7}* \frac{b-7}{8b+41} +\frac{7(b-7)}{b+7} =\frac{b(8b+41)}{b+7} *\frac{b-7}{8b+41} +\frac{7(b-7)}{b+7} =\\ \\ \frac{b(b-7)}{b+7} +\frac{7(b-7)}{b+7} =\frac{b^{2}-7b+7b-49 }{b+7} =\frac{b^{2}-49 }{b+7} =\frac{(b-7)(b+7)}{b+7} =b-7$