Question

Решить , мне нужно расписать их решения, умоляю

Answer 1

1) $\frac{x-2b}{x^{2} -2bx} =\frac{x-2b}{x(x-2b)} =\frac{x-2b}{x}$

2) $\frac{3x+a}{(x+a)^{2} } +\frac{x+3a}{(x+a)^{2} } =\frac{3x+a+x+3a}{(x+a)^{2} } =\frac{4x+4a}{(x+a)^{2} } =\frac{4(x+a)}{(x+a)^{2} } =\frac{4}{x+a }$

3) $\frac{3x}{y} -\frac{x}{y} =\frac{3x-x}{y} =\frac{2x}{y}$

4) $\frac{1}{a^{2} } +\frac{a-2}{a} =\frac{1+a(a-2)}{a^{2} } =\frac{1+a^{2}-2a }{a^{2} }$

5) $5x+\frac{1}{x} =\frac{5x^{2} +1}{x}$

6) $\frac{3a^{2} }{b^{2} } *\frac{b^{3} }{a} =3ab$

7) $\frac{a^{2}-b^{2}}{a+3b} :\frac{ab+b^{2} }{2a+6b} =\frac{(a-b)(a+b)}{a+3b} *\frac{a+3b}{b(a+b)} =\frac{a-b}{b}$

8)

$\frac{x^{-10} *x^{3} }{x^{-5} } =x^{-10+3-(-5)} =x^{-10+3+5} =x^{-2} \\\\x=\frac{1}{3} \\\\(\frac{1}{3} )^{-2} =3^{2} =9$

9) $(\frac{y}{y-5} -2y):\frac{11-2y}{y-5} =\frac{y-2y(y-5)}{y-5}*\frac{y-5}{11-2y} =\frac{y-2y^{2}+10y }{11-2y} =\frac{-2y^{2} +11y}{-(2y-11)} =\frac{-y(2y-11)}{-(2y-11)} =y$