Question

Пожалуйста отдаю все баллы​

Answer 1

8. (Г)

$\frac{4-m}{m-3} + \frac{2m-5}{3-m} = \frac{4-m}{m-3} + \frac{2m-5}{-(m-3)} = \frac{4-m}{m-3} - \frac{2m - 5}{m-3} = \frac{4-m-(2m-5)}{m-3} = \frac{4-m-2m+5}{m-3} = \frac{9-3m}{m-3} = -\frac{3m - 9}{m-3} = -\frac{3}{1} = -3$

9. (В)

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

10. (Г)

$\frac{2m+1}{3m-2} - \frac{3m^{2}+m-2 }{9m^{2} - 12m + 4 } = \frac{2m+1}{3m-2} -\frac{(3m-2)(m+1)}{(3m-2)^{2} } = \frac{2m+1}{3m-2} - \frac{m+1}{3m-2} = \frac{2m+1-(m+1)}{3m-2} = \frac{2m+1-m-1}{3m-2} = \frac{m}{3m-2}$

11. (Б)

$\frac{a-12}{a^{2}+4a }-\frac{a-4}{a}+\frac{a}{a+4}=\frac{a-12}{a(a+4) }-\frac{a-4}{a}+\frac{a}{a+4}=\frac{a-12}{a(a+4) }-\frac{(a-4)(a+4)}{a(a+4)}+\frac{a}{a+4}=\frac{a-12-a^{2}-4a+4a+16}{a(a+4)} + \frac{a}{a+4} = \frac{a+4-a^{2}}{a(a+4)} + \frac{a}{a+4}=\frac{a+4-a{2} }{a(a+4)} + \frac{a^{2} }{a(a+4)} = \frac{a+4-a^{2}+a^{2} }{a(a+4)} = \frac{a+4}{a(a+4)} = \frac{1}{a}$

12.(В)

$y=\frac{x^{2}-4x+4 }{x-2}; x^{2}-2*2x+2^{2}=(x-2)^{2}\\y=\frac{(x-2)^{2} }{x-2}; x-2\neq 0; x\neq 2\\\left \{ {{y=x-2} \atop {x \neq 2}} \right \\y\neq 2-2 = 0$