Question

Желательно быстро! решите хоть парочку, большое ) 2 3 2m+15 + + 2m+3 2m-3 4m^2-9 5 1 24 + - 2m+2 10m-10 m^2-1 1 2 1 - + m+3 m+2 m+1 x^2-7x+12 =0 x-4 47-x(3x+4)=2x-3(x-5)^2

Answer 1

$\frac{2}{2m+3} + \frac{3}{2m-3} + \frac{2m+15}{4m^{2}-9} = \frac{2}{2m+3} + \frac{3}{2m-3} + \frac{2m+15}{(2m-3)(2m+3)} = \\ \\ \frac{2(2m-3)+3(2m+3)+2m+15}{(2m-3)(2m+3)} = \frac{4m-6+6m+9+2m+15}{(2m-3)(2m+3)} = \\ \\ [tex] \frac{12m+18}{(2m-3)(2m+3)}= \frac{6(2m+3)}{(2m-3)(2m+3)} = \frac{6}{2m-3} \\ \\ \\ [tex] \frac{5}{2m+2} + \frac{1}{10m-10}- \frac{24}{ m^{2} -1} = \frac{5}{2(m+1)} + \frac{1}{10(m-1)}- \frac{24}{(m-1)(m+1)} = \\ \\ \frac{5*5(m-1)+m+1-24*10}{10(m-1)(m+1)} = \frac{25m-25+m+1-240}{10(m-1)(m+1)} = \frac{26m - 264}{10(m-1)(m+1)} ........[tex]= \frac{2(13m-132)}{10(m-1)(m+1)} = \frac{13m-132}{5(m-1)(m+1)}$

.
$\frac{1}{m+3}- \frac{2}{m+2} + \frac{1}{m+1}= \frac{(m+2)(m+1)-2(m+3)(m+1)+(m+3)(m+2)}{(m+3)(m+2)(m+1)} = \\ \\ \frac{ m^{2}+3m+2-2m^{2}-8m-6+m^{2} +5m+6 }{(m+3)(m+2)(m+1)}= \frac{2}{(m+3)(m+2)(m+1)}$

.
$\frac{ x^{2} -7x+12}{x-4} =0 \\ \\ x \neq 4$

x² - 7x+12=0
D=7² -4*12=1  (±1)
x1=(7+1)/2=4 - не подходит решению
х2=(7-1)/2=3

47-х(3х+4)=2х-3(х-5)²
47 - 3х² -4х=2х - 3х²+30х -75
36х=122
х=122:36
х=3 14/36=3 7/18