Question

Пожалуйста!!!!! Помогите!Алгебра
x/(x²-3) + (x-8)/x – (72+3x-7x²)/(x³-9)=

8/(10-2q) + (5-q)/(q²+5q) + (q³+5q²-15q+25)/(q³-25q)=

(4-m)/(m²-4m) - (m-4)/(12+3m) – (48-8m+11m²)/(48m-3m³)=

Answer 1

2)
$frac{8}{10-2q} + frac{5-q}{q^2+5q} + frac{q^3+5q^2-15q+25}{q^3-25q}= \ \ = frac{8}{2(5-q)} + frac{5-q}{q(q+5)} + frac{q^3+5q^2-15q+25}{(q^2-25)q} =$
$=- frac{8}{2(q-5)} + frac{5-q}{q(q+5)} + frac{q^3+5q^2-15q+25}{(q-5)(q+5)q}= \ \ = frac{-8q(q+5)+(5-q)cdot2(q-5)+2(q^3+5q^2-15q+25)}{2(q-5)(q+5)q}= \ \ =frac{-8q^2-40q-2q^2+50+2q^3+10q^2-30q+50}{2(q-5)(q+5)q}= \ \ =frac{2q^3-70q+100}{2(q-5)(q+5)q}= \ \= frac{q^3-35q+50}{q(q-5)(q+5)}= \ \=frac{(q-5)(q^2+5q-10)}{q(q-5)(q+5)}= \ \=frac{q^2+5q-10}{q(q+5)}$

3)
$frac{4-m}{m^2-4m} - frac{m-4}{12+3m} - frac{(48-8m+11m^2)}{48m-3m^3} =$
$=frac{4-m}{m(m-4)} - frac{m-4}{3(m+4)} - frac{(48-8m+11m^2)}{3m(16-m^2)} =$
$=frac{3(4-m)(m+4)}{3m(m-4)(m+4)} - frac{3m(m-4)(m-4)}{3m(m+4)(m-4)} +frac{(48-8m+11m^2)}{3m(m-4)(m+4)} =$
$=frac{3(4-m)(m+4)-3m(m-4)^2+(48-8m+11m^2)}{3m(m-4)(m+4)} =$
$=frac{48-3m^2-3m(m-4)^2+48-8m+11m^2}{3m(m-4)(m+4)} =$
$frac{96-8m+8m^2-3m(m-4)^2}{3m(m-4)(m+4)} =frac{8(m^2-m+12)-3m(m-4)^2}{3m(m-4)(m+4)} =$
$frac{8(m-4)(m+3)-3m(m-4)^2}{3m(m-4)(m+4)} =frac{(m-4)(8m+24-3m(m-4))}{3m(m-4)(m+4)} =frac{8m+24-3m^2+12m}{3m(m+4)} =$
$frac{-3m^2+20m+24}{3m(m+4)}$