Question

Помогите пожалуйста,зарание спасибо!

Answer 1

Ответ:

$\displaystyle 1)\ \ \frac{2b+5}{b-4}-\frac{6b-1}{3b+4}=\frac{6b^2+8b+15b+20-(6b^2-24b-b+4)}{(b-4)(3b+4)}=\\\\\\=\frac{48b+16}{(b-4)(3b+4)}=\frac{16(3b+1)}{(b-4)(3b+4)}\\\\\\2)\ \ \frac{m-1}{m^2-9}+\frac{2}{3m-9}=\frac{m-1}{(m-3)(m+3)}+\frac{2}{3(m-3)}=\\\\\\=\frac{3m-3+2(m+3)}{3(m-3)(m+3)}=\frac{5m+3}{3\, (m-3)(m+3)}=\frac{5m+3}{3\, (m^2-9)}$

$3)\ \ \displaystyle \frac{2x+16}{10-2x}\cdot \frac{25-x^2}{x^2-64}=\frac{2\cdot (x+8)\cdot (5-x)(x+5)}{2\cdot (5-x)\cdot (x-8)(x+8)}=\frac{x+5}{x-8}$

Answer 2

Ответ:

$\frac{2b + 5}{b - 4} - \frac{6b - 1}{3b + 4} \\ \\ \frac{(3b + 4) \times (2b + 5) - (b - 4) \times (6b - 1)}{(b - 4) \times (3b + 4)} \\ \\ \frac{6b {}^{2} + 15b + 8b + 20 - (6b {}^{2} - b - 24b + 4) }{3b {}^{2} + 4b - 12b - 16} \\ \\ \frac{6 {b}^{2} + 15b + 8b + 20 - (6b {}^{2} - 25b + 4)}{3 {b}^{2} - 8b - 16} \\ \\ \frac{6b {}^{2} + 15b + 8b + 20 - 6b {}^{2} + 25b - 4 }{3b {}^{2} - 8b - 16 } \\ \\ \boxed{ \frac{48b + 16}{3b {}^{2} - 8b - 16 } }$

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$\frac{m - 1}{m {}^{2} - 9 } + \frac{2}{3m - 9} \\ \\ \frac{m - 1}{(m - 3) \times (m + 3)} + \frac{2}{3 \times (m - 3)} \\ \\ \frac{3 \times (m - 1) + 2 \times (m + 3)}{3 \times (m + 3) \times (m - 3)} \\ \\ \frac{3m - 3 + 2m + 6}{3 \times (m {}^{2} - 9) } \\ \\ \boxed{ \frac{5m + 3}{3 {m}^{2} - 27 } }$

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$\frac{2x + 16}{10 - 2x} \times \frac{25 - x {}^{2} }{x {}^{2} - 64} \\ \\ \frac{2 \times (x + 8)}{2 \times (5 - x)} \times \frac{(5 - x) \times (5 + x)}{(x - 8) \times (x + 8)} \\ \\ \boxed{ \frac{5 + x}{x - 8} }$