Алгебра: Решить (см.прикреплённое фото)

Решить (см.прикреплённое фото)

Ответ:

dgrigoref

09.06.2020 11:32

$\left ( \frac{m^2-5m+25}{25m^2-1}*\frac{5m^2+m}{m^3+125}-\frac{m+5}{5m^2-m} \right ):\frac{5}{m^2+5m}-\frac{25m+22}{5-25m}=\\\\\\=\left ( \frac{ m^2-5m+25 }{ (5m -1)(5m+1)}*\frac {m (5m + 1)}{m^3+ 5^3}-\frac{m+5}{m(5m -1)} \right ):\frac{5} {(m +5)m}-\frac{25m+22}{5(1- 5m)}=\\\\\\ = \left ( \frac{ m^2-5m+25 }{ 5m -1 }*\frac {m }{(m + 5)(m^2-5m+25) }-\frac{m+5}{m(5m -1)} \right ):\frac{5} {(m +5)m}-\frac{25m+22}{5(1- 5m)}=$

$=\left ( \frac{ 1 }{ 5m -1 }*\frac {m }{ m + 5 }-\frac{m+5}{m( 5m-1)} \right )*\frac {(m +5)m}{5}+\frac{25m+22}{5( 5m-1)}= \\\\\\= \left ( \frac {m }{( 5m -1) (m + 5) }-\frac{m+5}{m(5m -1)} \right )*\frac {(m +5)m}{5}-\frac{25m+22}{5( 1-5m )}= \\\\\\=\left ( \frac {m *m -(m+5) (m+5)}{ m (5m -1) (m + 5) } \right )*\frac {(m +5)m}{5}+\frac{25m+22}{5( 5m-1)}= \\\\\\= \left ( \frac {m *m - (m^2+5m+5m+25 }{ m (5m -1) (m + 5) } \right )*\frac {(m +5)m}{5}+\frac{25m+22}{5( 5m-1)}=$

$= \left ( \frac {m^2- m^2-5m-5m-25 }{ m (5m -1) (m + 5) } \right )*\frac {(m +5)m}{5}-\frac{25m+22}{5( 1-5m )}= \\\\\\=\left ( \frac {-10m -25 }{ 5m -1 } \right )*\frac {1}{5}-\frac{25m+22}{5( 1-5m )}= \left ( \frac {-5(2m+5) }{ 5m -1 } \right )*\frac {1}{5} -\frac{25m+22}{5( 1-5m )}= \\\\\\= \frac { 2m+5 }{ 1-5m } -\frac{25m+22}{5( 1-5m )}= \frac { 5(2m+5)-25-22 }{ 5(1-5m )}= \frac { 10m+25-25m-22 }{ 5(1-5m )}=\\\\\\= \frac { 3-15m }{ 5(1-5m )} = \frac { 3(1- 5m) }{ 5(1-5m )}=\frac{3}{5}$