Question

добрые люди помогите, в математике вообще не шарю, буду очень благодарен​

Answer 1

Ответ:

Объяснение:

1)

$\frac{4}{16a^{3} b} - \frac{5}{12ab^{2} } = \\\frac{4}{4*4a^{3} b} - \frac{5}{4*3ab^{2} } = \\\frac{4*3b}{4*4a^{3} b*3b} - \frac{5*4a^{2} }{4*3ab^{2}*4a^{2} } = \\\frac{4*3b - 5*4a^{2} }{48a^{3} b^{2} } = \\\frac{12b - 20a^{2} }{48a^{3} b^{2} } =\\ \frac{2( 6b - 10a^{2} )}{48a^{3} b^{2} } = \\ \frac{2( 3b - 5a^{2} )}{24a^{3} b^{2} } =\\ \frac{3b - 5a^{2}}{12a^{3} b^{2} }$

2)

$\frac{a^{2} }{a-9} + \frac{81}{9-a} =\\\frac{a^{2} }{a-9} - \frac{81}{a-9} =\\\frac{a^{2} -81}{a-9} = \frac{(a -9)(a+9)}{a-9} = a+9$

3)

$\frac{4}{7a-7b} + \frac{a}{3a-3b} =\\\frac{4}{7(a-b)} + \frac{a}{3(a-b)} =\\\frac{4*3+ 7a}{21(a-b)} = \frac{12+ 7a}{21(a-b)}$

4)

$\frac{10c}{c^{2} -16} - \frac{5}{c+4} + \frac{7}{c-4} = \\\frac{10c}{(c -4)(c+4)} - \frac{5}{c+4} + \frac{7}{c-4} = \\\frac{10c}{(c -4)(c+4)} - \frac{5(c -4)}{(c -4)(c+4)} + \frac{7(c+4)}{(c -4)(c+4)} = \\\frac{10c - 5(c -4)+7(c+4)}{(c -4)(c+4)} = \\\frac{10c - 5c +20+7c+28}{(c -4)(c+4)} =\\\frac{12c+48}{(c -4)(c+4)} = \frac{12(c+4)}{(c -4)(c+4)} = \\\frac{12}{(c -4)}$