Предмет: Алгебра, автор: user05li

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Ответы

Автор ответа: govorukha

Объяснение:

А.

$\frac{2}{ {3x}^{2} {y}^{3} } - \frac{3}{ {5x}^{3} {y}^{2} } = \\ \\ \frac{2 \times 5x}{ {3x}^{2} {y}^{3} \times5x } - \frac{3 \times 3y}{ {5x}^{3} {y}^{2} \times 3y } = \\ \\ \frac{10x - 9y}{15 {x}^{3} {y}^{3} }$

Б.

$\frac{1}{y - 3} - \frac{6}{ {y}^{2} - 9} = \\ \\ \frac{1}{y - 3} - \frac{6}{(y - 3)(y + 3)} = \\ \\ \frac{1(y + 3)}{(y - 3)(y + 3)} - \frac{6}{(y - 3)(y + 3)} = \\ \\ \frac{y + 3 - 6}{(y - 3)(y + 3)} = \frac{y - 3}{(y - 3)(y + 3)} = \frac{1}{y + 3}$

В.

$\frac{3}{c - d} + \frac{4c - 4d}{ {c}^{2} - 2cd + {d}^{2} } = \\ \\ \frac{3}{c - d} + \frac{4c - 4d}{ {(c - d)}^{2} } = \\ \\ \frac{3(c - d)}{(c - d)( c- d)} + \frac{4(c - d)}{ {(c - d)}^{2} } = \\ \\ \frac{3(c - d)}{(c - d)^{2} } + \frac{4(c - d)}{ {(c - d)}^{2} } = \\ \\ \frac{3(c - d) + 4(c - d)}{(c - d)^{2} } = \frac{(3 + 4)(c - d)}{(c - d)^{2} } = \\ \\ \frac{7(c - d)}{(c - d)^{2} } = \frac{7}{c - d}$

Автор ответа: sangers1959

Объяснение:

$a)\ \frac{2}{3x^2y^3} -\frac{3}{5x^3y^2}=\frac{2*5x-3*3y}{15x^3y^3}=\frac{10x-9y}{15x^3y^3} . \\b)\ \frac{1}{y-3} -\frac{6}{y^2-9}=\frac{1}{y-3} -\frac{6}{(y-3)*(y+3)}=\frac{y+3-6}{(y-3)*(y+3)}=\frac{y-3}{(y-3)*(y+3)} =\frac{1}{y+3 } .\\c)\ \frac{3}{c-d} +\frac{4c-4d}{c^2-2cd+d^2}=\frac{3}{c-d}+\frac{4*(c-d)}{(c-d)^2}=\frac{3}{c-d} +\frac{4}{c-d} =\frac{7}{c-d} .$