Question

Алгебра. 8 класс.Помогите решить пожалуйста.Это очень срочноо!!! Буду очень благодарна.
Задание во вложении.

Answer 1

а)
$\frac{x - 2}{x} = \frac{3 - 2}{3} = \frac{1}{3}$
б)
$\frac{ {(t - 7)}^{2} }{2s} = \frac{(4 - 7)^{2} }{2 \times ( - 1)} = \frac{ {( - 3)}^{2} }{ - 2} = - \frac{9}{2} = - 4 \frac{1}{2}$
в)
$\frac{ {a}^{2} - {b}^{2} }{ {(a + b)}^{2} } = \frac{ {4}^{2} - {( - 2)}^{2} }{ {(4 - 2)}^{2} } = \frac{16 - 4}{ {2}^{2} } = \frac{12}{4} = 3$
г)
$\frac{y + 6}{y - 2} = \frac{4 + 6}{4 - 2} = \frac{10}{2} = 5$
д)
$\frac{x - 5}{ {(2y + 3)}^{2} } = \frac{2 - 5}{ {(2 \times ( - 2) + 3)}^{2} } = \frac{ - 3}{ {( - 1)}^{2} } = \frac{ - 3}{1} = - 3$
е)
$\frac{ {c}^{3} - dc}{ {c}^{2} d + {d}^{2} } = \frac{ {( - 2)}^{3} - 10 \times ( - 2) }{ {( - 2)}^{2} \times 10 + {10}^{2} } = \frac{ - 8 + 20}{4 \times 10 + 100} = \frac{12}{140} = \frac{3}{35}$
ж)
$\frac{ {x}^{2} + {y}^{2} }{ {x}^{4} - {y}^{4} } = \frac{ {x}^{2} + {y}^{2} }{( {x}^{2} + {y}^{2} )( {x}^{2} - {y}^{2} ) } = \frac{1}{ {x}^{2} - {y}^{2} } = \frac{1}{ {13}^{2} - {12}^{2} } = \frac{1}{169 - 144} = \frac{1}{25}$
з)
$\frac{ {m}^{4} - {n}^{4} }{{m}^{3} n - m {n}^{3} } = \frac{( {m}^{2} - {n}^{2} )( {m}^{2} + {n}^{2} ) }{ mn( {m}^{2} - {n}^{2} )} = \frac{ {m}^{2} + {n}^{2} }{mn} = \frac{ {2}^{2} + {( - 1)}^{2} }{ 2\times ( - 1)} \frac{4 + 1}{ - 2} = - \frac{5}{2} = - 2 \frac{1}{2}$