Question

Помогите пожалуйста выполнить сложение и вычитание дробей

Answer 1

а)

$\frac{7 + 3x}{x} + \frac{10 - 3y}{ y} = \frac{(7 + 3x) \times y}{x \times y} + \frac{(10 - 3y) \times x}{y \times x} = \frac{7y + 3xy}{xy} + \frac{10x - 3xy}{xy} = \frac{7y + 3xy + 10x - 3xy}{xy} = \frac{7y + 10x}{xy}$

б)

$\frac{a + 2b}{b} - \frac{2a - b}{a} = \frac{(a + 2b) \times a}{b \times a} - \frac{(2a - b) \times b}{a \times b} = \frac{ {a}^{2} + 2ab }{ab} - \frac{2ab - {b}^{2} }{ab} = \frac{( {a}^{2} + 2ab) - (2ab - {b}^{2}) }{ab} = \frac{ {a}^{2} + 2ab - 2ab + {b}^{2} }{ab} = \frac{ {a}^{2} + {b}^{2} }{ab}$

в)

$\frac{a}{a + c} - \frac{a}{c} = \frac{a \times c}{(a + c) \times c} - \frac{a \times (a + c)}{c \times (a + c)} = \frac{ac}{ac + {c}^{2} } - \frac{ {a}^{2} + ac}{ac + {c}^{2} } = \frac{ac - ( {a}^{2} + ac)}{ac + {c}^{2} } = \frac{ac - {a}^{2} - ac}{ac + {c}^{2} } = \frac{ - {a}^{2} }{ac + {c}^{2} } = - \frac{ {a}^{2} }{ac + {c}^{2} }$

г)

$\frac{2}{z - 1} + \frac{2}{z + 1} = \frac{2 \times (z + 1)}{(z - 1) \times (z + 1)} + \frac{2 \times (z - 1)}{(z + 1) \times (z - 1)} = \frac{2z + 2}{ {z}^{2} - 1 } + \frac{2z - 2}{ {z}^{2} - 1 } = \frac{2z + 2 + 2z - 2}{ {z}^{2} - 1 } = \frac{4z}{ {z}^{2} - 1 }$

д)

$\frac{2a}{2a + 1} - \frac{3a}{3a + 2} = \frac{2a \times (3a + 2)}{(2a + 1) \times (3a + 2)} - \frac{3a \times (2a + 1)}{(3a + 2) \times (2a + 1)} = \frac{6 {a}^{2} + 4a }{6 {a}^{2} +4a + 3a + 2} - \frac{6 {a}^{2} + 3a}{6 {a}^{2} + 3a + 4a + 2} = \frac{6 {a}^{2} + 4a}{6 {a}^{2} + 7a + 2} - \frac{6 {a}^{2} + 3a }{6 {a}^{2} + 7a + 2} = \frac{(6 {a}^{2} + 4a) - (6 {a}^{2} + 3a) }{6 {a}^{2} + 7a + 2 } = \frac{6 {a}^{2} + 4a - 6 {a}^{2} - 3a }{6 {a}^{2} + 7a + 2} = \frac{a}{6 {a}^{2} + 7a + 2}$

е)

$\frac{x + 1}{x} - \frac{ x}{x - 1} = \frac{(x + 1) \times (x - 1)}{x \times (x - 1)} - \frac{x \times x}{(x - 1) \times x} = \frac{ {x}^{2} - 1}{ {x}^{2} - x } - \frac{ {x}^{2} }{ {x}^{2} - x } = \frac{{x}^{2} - 1 - {x}^{2} }{ {x}^{2} - x} = \frac{ - 1}{ {x}^{2} - x} = - \frac{1}{ {x}^{2} - x }$