Question

Помогите решить пожалуйста алгебру, задание на время!!!!!!!!!!!!!

Answer 1

1 вариант.

1.

а)

$\frac{4b + 9}{6b} + \frac{2b - 3}{6b} = \frac{4b + 9 - 3 + 2b}{6b} = \frac{6b + 6}{6b} = \frac{6(b + 1)}{6b} = \frac{b + 1}{b}$

б)

$\frac{q + 2}{q - 2} - \frac{6 - q}{q - 2} = \frac{q + 2 - 6 + q}{q - 2} = \frac{2q - 4}{q - 2} = \frac{2(q - 2)}{q - 2} = 2$

в)

$\frac{a}{a - 3} + \frac{3}{3 - a} = \frac{a}{a - 3} - \frac{3}{a - 3} = \frac{a - 3}{a - 3} = 1$

г)

$\frac{ {m}^{2} }{m - n} - \frac{ {n}^{2} }{m - n} = \frac{ {m}^{2} - {n}^{2} }{m - n} = \frac{(m - n)(m + n)}{m - n} = m + n$

2.

$\frac{ {x}^{2} + 1}{ {(x - 3)}^{2} } - \frac{5x}{ {(x - 3)}^{2} } + \frac{8 - x}{ {(x - 3)}^{2} } = 1$

$\frac{ {x}^{2} + 1 - 5x + 8 - x}{ {(x - 3)}^{2} } = 1$

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

$\frac{ {(x - 3)}^{2} }{ {(x - 3)}^{2} } = 1$

$1 = 1$

2 вариант.

1.

а)

$\frac{3c + a}{4c} - \frac{a - 7c}{4} = \frac{3c + a - a + 7c}{4c} = \frac{10c}{4c} = \frac{5}{2} = 2.5$

б)

$\frac{2p}{p + 3} + \frac{3 - p}{p + 3} = \frac{2p + 3 - p}{p + 3} = \frac{p + 3}{p + 3} = 1$

в)

$\frac{ {x}^{2} }{x - 4} + \frac{4x}{4 - x} = \frac{ {x}^{2} }{ x- 4} - \frac{4x}{x - 4} = \frac{ {x}^{2} - 4x}{x - 4} = \frac{x(x - 4) }{x - 4} = x$

г)

$\frac{m}{ {m}^{2} - {n}^{2} } - \frac{n}{ {m}^{2} - {n}^{2}} = \frac{m - n}{ {m}^{2} - {n}^{2}} = \frac{m - n}{(m - n)(m + n)} = \frac{1}{m + n}$

2.

$\frac{60 + y}{ {(y + 8)}^{2}} + \frac{15y}{ {(y + 8)}^{2}} + \frac{4 + {y}^{2} }{ {(y + 8)}^{2}} = 1$

$\frac{60 + y + 15y + 4 + {y}^{2} }{ {(y + 8)}^{2}} = 1$

$\frac{ {y}^{2} + 16y + 64 }{ {(y + 8)}^{2}} = 1$

$\frac{ {(y + 8)}^{2}}{ {(y + 8)}^{2}} = 1$

$1 = 1$