Question

Помогите, пожалуйста, с алгеброй!!!

Answer 1

#2. $(3x - \frac{6x}{x + 5} ) \div \frac{9x + 27}{8x + 40} = \frac{8x}{3}$

1) $3x - \frac{6x}{x + 5} = \frac{3x \times (x + 5)}{x + 5} - \frac{6x}{x + 5} = \frac{3 {x}^{2} + 15x - 6x}{x + 5} = \frac{3 {x}^{2} + 9x}{x + 5}$

2) $\frac{3 {x}^{2} + 9x}{x + 5} \div \frac{9x + 27}{8x + 40} = \frac{3 {x}^{2} + 9x}{x + 5} \times \frac{8x + 40}{9x + 27} = \frac{(3 {x}^{2} + 9x) \times (8x + 40)}{(x + 5) \times (9x + 27)} = \frac{3x(x + 3) \times 8(x + 5)}{(x + 5) \times 9(x + 3)} = \frac{3x \times 8}{9} = \frac{8x}{3}$

#3. $\frac{2a}{a - 5} - \frac{a + 7}{4a - 20} \times \frac{200}{ {a}^{2} + 7a } = \frac{2a + 10}{a}$

1) $\frac{a + 7}{4a - 20} \times \frac{200}{ {a}^{2} + 7a} = \frac{(a + 7) \times 200}{4(a - 5) \times a(a + 7)} = \frac{50}{a(a - 5)}$

2) $\frac{2a}{a - 5} - \frac{50}{a(a - 5)} = \frac{2a \times a}{(a - 5) \times a} - \frac{50}{a(a - 5)} = \frac{2 {a}^{2} - 50 }{a(a - 5)} = \frac{2( {a}^{2} - 25)}{a(a - 5)} = \frac{2(a - 5)(a + 5)}{a(a - 5)} = \frac{2(a + 5)}{a} = \frac{2a + 10}{a}$

#4. $( \frac{4c}{c - 4} - \frac{3c}{ {c}^{2} - 8c + 16} ) \div \frac{4c - 19}{ {c}^{2} - 16} - \frac{4c + 16}{c - 4} = c + 4$

1) $\frac{4c}{c - 4} - \frac{3c}{ {c}^{2} - 8c + 16} = \frac{4c}{c - 4} - \frac{3c}{ {(c - 4)}^{2} } = \frac{4c \times (c - 4)}{(c - 4) \times (c - 4)} - \frac{3c}{ {(c - 4)}^{2} } = \frac{4c(c - 4) - 3c}{ {(c - 4)}^{2} } = \frac{4 {c}^{2} - 16c - 3c }{ {(c - 4)}^{2} } = \frac{4 {c}^{2} - 19c }{ {(c - 4)}^{2} }$

2) $\frac{4 {c}^{2} - 19c }{ {(c - 4)}^{2} } \div \frac{4c - 19}{ {c}^{2} - 16 } = \frac{c(4c - 19)}{(c - 4)(c - 4)} \div \frac{4c - 19}{(c - 4)(c + 4)} = \frac{c(4c - 19)}{(c - 4)(c - 4)} \times \frac{(c - 4)(c + 4)}{4c - 19} = \frac{c \times (c + 4)}{c - 4} = \frac{ {c}^{2} + 4c }{c - 4}$

3) $\frac{ {c}^{2} + 4c }{c - 4} - \frac{4c + 16}{c - 4} = \frac{ {c}^{2} + 4c - 4c - 16 }{c - 4} = \frac{ {c}^{2} - 16}{c - 4} = \frac{(c - 4)(c + 4)}{c - 4} = c + 4$

#5. $( \frac{ {n}^{2} }{ {m}^{3} - m {n}^{2} } + \frac{1}{m - n} ) \div ( \frac{m}{mn - {n}^{2} } - \frac{m + n}{mn - {m}^{2} } ) = \frac{n}{m}$

1) $\frac{ {n}^{2} }{ {m}^{3} - m {n}^{2} } + \frac{1}{m - n} = \frac{ {n}^{2} }{m( {m}^{2} - {n}^{2} ) } + \frac{1}{m - n} = \frac{ {n}^{2} }{m(m - n)(m + n)} + \frac{1}{m - n} = \frac{ {n}^{2} }{m(m - n)(m + n)} + \frac{1 \times m(m + n)}{(m - n) \times m(m + n)} = \frac{ {n}^{2} + m(m + n)}{m(m - n)(m + n)} = \frac{ {n}^{2} + {m}^{2} + mn}{ {m}^{3} - m {n}^{2} }$

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

3) $\frac{ {n}^{2} + {m}^{2} + mn}{ {m}^{3} - m {n}^{2} } \div \frac{ {m}^{2} + mn + {n}^{2} }{ {m}^{2}n - m {n}^{2} } = \frac{ {n}^{2} + {m}^{2} + mn}{ {m}^{3} - m {n}^{2} } \times \frac{{m}^{2}n - m {n}^{2} }{{m}^{2} + mn + {n}^{2}} = \frac{ {m}^{2} n - m {n}^{2} }{ {m}^{3} - m {n}^{2} } = \frac{n( {m}^{2} - mn)}{m( {m}^{2} - mn)} = \frac{n}{m}$

#6. $( \frac{ {b}^{2} + 9 }{ {b}^{2} - 9} + \frac{b}{b + 3} + \frac{b}{3 - b} ) \div \frac{ {b}^{2} - 3b}{ {(b + 3)}^{2} } = \frac{b + 3}{b}$

1) $\frac{ {b}^{2} + 9 }{ {b}^{2} - 9} = \frac{ {b}^{2} + 9}{ {b}^{2} - {3}^{2} } = \frac{ {b}^{2} + 9}{(b - 3)(b + 3)}$,

$\frac{b}{b + 3}$ ,

$\frac{b}{3 - b} = \frac{b}{ - (b - 3)} = - \frac{b}{b - 3}$.

Общий знаменатель: $(b - 3)(b + 3)$

Приводим: $\frac{ {b}^{2} + 9}{(b - 3)(b + 3)} + \frac{b \times (b - 3)}{(b + 3) \times (b - 3)} - \frac{b \times (b + 3)}{(b - 3) \times (b + 3)} = \frac{ {b}^{2} + 9 + {b}^{2} - 3b - {b}^{2} - 3b}{(b - 3)(b + 3)} = \frac{ {b}^{2} - 6b + 9}{(b - 3)(b + 3)} = \frac{ {(b - 3)}^{2} }{(b - 3)(b + 3)} = \frac{b - 3}{b + 3}$

2) $\frac{b - 3}{b + 3} \div \frac{ {b}^{2} - 3b}{ {(b + 3)}^{2} } = \frac{b - 3}{b + 3} \times \frac{ {(b + 3)}^{2} }{ {b}^{2} - 3b} = \frac{(b - 3) \times (b + 3)(b + 3)}{(b + 3) \times b(b - 3)} = \frac{b + 3}{b}$