Question

Вирішіть це будь ласка!

Answer 1

$№1$

$1) \: \: \: \: \: \frac{6m}{26} + \frac{7m}{26} = \frac{13m}{26}$

$2) \: \: \: \: \: \frac{14a}{9b} - \frac{5a}{9b} = \frac{9a}{9b} = \frac{a}{b}$

$3) \: \: \: \: \: \frac{4b - 15c}{18a} + \frac{2b + 3c}{18a} = \frac{2b - 12c}{18a} = \frac{2(b - 6c)}{18a} = \frac{b - 6c}{9a}$

$4) \: \: \: \: \: \frac{8m - 5n}{mn} - \frac{2m - 5n}{mn} = \frac{6m}{mn} = \frac{6}{n}$

$5) \: \: \: \: \: \frac{2y}{ {y}^{2} - 49 } - \frac{14}{ {y}^{2} - 49 } = \frac{2(y - 7)}{(y - 7)(y + 7)} = \frac{2}{y + 7}$

$6) \: \: \: \: \: \frac{ {x}^{2} + 12x }{25 - {x}^{2} } - \frac{2x - 25}{25 - {x}^{2} } = \frac{ {x}^{2} - 10x + 25}{(5 - x)(5 + x)} = \frac{ {(x - 5)}^{2} }{(5 - x)(5 + x)} = \frac{x - 5}{x + 5}$

$№2$

$1) \: \: \: \: \: \frac{4a + 8b}{4a} = \frac{4(a + 2b)}{4a} = \frac{a + 2b}{a}$

$2) \: \: \: \: \: \frac{5x - 10y}{3x - 6y} = \frac{5(x - 2y)}{3(x - 2y)} = \frac{5}{3} = 1 \frac{2}{3}$

$3) \: \: \: \: \: \frac{ {x}^{2} - 25 }{2x - 10} = \frac{(x - 5)(x + 5)}{2(x - 5)} = \frac{x + 5}{2}$

$4) \: \: \: \: \: \frac{6 {x}^{2} - 3x}{4 - 8x} = \frac{3x(2x - 1)}{4(1 - 2x)} = \frac{ - 3x(1 - 2x)}{4(1 - 2x)} = - \frac{3x}{4}$

$5) \: \: \: \: \: \frac{ {m}^{2} - 16 }{ {m}^{2} + 8m + 16 } = \frac{(m - 4)(m + 4) }{ {(m + 4)}^{2} } = \frac{m - 4}{m + 4}$

$6) \: \: \: \: \: \frac{ {b}^{5} - {b}^{3} }{ {b}^{2} - {b}^{4} } = \frac{ {b}^{3}( {b}^{2} - 1) }{ {b}^{2} (1 - {b}^{2}) } = \frac{ - {b}^{3}(1 - {b}^{2} ) }{ {b}^{2} (1 - {b}^{2}) } = - \frac{ {b}^{3} }{ {b}^{2} } = - b$

$№3$

$1) \: \: \: \: \: \frac{a - 2}{a - 1} - \frac{a}{1 - a} = \frac{(1 - a)(a - 2) - a(a - 1)}{(a - 1)(1 - a)} = \frac{a - 2 - {a}^{2} + 2a - {a}^{2} + a }{a - {a}^{2} - 1 + a } = \frac{ - 2 {a}^{2} + 4a - 2 }{ - {a}^{2} + 2a - 1} = \frac{2( - {a}^{2} + 2a - 1) }{ - {a}^{2} + 2a - 1 } = 2$

$2) \: \: \: \: \: \frac{3y + 7}{4 - y} + \frac{y + 15}{y - 4} = \frac{(3y + 7)(y - 4) + (y + 15)(4 - y)}{(4 - y)(y - 4)} = \frac{3 {y}^{2} - 12y + 7y - 28 + 4y - {y}^{2} + 60 - 15y}{4y - 16 - {y}^{2} + 4y } = \frac{2 {y}^{2} - 16y + 32 }{ - {y}^{2} + 8y - 16 } = \frac{2( {y}^{2} - 8y + 16) }{ - ( {y}^{2} - 8y + 16) } = - 2$

$3) \: \: \: \: \: \frac{ {(2a - 3)}^{2} }{9a - 27} + \frac{( {a - 6)}^{2} }{27 - 9a} = \frac{(2a { - 3)}^{2}(27 - 9a) + {(a - 6)}^{2}(9a - 27) }{(9a - 27)(27 - 9a)} = \frac{( {4a}^{2} - 12a + 9)(27 - 9a) + ( {a}^{2} - 12a + 36)(9a - 27) }{243a - 81 {a}^{2} - 729 + 243a } = \frac{108 {a}^{2} - 36 {a}^{3} - 324a + 108 {a}^{2} + 243 - 81a + 9 {a}^{3} - 27 {a}^{2} - 108 {a}^{2} + 324 + 324a - 972 }{ - 81 {a}^{2} + 486a - 729} = \frac{ - 27 {a}^{3} + 81 {a}^{2} - 81a - 405}{ - 81 {a}^{2} + 486a - 729} = \frac{ - 27( {a}^{3} - 3 {a}^{2} + 3a + 15)}{ - 27(3 {a}^{2} - 18a + 27)} = \frac{ {a}^{3} - 3 {a}^{2} + 3a + 15 }{3 {a}^{2} - 18a + 27 } = \frac{ a {}^{2}(a - 3) + 3(a + 5) }{3( {a}^{2} - 6a + 9) } = \frac{ {a}^{2}(a - 3) + 3(a + 5) }{3( {a - 3)}^{2} }$

$4) \: \: \: \: \: \frac{25 - 3x}{ {(x - 5)}^{2} } - \frac{(7x - {x)}^{2} }{ {(5 - x)}^{2} } = \frac{25 - 3x}{ {( x - 5)}^{2} } - \frac{7x - {x}^{2} }{( - (x { - 5))}^{2} } = \frac{25 - 3x}{ {(x - 5)}^{2} } - \frac{7x - {x}^{2} }{ {(x - 5)}^{2} } = \frac{25 - 3x - (7x - {x}^{2} )}{ {(x - 5)}^{2} } = \frac{25 - 3x - 7x + {x}^{2} }{ {(x - 5)}^{2} } = \frac{25 - 10 + {x}^{2} }{ {(x - 5)}^{2} } = \frac{ {(5 - x)}^{2} }{ {(x - 5)}^{2} } = \frac{( - (x - 5) {)}^{2} }{ {(x - 5)}^{2} } = \frac{ {(x - 5)}^{2} }{ {(x - 5)}^{2} } = 1$