Предмет: Алгебра, автор: skylinepasha2019

Вирішіть це будь ласка!

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Ответы

Автор ответа: MuhammadGulu
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

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