Алгебра, вопрос задал dimon6071 , 6 лет назад

решите пж 1 и 2 вариант даю 100 балов​

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Ответы на вопрос

Ответил Ляляляля109
0

1 вариант

1)

а)

 \frac{2}{3}  +  \frac{4}{11}  =  \frac{22}{33}  +  \frac{12}{33}  =  \frac{34}{33}  = 1 \frac{1}{33}

б)

 \frac{3x}{5}  -  \frac{2y}{7}  =  \frac{21x}{35}  -  \frac{10y}{35}  =  \frac{21x - 10y}{35}

2.

 \frac{1}{x + 2}  +  \frac{4}{ {x}^{2} - 4 }  =  \frac{1}{x + 2}  +  \frac{4}{(x - 2)(x + 2)}  =  \frac{x - 2}{(x - 2)(x + 2)}  +  \frac{4}{(x - 2)(x + 2)}  =  \frac{x - 2 + 4}{(x - 2)(x + 2)}   =  \\  \\  =  \frac{x + 2}{(x - 2)(x + 2)}  =  \frac{1}{x - 2}

3.

 \frac{a - 3}{ {a}^{2} + 3a + 9 }  +  \frac{9a}{ {a}^{3} - 27 }  -  \frac{1}{a - 3}  =  \frac{a - 3}{ {a}^{2}  + 3a + 9}  +  \frac{9a}{(a - 3)( {a}^{2}  + 3a + 9)}  -  \frac{1}{a - 3}  =   \\  \\  = \frac{ {(a - 3)}^{2} }{(a - 3)( {a}^{2}  + 3a + 9)}  +  \frac{9a}{(a - 3)( {a}^{2}  + 3a + 9)}  -  \frac{ {a}^{2} + 3a + 9 }{(a - 3)( {a}^{2}  + 3a + 9)}  =  \frac{ {a}^{2}  - 6a + 9 + 9a -  {a}^{2}  - 3a - 9}{(a - 3)( {a}^{2}  + 3a + 9)}  =  \\  \\  =  \frac{0}{ {a}^{3} - 27 }  = 0

2 вариант

1.

а)

 \frac{1}{7}  +  \frac{3}{14}  =  \frac{2}{14}  +  \frac{3}{14}  =  \frac{5}{14}

б)

 \frac{2x}{3}  -  \frac{3y}{5}  =  \frac{10x}{15}  -  \frac{9y}{15}  =  \frac{10x - 9y}{15}

2.

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

3.

 \frac{a - 4}{ {a}^{2}  + 4a + 16}  +  \frac{12a}{ {a}^{3} - 64 }  -  \frac{1}{a - 4}  =  \frac{a - 4}{ {a}^{2} + 4a + 16 }  +  \frac{12a}{(a - 4)( {a}^{2} + 4a + 16) }  -  \frac{1}{a - 4}  = \\  \\  =   \frac{ {(a - 4)}^{2} }{(a - 4)( {a}^{2}  + 4a + 16) }  +  \frac{12a}{(a - 4)( {a }^{2}  + 4a + 16)}  -  \frac{ {a}^{2} + 4a + 16 }{(a - 4)( {a}^{2} + 4a + 16) }  =  \frac{ {a}^{2} - 8a + 16 + 12a -  {a}^{2}  - 4a - 16 }{(a - 4)( {a}^{2}  + 4a + 16}  =  \\  \\  =  \frac{0}{ {a}^{3} - 64 }  = 0

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