Алгебра: Все на картинке. даю, выручайте годовая контр))

Все на картинке. даю, выручайте годовая контр))

Ответ:

alexandrapavlov1

27.05.2020 21:30

ответ прикреплен Все на картинке. 100 балов даю, выручайте годовая контр)) ">

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Ответ:

kiskaignatushkina

27.05.2020 21:30

$1. \\\\ a) \; \; \displaystyle \frac{m+2}{3-m} + \frac{m-1}{m-3} = -\frac{m+2}{m-3} + \frac{m-1}{m-3} = \frac{-m-2+m-1}{m-3} = \frac{-3}{m-3} = -\frac{3}{m-3} \\\\</p<p\displaystyle b) \; \; (y-3)*(\frac{4}{y}+\frac{2}{y-3}) = (y-3)*\frac{4(y-3)+2y}{y*(y-3)} = \frac{4y-12+2y}{y} = \frac{6y-12}{y} \\\\\\</p<p2. \\\\</p<p\displaystyle a) \; \; (2x+7)(7-2x) =49+7x \\\\</p<p \displaystyle 49 - 4x^2 = 49 + 7x \\\\</p<p \displaystyle -4x^2=7x \\\\</p<p \displaystyle-4x^2-7x = 0 \\\\</p<p \displaystyle -x(4x+7) = 0 \;\; \mid \; \; *(-1) \\\\</p<p \displaystyle x(4x+7)=0 \\\\</p<p \displaystyle x=0 \; \; \; ili \; \; \; 4x + 7 = 0 \Rightarrow x = -\frac{7}{4} \\\\</p<p \displaystyle b) \; \; \sqrt{x} - 4 = 0 \\\\</p<p \displaystyle \sqrt{x} = 4 \\\\</p<p\displaystyle x = 16 \; \; \; \; (\sqrt{16} - 4 = 4-4 = 0) \\\\</p<pc) \; \; \displaystyle \frac{y}{y-3}-\frac{5}{y+3}=\frac{18}{y^2-9} \; \; \; ; \; \; \; y\neq 3 \; , \; y\neq - 3 \\\\</p<p\displaystyle \frac{y}{y-3}-\frac{5}{y+3} - \frac{18}{y^2-9} = 0 \\\\</p<p\displaystyle \frac{y}{y-3}-\frac{5}{y+3} - \frac{18}{(y-3)(y+3)} = 0 \\\\</p<p\displaystyle \frac{y(y+3)-5(y-3)-18}{(y-3)(y+3)} = 0 \\\\</p<p\displaystyle \frac{y^2+3y-5y+15-18}{(y-3)(y+3)} = 0 \\\\</p<p\displaystyle \frac{y^2-2y-3}{(y-3)(y+3)} = 0 \\\\</p<p\displaystyle y^2-2y-3 = 0 \\\\ </p<p\displaystyle D = (-2)^2 - 4*(-3) = 4 + 12 = 16 \\\\</p<p\displaystyle x_{1} = \frac{2 - 4}{2} = \frac{-2}{2} = -1 \\\\</p<p\displaystyle x_2 = \frac{2 + 4 }{2} = \frac{6 }{2} = 3 \; (postoronniy \; \; koren) \\\\</p<p\displaystyle d) \; \; x^2+3x-10 = 0 \\\\</p<p\displaystyle D = 3^2 - 4 *(-10) = 9 + 40 = 49 \\\\</p<p\displaystyle x_1 = \frac{-3-7}{2} = \frac{-10}{2} = -5 \\\\</p<p\displaystyle x_2 = \frac{-3+7}{2} = \frac{4}{2} = 2 \\\\\\</p<p\displaystyle 3. \; \; \displaystyle \Bigg(\frac{x}{y^2} - \frac{1}{x}\Bigg) \div \Bigg( \frac{1}{x}+\frac{1}{y} \Bigg) = \frac{x^2-y^2}{xy^2} \div \frac{y+x}{xy} = \frac{(x-y)(x+y)}{xy^2} * \frac{xy}{y+x} = \frac{x-y}{y}$