Question

Умоляю помогите пожалуйста!​

Answer 1

Ответ:

$\dfrac{3x}{x-5};$

$y-2;$

$-4;$

Объяснение:

$\bigg (3x-\dfrac{3x}{x-4} \bigg ): \bigg (x-\dfrac{6x-25}{x-4} \bigg )=\dfrac{3x(x-4)-3x}{x-4}:\dfrac{x(x-4)-(6x-25)}{x-4}=$

$=\dfrac{3x^{2}-12x-3x}{x-4}:\dfrac{x^{2}-4x-6x+25}{x-4}=\dfrac{3x^{2}-15x}{x-4}:\dfrac{x^{2}-10x+25}{x-4}=$

$=\dfrac{3x^{2}-15x}{x-4} \cdot \dfrac{x-4}{x^{2}-10x+25}=\dfrac{(3x^{2}-15x) \cdot (x-4)}{(x-4) \cdot (x^{2}-10x+25)}=\dfrac{3x^{2}-15x}{x^{2}-10x+25}=$

$=\dfrac{3x(x-5)}{(x-5)^{2}}=\dfrac{3x}{x-5};$

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$\bigg (y+2+\dfrac{8}{y-2} \bigg ): \dfrac{y^{2}+4}{4-4y+y^{2}}=\dfrac{(y+2) \cdot (y-2)+8}{y-2} \cdot \dfrac{y^{2}-4y+4}{y^{2}+4}=$

$=\dfrac{y^{2}-4+8}{y-2} \cdot \dfrac{(y-2)^{2}}{y^{2}+4}=\dfrac{(y^{2}+4) \cdot (y-2)^{2}}{(y-2) \cdot (y^{2}+4)}=y-2;$

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$\bigg (\dfrac{b-3}{7b-4}-\dfrac{b-3}{b-4} \bigg ) \cdot \dfrac{7b-4}{9b-3b^{2}}+\dfrac{b^{2}-14}{4-b}+b=\dfrac{(b-3) \dot (b-4)-(b-3) \cdot (7b-4)}{(7b-4) \cdot (b-4)} \cdot$

$\cdot \dfrac{7b-4}{3b(3-b)}+\dfrac{b^{2}-14}{4-b}+b=\dfrac{b^{2}-4b-3b+12-(7b^{2}-4b-21b+12)}{(7b-4) \cdot (b-4)} \cdot \dfrac{7b-4}{3b(3-b)}+$

$+\dfrac{b^{2}-14}{4-b}+b=\dfrac{b^{2}-7b^{2}-7b+25b+12-12}{(7b-4) \cdot (b-4)} \cdot \dfrac{7b-4}{3b(3-b)}+\dfrac{b^{2}-14}{4-b}+b=$

$=\dfrac{18b-6b^{2}}{(7b-4) \cdot (b-4)} \cdot \dfrac{7b-4}{3b(3-b)}+\dfrac{b^{2}-14}{4-b}+b=\dfrac{6b(3-b) \cdot (7b-4)}{(7b-4) \cdot (b-4) \cdot 3b(3-b)}+$

$+\dfrac{b^{2}-14}{4-b}+b=\dfrac{2}{b-4}+\dfrac{(b^{2}-14) \cdot (-1)}{(4-b) \cdot (-1)}+b=\dfrac{2}{b-4}+\dfrac{14-b^{2}}{b-4}+b=$

$=\dfrac{2+14-b^{2}}{b-4}+b=\dfrac{16-b^{2}}{b-4}+b=\dfrac{(4-b)(4+b)}{b-4}+b=\dfrac{-(b-4)(4+b)}{b-4}+b=$

$=-(4+b)+b=-4-b+b=-4;$