Question

номер 3 и номер 4
100 балов ​

Answer 1

Ответ:

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Answer 2

$\displaystyle\bf\\1)\\\\\frac{5x-6}{6x^{2} } -\frac{4-9x}{9x^{3} } =\frac{(5x-6)\cdot3x-(4-9x)\cdot 2}{18x^{3} } =\\\\\\=\frac{15x^{2} -18x-8+18x}{18x^{3} } =\frac{15x^{2} -8}{18x^{3} } \\\\\\2)\\\\\frac{42}{b^{2} +7b} -\frac{6}{b} =\frac{42}{b(b +7)} -\frac{6}{b} =\frac{42-6\cdot(b+7)}{b(b+7)} =\\\\\\=\frac{42-6b-42}{b(b +7)} =-\frac{6b}{b(b+7)} =-\frac{6}{b+7} \\\\\\3)\\\\\frac{c^{2} }{c^{2} -16} -\frac{c}{c+4} =\frac{c^{2} }{(c+4)(c-4)}-\frac{c}{c+4} =$

$\displaystyle\bf\\=\frac{c^{2}-c\cdot(c-4) }{c^{2}-16 } =\frac{c^{2} -c^{2} +4c}{c^{2}-16 } =\frac{4c}{c^{2}-16 } \\\\\\4)\\\\3y-\frac{18y^{2} }{6y+1} =\frac{3y\cdot(6y+1)-18y^{2} }{6y+1} =\frac{18y^{2}+3y-18y^{2} }{6y+1}=\\\\\\=\frac{3y}{6y+1} \\\\\\\\1)\\\\\frac{y+6}{4y+8}-\frac{y+2}{4y-8} +\frac{5}{y^{2} -4} =\frac{y+6}{4(y+2)}-\frac{y+2}{4(y-2)} +\frac{5}{(y -2)(y+2)} =\\\\\\=\frac{(y+6)(y-2)-(y+2)(y+2)+5\cdot 4}{4(y-2)(y+2)} =$

$\displaystyle\bf\\=\frac{y^{2}-2y+6y-12-y^{2} -4y-4+20 }{4(y-2)(y+2)} =\frac{4}{4(y^{2}-4 )} =\frac{1}{y^{2}-4 } \\\\\\2)\\\\\frac{6b^{3} +48b}{b^{3} +64} -\frac{3b^{2} }{b^{2} -4b+16} =\frac{6b^{3} +48b}{(b+4)(b^{2} -4b+16)} -\frac{3b^{2} }{b^{2} -4b+16} =\\\\\\=\frac{6b^{3} +48b-3b^{2} \cdot(b+4)}{(b+4)(b^{2}-4b+16) } =\frac{6b^{3} +48b-3b^{3} -12b^{2} }{(b+4)(b^{2}-4b+16) } =\\\\\\=\frac{3b^{3}-12b^{2} +48b }{(b+4)(b^{2} -4b+16)} =\frac{3b(b^{2} -4b+16)}{(b+4)(b^{2} -4b+16)} =\frac{3b}{b+4}$