Question

Даю 50 баллов! плиз пошаговое решение трёх заданий по алгебре) буду очень благодарна^^

Answer 1

$\displaystyle\bf\\1)\\\\\frac{3x+2}{x^{2} +2x} +\frac{x-2}{2x+4} =\frac{3x+2}{x\cdot(x+2)} +\frac{x-2}{2\cdot(x+2)} =\\\\\\=\frac{(3x+2)\cdot 2+(x-2)\cdot x}{2x(x+2)} =\frac{6x+4+x^{2} -2x}{2x(x+2)} =\\\\\\=\frac{x^{2}+4x+4 }{2x(x+2)}=\frac{(x+2)^{2} }{2x(x+2)} =\frac{x+2}{2x} \\\\\\x=2,5\\\\\\\frac{x+2}{2x} =\frac{2,5+2}{2\cdot 2,5} =\frac{4,5}{5} =0,9\\\\\\Otvet \ : \ \frac{x+2}{2x} \ \ ; \ \ 0,9$

$\displaystyle\bf\\2)\\\\\frac{1}{c^{2} -10cd+25d^{2} } -\frac{1}{c^{2} -25d^{2} } +\frac{1}{c^{2} +10cd+25d^{2} } =\\\\\\=\frac{1}{(c-5d)^{2} } -\frac{1}{(c -5d)(c+5d)} +\frac{1}{(c+5d)^{2} } =\\\\\\=\frac{(c+5d)^{2} -(c^{2} -25d^{2})+(c-5d)^{2} }{(c-5d)^{2} (c+5d)^{2} } =\\\\\\=\frac{c^{2} +10cd+25d^{2} -c^{2} +25d^{2} +c^{2} -10cd+25d^{2} }{(c-5d)^{2} (c+5d)^{2} } =\\\\\\=\frac{c^{2} +75d^{2} }{(c-5d)^{2}(c+5d)^{2} }$

Все члены дроби положительны .

$\displaystyle\bf\\3)\\\\\frac{2d+1}{2cd} -\frac{c-1}{c^{2} } =\frac{(2d+1)\cdot c-(c-1)\cdot2d}{2c^{2} d} =\\\\\\=\frac{2cd+c-2cd+2d}{2c^{2} d} =\frac{2d+c}{2c^{2} d} \\\\4)\\\\\frac{k^{2} }{k+6} +6-k=\frac{k^{2}+(6-k)\cdot(k+6) }{k+6} =\\\\\\=\frac{k^{2} +36-k^{2} }{k+6} =\frac{36}{k+6} \\\\5)\\\\-\frac{6}{c^{2}+12c+36 } +\frac{1}{c+6} =\frac{1}{c+6} -\frac{6}{(c+6)^{2} } =\\\\\\=\frac{c+6-6}{(c+6)^{2} } =\frac{c}{(c+6)^{2} }$