Question

80 баллов задание в файле

Answer 1

$\displaystyle\bf\\1)\\\\\frac{4x^{2} -2}{x-9} =\frac{4x+1}{x-9} \\\\4x^{2} -2=4x+1 \ \ ; \ \ x-9\neq 0\\\\4x^{2} -2-4x-1=0 \ \ ; \ \ x\neq 9\\\\4x^{2} -4x-3=0\\\\D=(-4)^{2} -4\cdot4\cdot(-3)=16+48=64=8^{2} \\\\x_{1} =\frac{4-8}{8} =-0,5\\\\x_{2} =\frac{4+8}{8} =1,5\\\\Otvet:-0,5 \ ; \ 1,5$

$\displaystyle\bf\\2)\\\\\frac{3x^{2} -7x}{x-1} =\frac{2x^{2} +2}{1-x} \\\\\\\frac{3x^{2} -7x}{x-1} =\frac{-(2x^{2} +2)}{x-1} \\\\3x^{2} -7x=-2x^{2} -2 \ \ ; \ \ x-1\neq 0\\\\3x^{2} -7x+2x^{2} +2=0 \ \ ; \ \ x\neq 1\\\\5x^{2} -7x+2=0\\\\D=(-7)^{2} -4\cdot 5\cdot 2=49-40=9=3^{2} \\\\x_{1} =\frac{7-3}{10}=0,4\\\\x_{2} =\frac{7+3}{10} =1-neyd\\\\Otvet:0,4$

$\displaystyle\bf\\3)\\\\\frac{3}{a+2}+1=\frac{4}{a^{2} +4a+4} \\\\\\ \frac{3}{a+2}+1-\frac{4}{a^{2} +4a+4}=0\\\\\\\frac{3}{a+2}+1-\frac{4}{(a+2)^{2} }=0\\\\\\\frac{3\cdot(a+2)+a^{2} +4a+4-4}{(a+2)^{2} } =0\\\\\\\frac{3a+6+a^{2} +4a}{(a+2)^{2} } =0\\\\\\\frac{a^{2} +7a+6}{(a+2)^{2} } =0\\\\a^{2} +7a+6=0 \ \ ; \ \ a+2\neq 0 \ \ ; \ \ a\neq -2\\\\D=7^{2}-4\cdot6=49-24=25=5^{2} \\\\x_{1} =\frac{-7-5}{2} =-6\\\\x_{2} =\frac{-7+5}{2} =-1\\\\Otvet:-6 \ ; \ -1$