Question

разложите трехчлен пожалуйста
$\frac{3 {x}^{2} { -x - 2}^{} }{9 {x}^{2} - 4 }$
$\frac{4 {a}^{2} + 12a + 9 }{2 {a}^{2} + a - 3 }$
​

Answer 1

$\displaystyle\bf\\1)\\\\3x^{2} -x-2=0\\\\D=(-1)^{2} -4\cdot 3\cdot (-2)=1+24=25=5^{2} \\\\\\x_{1} =\frac{1-5}{6} =-\frac{4}{6} =-\frac{2}{3} \\\\\\x_{2} =\frac{1+5}{6} =1\\\\\\3x^{2} -x-2=3\cdot\Big(x+\frac{2}{3}\Big)\Big(x-1\Big) =(3x+2)(x-1)\\\\\\\\9x^{2} -4=(3x)^{2} -2^{2} =(3x+2)(3x-2)\\\\\\\\\frac{3x^{2} -x-2}{9x^{2} -4} =\frac{(3x+2)(x-1)}{(3x+2)(3x-2)} =\frac{x-1}{3x-2}$

$\displaystyle\bf\\2)\\\\4a^{2}+12a+9=(2a)^{2} +2\cdot 2a\cdot 3+3^{2}=(2a+3)^{2} \\\\\\2a^{2} +a-3=0\\\\D=1^{2} -4\cdot 2\cdot(-3)=1+24=25=5^{2} \\\\\\a_{1} =\frac{-1-5}{4}=-\frac{6}{4} =-1,5\\\\\\a_{2} =\frac{-1+5}{4} =\frac{4}{4} =1\\\\\\2a^{2} +a-3=2\cdot(a+1,5)(a-1)=(2a+3)(a-1)\\\\\\\\\frac{4a^{2} +12a+9}{2a^{2}+a-3 } =\frac{(2a+3)^{2} }{(2a+3)(a-1)} =\frac{2a+3}{a-1}$