Question

Решите два уравнения пожалуйста

Answer 1

$1)(\sqrt{x+2\sqrt{x-1} }-\sqrt{x-2\sqrt{x-1} })^{2} =3^{2}\\\\x+2\sqrt{x-1}-2\sqrt{(x+2\sqrt{x-1})(x-2\sqrt{x-1}})+x-2\sqrt{x-1}=9\\\\2x-2\sqrt{x^{2}-4(x-1) }=9\\\\2\sqrt{x^{2}-4x+4 } =2x-9\\\\\sqrt{(x-2)^{2} }=x-4,5\\\\|x-2|=x-4,5\\\\1)\left \{ {x-2=x-4,5} \atop {x-4,5\geq0 }} \right.\\\\\left \{ {{0*x=-2,5} \atop {x\geq4,5 }} \right.$

Решений нет

$2)\left \{ {{x-2=4,5-x} \atop {x-4,5\geq0 }} \right.\\\\\left \{ {{y=2} \atop {x=2}} \right. \left \{ {{x+x=6,5} \atop {x\geq4,5 }} \right. \\\\\left \{ {{2x=6,5} \atop {x\geq4,5 }} \right. \\\\\left \{ {{x=3,25} \atop {x\geq4,5 }} \right.$

Решений нет

Ответ : решений нет

$4^{x+1}-13*6^{x}+9^{x+1}=0\\\\4*4^{x}-13*6^{x}+9*9^{x}=0|:9^{x}>0 \\\\4*(\frac{4}{9})^{x}-13*(\frac{6}{9})^{x}+9*(\frac{9}{9})^{x}=0\\\\4*(\frac{2}{3})^{2x} -13*(\frac{2}{3})^{2}+9=0\\\\(\frac{2}{3})^{x}=m,m>0$

$4m^{2}-13m+9=0\\\\D=(-13)^{2}-4*4*9=169-144=25=5^{2}\\\\m_{1}=\frac{13+5}{8}=\frac{9}{4}\\\\m_{2}=\frac{13-5}{8}=1$

$(\frac{2}{3})^{x} =\frac{9}{4} \\\\(\frac{2}{3})^{x}=(\frac{2}{3})^{-2}\\\\x_{1}=-2\\\\(\frac{2}{3})^{x}=1\\\\(\frac{2}{3})^{x}=(\frac{2}{3})^{o}\\\\x_{2}=0$

Ответ : - 2 , 0