Question

Решите пожалуйста уравнение срочно!

Answer 1

Ответ:

$x=2$

Объяснение:

$x^{2} -4x-\sqrt{2x^{2} -8x+12}+6=0\\-\sqrt{2x^{2} -8x+12} =x^{2} +4x-6\\\sqrt{2x^{2} -8x+12} =x^{2}-4x+6\\2x^{2} -8x+12=(x^{2} -4x+6)^{2}\\2(x^{2} -4x+6)=(x^{2} -4x+6)^{2}=0 \\(x^{2} -4x+6)(2-x^{2} +4x-6)=0\\(x^{2} -4x+6)(-4-x^{2} +4x)=0\\x^{2}-4x+6=0\\-4-x^{2} +4x=0\\x=2$

Answer 2

$x^{2}-4x-\sqrt{2x^{2}-8x+12 }+6=0\\\\\frac{1}{2} (2x^{2}-8x+12)-\sqrt{2x^{2}-8x+12} =0\\\\\sqrt{2x^{2}-8x+12}=m,m\geq0\Rightarrow2x^{2}-8x+12= m^{2}\\\\\frac{1}{2}m^{2}-m=0|*2\\\\m^{2}-2m=0\\\\m(m-2)=0\\\\m_{1} =0\\\\m_{2} =2\\\\1)\sqrt{2x^{2}-8x+12 } =0\\\\2x^{2}-8x+12=0\\\\x^{2} -4x+6=0\\\\D=(-4)^{2} -4*6=16-24=-8<0-kornei...net\\\\2)\sqrt{2x^{2}-8x+12 } =2\\\\2x^{2}-8x+12=4\\\\2x^{2}-8x+8=0\\\\x^{2}-4x+4=0\\\\(x-2)^{2}=0\\\\x-2=0\Rightarrow \boxed{x=2}$