Question

помогите срочно! нужны номера 4 5 и 6

Answer 1

$4)2\sqrt{(x-2)(4-x)}=x-2\\\\\left \{ {{(2\sqrt{(x-2)(4-x)})^{2}}=(x-2)^{2} \atop {x-2\geq0 }} \right.\\\\\left \{ {{4(x-2)(4-x)=x^{2} -4x+4} \atop {x\geq2 }} \right.\\\\\left \{ {{4(4x-x^{2}-8+2x)=x^{2}-4x+4} \atop {x\geq2 }} \right. \\\\\left \{ {{-4x^{2}+24x-32=x^{2}-4x+4} \atop {x\geq2}} \right. \\\\\left \{ {{5x^{2}-28x+36=0 } \atop {x\geq2}} \right.\\\\5x^{2}-28x+36=0\\\\\frac{D}{4}=(-\frac{28}{2})^{2}-5*36=196-180=16=4^{2}\\\\x_{1}=\frac{14-4}{5}=2$

$x_{2}=\frac{14+4}{5}=3,6\\\\Otvet:\boxed{2;3,6}$

$5)\sqrt{3x+1}+\sqrt{x}=11\\\\ODZ:\\\left \{ {{3x+1\geq0 } \atop {x\geq0 }} \right. \\\\\left \{ {{x\geq-\frac{1}{3}} \atop {x\geq0 }} \right.\Rightarrow x\geq0\\\\(\sqrt{3x+1} )^{2}=(11-\sqrt{x})^{2} \\\\3x+1=121-22\sqrt{x}+x\\\\2x+22\sqrt{x}-120=0\\\\x+11\sqrt{x}-60=0\\\\\sqrt{x}=m,m\geq0\\\\m^{2}+11m-60=0\\\\m_{1}=4\\\\m_{2}=-15<0-neyd\\\\\sqrt{x}=4\\\\x=16\\\\Otvet:\boxed{16}$

$6)\sqrt{2x^{2}-2x }=3-\sqrt{x^{2}-x-1} \\\\(\sqrt{2x^{2}-2x })^{2} =(3-\sqrt{x^{2}-x-1})^{2}\\\\2x^{2}-2x=9-6\sqrt{x^{2}-x-1 }+x^{2}-x-1\\\\6\sqrt{x^{2}-x-1 }=-x^{2}+x+8\\\\6\sqrt{x^{2}-x-1 }=-(x^{2}-x-1)+7\\\\\sqrt{x^{2}-x-1 }=m,m\geq 0\\\\6m=-m^{2}+7\\\\m^{2}+6m-7=0\\\\m_{1} =1\\\\m_{2}=-7<0-neyd\\\\\sqrt{x^{2}-x-1 } =1\\\\(\sqrt{x^{2}-x-1 })^{2} =1^{2}\\\\x^{2} -x-1=1\\\\x^{2}-x-2=0\\\\x_{1} =-1\\\\x_{2}=2\\\\Proverka:$

$1)\sqrt{2*(-1)^{2} -2*(-1)}=3-\sqrt{(-1)^{2}-(-1)-1 }\\\\\sqrt{4}=3-\sqrt{1}\\\\2=2-verno\\\\2)\sqrt{2*2^{2}-2*2 }=3-\sqrt{2^{2}-2-1 }\\\\\sqrt{4}=3-\sqrt{1}\\\\2=2-verno\\\\Otvet:\boxed{-1;2}$