Показательное уравнение Алгебра 9-10 класс 15 БАЛЛОВ

Question 1

Question 2

0 \\\\2^{\sqrt{x}-1 }=m^{2}\\\\3m^{2}-8m+4=0\\\\D=(-8)^{2}-4*3*4=64-48=16=4^{2}\\\\m_{1}=\frac{8+4}{6}=2\\\\m_{2}=\frac{8-4}{6}=\frac{2}{3}" alt="3*2^{\frac{x-1}{\sqrt{x} +1} }-8*2^{\frac{\sqrt{x}-1 }{2} }+4=0\\\\3*2^{\frac{(\sqrt{x} -1)(\sqrt{x}+1) }{\sqrt{x}+1 } }-8*2^{\frac{\sqrt{x}-1 }{2} }+4=0\\\\3*2^{\sqrt{x}-1 }-8*2^{\frac{\sqrt{x}-1 }{2} }+4=0\\\\2^{\frac{\sqrt{x}-1 }{2} }=m,m>0 \\\\2^{\sqrt{x}-1 }=m^{2}\\\\3m^{2}-8m+4=0\\\\D=(-8)^{2}-4*3*4=64-48=16=4^{2}\\\\m_{1}=\frac{8+4}{6}=2\\\\m_{2}=\frac{8-4}{6}=\frac{2}{3}" align="absmiddle" class="latex-formula">

$2^{\frac{\sqrt{x} -1}{2} }=2\\\\\frac{\sqrt{x}-1 }{2}=1\\\\\sqrt{x}-1=2\\\\\sqrt{x}=3\\\\x_{1} =9\\\\2^{\frac{\sqrt{x}-1 }{2} }=\frac{2}{3}\\\\\frac{\sqrt{x}-1 }{2}=log_{2}\frac{2}{3}\\\\\frac{\sqrt{x}-1 }{2}=1-log_{2}3\\\\\sqrt{x}-1=2-log_{2}9\\\\\sqrt{x}=3-log_{2}9<0$

Ответ : 9

Universalka_zn · Answer 1 · 2020-05-24T14:54:34+0000

0 \\\\2^{\sqrt{x}-1 }=m^{2}\\\\3m^{2}-8m+4=0\\\\D=(-8)^{2}-4*3*4=64-48=16=4^{2}\\\\m_{1}=\frac{8+4}{6}=2\\\\m_{2}=\frac{8-4}{6}=\frac{2}{3}" alt="3*2^{\frac{x-1}{\sqrt{x} +1} }-8*2^{\frac{\sqrt{x}-1 }{2} }+4=0\\\\3*2^{\frac{(\sqrt{x} -1)(\sqrt{x}+1) }{\sqrt{x}+1 } }-8*2^{\frac{\sqrt{x}-1 }{2} }+4=0\\\\3*2^{\sqrt{x}-1 }-8*2^{\frac{\sqrt{x}-1 }{2} }+4=0\\\\2^{\frac{\sqrt{x}-1 }{2} }=m,m>0 \\\\2^{\sqrt{x}-1 }=m^{2}\\\\3m^{2}-8m+4=0\\\\D=(-8)^{2}-4*3*4=64-48=16=4^{2}\\\\m_{1}=\frac{8+4}{6}=2\\\\m_{2}=\frac{8-4}{6}=\frac{2}{3}" align="absmiddle" class="latex-formula">

$2^{\frac{\sqrt{x} -1}{2} }=2\\\\\frac{\sqrt{x}-1 }{2}=1\\\\\sqrt{x}-1=2\\\\\sqrt{x}=3\\\\x_{1} =9\\\\2^{\frac{\sqrt{x}-1 }{2} }=\frac{2}{3}\\\\\frac{\sqrt{x}-1 }{2}=log_{2}\frac{2}{3}\\\\\frac{\sqrt{x}-1 }{2}=1-log_{2}3\\\\\sqrt{x}-1=2-log_{2}9\\\\\sqrt{x}=3-log_{2}9<0$

Ответ : 9