Ответ:
б)
на фото...................
Пользуемся свойствами степени: [tex]a^{x-y}=a^{x}\cdot a^{-y}=\dfrac{a^{x}}{a^{y}}\ \ ,\ \ a^{-1}=\dfrac{1}{a}[/tex] .
[tex]\Big(\dfrac{1}{2}\Big)^{x}+\Big(\dfrac{1}{2}\Big)^{x-2}=5\\\\\\\Big(\dfrac{1}{2}\Big)^{x}+\Big(\dfrac{1}{2}\Big)^{x}\cdot \Big(\dfrac{1}{2}\Big)^{-2}=5\\\\\\\Big(\dfrac{1}{2}\Big)^{x}\cdot \Big(1+\Big(\dfrac{1}{2}\Big)^{-2}\Big)=5\\\\\\\Big(\dfrac{1}{2}\Big)^{x}\cdot (1+4)=5\\\\\\\Big(\dfrac{1}{2}\Big)^{x}\cdot 5=5\\\\\\\Big(\dfrac{1}{2}\Big)^{x}=1\\\\\\\Big(\dfrac{1}{2}\Big)^{x}=\Big(\dfrac{1}{2}\Big)^{0}\\\\\\x=0[/tex]
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Answers & Comments
Ответ:
б)
на фото...................
Ответ:
Пользуемся свойствами степени: [tex]a^{x-y}=a^{x}\cdot a^{-y}=\dfrac{a^{x}}{a^{y}}\ \ ,\ \ a^{-1}=\dfrac{1}{a}[/tex] .
[tex]\Big(\dfrac{1}{2}\Big)^{x}+\Big(\dfrac{1}{2}\Big)^{x-2}=5\\\\\\\Big(\dfrac{1}{2}\Big)^{x}+\Big(\dfrac{1}{2}\Big)^{x}\cdot \Big(\dfrac{1}{2}\Big)^{-2}=5\\\\\\\Big(\dfrac{1}{2}\Big)^{x}\cdot \Big(1+\Big(\dfrac{1}{2}\Big)^{-2}\Big)=5\\\\\\\Big(\dfrac{1}{2}\Big)^{x}\cdot (1+4)=5\\\\\\\Big(\dfrac{1}{2}\Big)^{x}\cdot 5=5\\\\\\\Big(\dfrac{1}{2}\Big)^{x}=1\\\\\\\Big(\dfrac{1}{2}\Big)^{x}=\Big(\dfrac{1}{2}\Big)^{0}\\\\\\x=0[/tex]