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\(FeO:\%m_{Fe}=\left(56.100\right):\left(56+16\right)=77,78\%\)
\(Fe_2O_3:\%m_{Fe}=\left(56.2.100\right):\left(56.2+16.3\right)=70\%\)
\(Fe_3O_4:\%m_{Fe}=\left(56.3.100\right):\left(56.3+16.4\right)=72,71\%\)
\(FeCO_3:\%m_{Fe}=\left(56.100\right):\left(56+12+16.3\right)=48,28\%\)
a, \(P=\left(\frac{1}{x-1}+\frac{x}{x^2-1}\right):\left(\frac{x}{x^2-1}\right)\)
\(=\left(\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{x}{x^2-1}\right)\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)}{x}=\frac{2x+1}{x}\)
b, Ta có : \(\frac{2x+1}{x}=2\Leftrightarrow2x+1=2x\Leftrightarrow0\ne-1\)Vậy PT vô nghiệm
mk chưa học
\(\%m_{Fe\left(FeO\right)}=\frac{56}{56+16}\cdot100\%\approx77,78\%\)
\(\%m_{Fe\left(Fe_2O_3\right)}=\frac{56\cdot2}{56\cdot2+16\cdot3}\cdot100\%=70\%\)
\(\%m_{Fe\left(Fe_3O_4\right)}=\frac{56\cdot3}{56\cdot3+16\cdot4}\cdot100\%\approx72,41\%\)
\(\%m_{Fe\left(FeCO_3\right)}=\frac{56}{56+12+16\cdot3}\cdot100\%\approx48,28\%\)