PTHH: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

\(n_{Fe_2O_3}=\dfrac{3,2}{160}=0,02\left(mol\right)\)

\(n_{Fe\left(lt\right)}=2.n_{Fe_2O_3}=2.0,02=0,04\left(mol\right)\)

\(m_{Fe\left(lt\right)}=0,04.56=2,24\left(g\right)\)

\(H=\dfrac{m_{tt}}{m_{lt}}.100\%=\dfrac{1,792}{2,24}.100\%=80\%\)

1) b) \(A=x^4+2x^3+3x^2+2x+1\)

\(A=\left(x^2+x\right)^2+\left(x+1\right)^2+x^2\)

Min_A=0 khi x=-1

3) c) \(A=\dfrac{-8}{x^2-2x+5}\)

\(B=x^2-2x+5=\left(x-1\right)^2+4\)

\(B\ge4\Rightarrow A\ge-2\)

Vậy: Min_A =-2 khi x=1

b) \(A=\dfrac{5}{x^2-6x+10}\)

\(B=x^2-6x+10=\left(x-3\right)^2+1\)

\(B\ge1\Rightarrow A\ge5\)

Vậy Min_A = 5 khi x=3

3) a) \(A=\dfrac{1}{x^2+x+1}\)

\(B=x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

\(\Rightarrow B\ge\dfrac{3}{4}\Rightarrow A\ge\dfrac{4}{3}\)

Vậy Min_A là \(\dfrac{4}{3}\) khi x=-0,5

2) b) Đặt biểu thức là A

\(A=-x^2+2xy-4y^2+2x+10y-8\)

\(A=-\left(x^2-2xy+4y^2-2x-10y+8\right)\)

\(A=-\left[\left(x-y\right)^2+3y^2-2\left(x-y\right)-12y+8\right]\)

\(A=-\left[\left(x-y-1\right)^2+\left(\sqrt{3y}-2\sqrt{3}\right)^2\right]-12+8\)

Vậy Min_A là 4

1) a) Đặt biểu thức là A

\(A=2x^2+4y^2-4xy-4x-4y+2017\)

\(A=\left(x-2y\right)^2+x^2-4x-4y+2017\)

\(A=\left(x-2y\right)^2+2\left(x-2y\right)+x^2-6x+2017\)

\(A=\left(x-2y-1\right)^2+\left(x+3\right)^2+2008\)

Vậy: Min_A=2008 khi x=-3; y=-2

\(A=x^2-2xy+y^2+x^2+4x+2y+5\)

\(A=\left(x-y\right)^2-2\left(x-y\right)+x^2+6x+5\)

\(A=\left(x-y\right)^2-2\left(x-y\right).1+1^2+x^2+6x+4\)

\(A=\left(x-y-1\right)^2+\left(x+3\right)^2-5\)

Vậy: Min_A = -5 khi............

\(2xFe+yO_2\rightarrow2Fe_xO_y\)

Theo PTHH, ta có: \(\dfrac{3,46}{2x.56}=\dfrac{4,64}{2\left(56x+16y\right)}\)

\(3,46.2\left(56x+16y\right)=4,64.2x.56\)

\(387,52x+110,72y=519,68x\)

\(110,72y=132,16x\)

\(\dfrac{x}{y}=\dfrac{110,72}{132,16}\simeq\)\(\dfrac{3}{4}\)

CT: Fe₃O₄

Đề cho số không xấp xỉ lắm........................