$Fe_2O_3 + 3H_2SO_4 \to Fe_2(SO_4)_3 + 3H_2O$

Theo PTHH:  $n_{H_2O} = n_{H_2SO_4} = 0,5.0,1 = 0,05(mol)$
Bảo toàn khối lượng : 

$m_{muối} = 2,81 + 0,05.98 - 0,05.2 = 7,61(gam)$

a: \(A=\dfrac{2x^2+x^2-1-2x^2+2x+1}{x\left(x+1\right)}=\dfrac{x^2+2x}{x\left(x+1\right)}=\dfrac{x+2}{x+1}\)

b: Ta có: \(x^2-2x=0\)

=>x=2

Thay x=2 vào A, ta được:

\(A=\dfrac{2+2}{2+1}=\dfrac{4}{3}\)

(a)

\(A=\dfrac{2x}{x+1}+\dfrac{x-1}{x}-\dfrac{2x^2-2x-1}{x^2+x}\\ =\dfrac{2x}{x+1}+\dfrac{x-1}{x}-\dfrac{2x^2-2x-1}{x\left(x+1\right)}=\dfrac{2x^2}{x\left(x+1\right)}+\dfrac{x^2-1}{x\left(x+1\right)}-\dfrac{2x^2-2x-1}{x\left(x-1\right)}\)

\(=\dfrac{2x^2+x^2-1-2x^2+2x+1}{x\left(x+1\right)}=\dfrac{x^2+2x+1}{x\left(x+1\right)}=\dfrac{\left(x+1\right)\left(x+1\right)}{x\left(x+1\right)}=\dfrac{x+1}{x}\)

(b)

\(x^2-2x=0\\ x\left(x-2\right)=0\)

=>x=0 hoặc x=2 mà đk x khác 0 nên thay x=2 vào bt A , ta có:

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

1.He doesn't work in the city anymore
2.They go to work on foot
3.Mai goes to her English class by bike
4.My bike is different from your bike
5.Lan's style is not the same as Minh's

mắ nhìn cún sách ám ảnh vải sau có key với giải thích chi tiết á b

(a) \(A=\dfrac{3}{x-2}\in Z\)

\(\Rightarrow\left(x-2\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x=4\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{-2;0;2;4\right\}.\)

(b) \(B=-\dfrac{11}{2x-3}\in Z\)

\(\Rightarrow\left(2x-3\right)\inƯ\left(11\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=1\\2x-3=-1\\2x-3=11\\2x-3=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=7\\x=-4\end{matrix}\right.\)

Vậy: \(x\in\left\{-4;1;2;7\right\}.\)

(c) \(C=\dfrac{x+3}{x+1}=\dfrac{\left(x+1\right)+2}{x+1}=1+\dfrac{2}{x+1}\in Z\Rightarrow\dfrac{2}{x+1}\in Z\)

\(\Rightarrow\left(x+1\right)\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

\(\Rightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\\x+1=2\\x+1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=1\\x=-3\end{matrix}\right.\)

Vậy: \(x\in\left\{-3;-2;0;1\right\}.\)

(d) \(D=\dfrac{2x+10}{x+3}=\dfrac{2\left(x+3\right)+4}{x+3}=2+\dfrac{4}{x+3}\in Z\Rightarrow\dfrac{4}{x+3}\in Z\)

\(\Rightarrow\left(x+3\right)\inƯ\left(4\right)=\left\{\pm1;\pm2\pm4\right\}\)

\(\Rightarrow x\in\left\{-2;-4;-1;-5;1;-7\right\}\)