số mol H₂ là: n_H2= \(\dfrac{6,72}{22,4}=0,3mol\)

pthh 2M + 3H₂SO4 -> M₂(SO₄)₃+ 3H₂

2mol 3mol

0,2mol <-- 0,3 mol

M_M= \(\dfrac{5,4}{0,2}\)

= 27g/mol

=> chất đó là Al

theo đề ta có n+e+p=48

<=> 2p + n = 48 (10

và có 2p = 2 n

<=> n = p (2)

từ 1 và 2 => 3p = 48 <=> p = n = e = 16

b.

\(\dfrac{x+2}{x-1}>0\)

\(1-\dfrac{3}{x-1}>0\)

\(\dfrac{3}{x-1}< 1\)

\(x-1< 3\)

\(x< 4\)

mấy cái kia tương tự

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

\(\dfrac{\left(x^2-x-2x+2\right)\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(x^2-2x-2x+4\right)}\)

\(\dfrac{\left[x\left(x-1\right)-2\left(x-1\right)\right]\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left[x\left(x-2\right)-2\left(x-2\right)\right]}\)

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