Bạn cho mình xem đề đi bạn

Câu 3:

\(n_{CuO}=\dfrac{6,4}{80}=0,08(mol)\\ n_{HCl}=\dfrac{36,5.20}{100.36,5}=0,2(mol)\\ CuO+2HCl\to CuCl_2+H_2\\ \dfrac{n_{CuO}}{1}<\dfrac{n_{HCl}}{2}\Rightarrow HCl\text{ dư}\\ \Rightarrow n_{HCl(dư)}=0,2-0,08.2=0,04(mol)\\ n_{CuCl_2}=n_{H_2}=0,08(mol);n_{HCl(p/ứ)}=0,16(mol)\\ \Rightarrow C\%_{CuCl_2}=\dfrac{0,08.135}{6,4+36,5-0,08.2}.100\%=25,27\%\\ C\%_{HCl(dư)}=\dfrac{0,04.36,5}{6,4+36,5-0,08.2}.100\%=3,42\%\)

in 

I think .-.

Sao bn ko copy ảnh trong phần câu hỏi luôn ik ❓

28: 

\(m\perp n\)

\(n\perp p\)

Do đó: m//p

=>Chọn B

29:

m//n

\(p\perp n\)

Do đó: \(m\perp p\)

=>Chọn A

3: \(\left(3x+5\right)\left(2x-7\right)\)

\(=6x^2-21x+10x-35\)

\(=6x^2-11x-35\)

4: \(\left(5x-2\right)\left(3x+4\right)\)

\(=15x^2+20x-6x-8\)

\(=15x^2+14x-8\)

a: \(=\dfrac{x^3+2x+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)

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

b: \(=\dfrac{x^2-2x-3+x^2+2x-3+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x+3}\)

c: \(=\dfrac{6-7+x}{3\left(x-1\right)}=\dfrac{x-1}{3\left(x-1\right)}=\dfrac{1}{3}\)

d: \(=\dfrac{x^3+2x+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^3-x^2+3x-3}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+3}{x^2+x+1}\)