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s mình tìm k ra nhỉ

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

PTHH: 2Al + 6H₂SO₄ --> Al₂(SO₄)₃ + 3SO₂ + 6H₂O

_____0,2-------------------------------->0,3

=> V_SO2 = 0,3.22,4 = 6,72 (l)

\(PTPU:2Al+6H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+6H_2O+3SO_2\)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ Theo.pt:n_{SO_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\\ \Rightarrow V_{SO_2}=0,3.22,4=6,72\left(l\right)\)

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

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b,=\(\dfrac{x+1}{x+3}+\dfrac{x-1}{x-3}+\dfrac{2x-2x^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x+1\right)\left(x-3\right)+\left(x-1\right)\left(x+3\right)+2x-2x^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+x-3x-3+x^2-x+3x-3+2x-2x^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+2x-6}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+2}{x+3}\)

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

11)

\(\dfrac{2x}{x+2}\) \(\times\) \(\dfrac{x^{2^{ }}-4}{4}\) - \(\dfrac{x}{2}\) 

= \(\dfrac{2x}{x+2}\) \(\times\) \(\dfrac{x^2-2^2}{4}\) - \(\dfrac{x}{2}\) 

= \(\dfrac{2x}{x+2}\) \(\times\) \(\dfrac{\left(x-2\right)\left(x+2\right)}{4}\) - \(\dfrac{x}{2}\) 

= \(\dfrac{x\left(x-3\right)}{2}\)

e) Ta có: \(x^3-4x-14x\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-14x\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x+2-14\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=12\end{matrix}\right.\)

e)x³-4x+14x(x-2)=0

⇔ x(x²-4)+14x(x-2)=0

⇔ x(x-2)(x+2)+14x(x-2)=0

⇔ (x-2)(x²+2x+14x)=0

⇔ x(x-2)(x+16)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\\x+16=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\\x=-16\end{matrix}\right.\)

g)x²(x+1)-x(x+1)+x(x-1)=0

⇔ (x+1)(x²-x)+x(x-1)=0

⇔ x(x+1)(x-1)+x(x-1)=0

⇔ x(x-1)(x+2)=0

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