\(x^2-25-4xy+4y^2\)

\(=\left(x^2-4xy+4y^2\right)-25\)

\(=\left[x^2-2\cdot x\cdot2y+\left(2y\right)^2\right]-25\)

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

\(=\left(x-2y-5\right)\cdot\left(x-2y+5\right)\)

\(\frac{x^2+y^2}{xy}=\frac{10}{3}\Rightarrow3x^2+3y^2-10xy=0\)

\(\Rightarrow\left(3x^2-9xy\right)-\left(xy-3y^2\right)=0\Rightarrow3x\left(x-3y\right)-y\left(x-3y\right)=0\)

\(\Rightarrow\left(x-3y\right)\left(3x-y\right)=0\Rightarrow3x-y=0\left(y>x>0\Rightarrow x-3y< 0\right)\Rightarrow3x=y\)

\(M=\frac{x-y}{x+y}=\frac{x-3x}{x+3x}=\frac{-2x}{4x}=-\frac{1}{2}\)

\(x^4+x^3y-xy^3-y^4=x^3\left(x+y\right)-y^3\left(x+y\right)\)

\(=\left(x+y\right)\left(x^3-y^3\right)=\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)

\(x^4+x^3y-xy^3-y^4\)

\(=x^3.\left(x+y\right)-y^3.\left(x+y\right)\)

\(=\left(x^3-y^3\right).\left(x+y\right)\)

Cách làm khá dễ ạ

Đầu tiên ta đặt như thế này:

\(aFe+bH_2SO_4\rightarrow cFe_2\left(SO_4\right)_3+dSO_2+eH_2O\)

\(\hept{\begin{cases}a=2c\\b=e;b=3c+d\\4b=12c+2d+e\end{cases}}\)cho \(e=1\Rightarrow b=1\)

\(\hept{\begin{cases}b=3c+d\\4b=12c+2d+e\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}1=3c+d\\3=12c+2d\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}c=\frac{1}{6}\\d=\frac{1}{2}\end{cases}}\)

\(\Rightarrow a=\frac{1}{3};b=1;c=\frac{1}{6};d=\frac{1}{2};e=1\)

ta có \(\frac{1}{3};1;\frac{1}{6};\frac{1}{2};1\)

Quy đồng ta được \(\Rightarrow\frac{2}{6};\frac{6}{6};\frac{1}{6};\frac{3}{6};\frac{6}{6}\)

Vậy \(\Rightarrow2Fe+6H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)

=.= mệt qué .V

Tớ cảm ơn cậu rất nhiều ak >.< mong lần sau cậu giúp đỡ tớ nhé