\(\text{a) Ta có:}xy=1\Rightarrow\hept{\begin{cases}2xy=2\\-2xy=-2\end{cases}}\)

\(\text{Ta lại có: }x^2+y^2=2\Rightarrow\hept{\begin{cases}x^2+y^2+2xy=2+2=4\\x^2+y^2-2xy=2-2=0\end{cases}\Rightarrow\hept{\begin{cases}\left(x+y\right)^2=4\\\left(x-y\right)^2=0\end{cases}\Rightarrow}\hept{\begin{cases}x+y=\pm2\\x-y=0\end{cases}}}\)

\(\text{b) Ta có: }x+y=5\)

\(\Rightarrow\left(x+y\right)^2=25\)

\(\Rightarrow x^2+2xy+y^2=25\)

\(\Rightarrow x^2+4+y^2=25\)

\(\Rightarrow x^2+y^2=21\)

\(\text{b) Ta có: }x^2+y^2=21\)

\(\Rightarrow x^2-2xy+y^2=21-2xy\)

\(\Rightarrow\left(x-y\right)^2=21-4\)

\(\Rightarrow\left(x-y\right)^2=17\)

\(\Rightarrow x-y=\pm\sqrt{17}\)

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

\(\Leftrightarrow\frac{x}{\left(x+1\right)\left(x-1\right)}+\frac{3}{\left(x-3\right)\left(x+1\right)}=\frac{x}{\left(x-3\right)\left(x-1\right)}\)

\(\Leftrightarrow x\left(x-3\right)+3\left(x-1\right)=x\left(x+1\right)\)

\(\Leftrightarrow x^2-3=x^2+x\)

\(\Leftrightarrow-3=x\)

\(\Leftrightarrow x=-3\)

Vậy: nghiệm phương trình là -3

\(3,\text{ }\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16=0\)

\(\Rightarrow\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)=0-16\)

\(\Rightarrow\text{ Có lẻ thừa số âm }\)

Mà \(\left(x+8\right)>\left(x+6\right)>\left(x+4\right)>\left(x+2\right)\)

Ta có hai trường hợp : 

\(TH\text{ }1\text{ :}\) Có một thừa số âm

\(\Rightarrow\text{ }\left(x+2\right)< 0\)

\(\Rightarrow\text{ }x< -2\)

\(TH\text{ }2\text{ : }\) Có 3 thừa số âm

\(\Rightarrow\text{ }\hept{\begin{cases}\left(x+2\right)< 0\\\left(x+4\right)< 0\\\left(x+6\right)< 0\end{cases}}\)                \(\Rightarrow\text{ }\left(x+2\right)< 0\text{ }\Rightarrow\text{ }x< -2\)

Si thì thôi nha ! Mong bạn thông cảm !

\(3\left(x+2\right)+2x+4=3\left(x+2\right)+2\left(x+2\right)=5\left(x+2\right)\)

\(4\left(x-2\right)-x+2=4\left(x-2\right)-\left(x-2\right)=3\left(x-2\right)\)

\(3\left(x+2\right)+2x+4\)

\(=3\left(x+2\right)+2\left(x+2\right)\)

\(=\left(x+2\right).5\)

\(4\left(x-2\right)-x+2\)

\(=4\left(x-2\right)-\left(x-2\right)\)

\(=3.\left(x-2\right)\)

~ Rất vui vì giúp đc bn ~