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Với \(y\ge0\)hệ tương đương với: 

\(\hept{\begin{cases}2x+3y=13\\3x-y=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}\)(tm) 

Với \(y< 0\)hệ tương đương với: 

\(\hept{\begin{cases}2x-3y=13\\3x-y=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{4}{7}\\y=-\frac{33}{7}\end{cases}}\)(tm) 

\(a^4+3a^2+2a-6=0\)

\(\Leftrightarrow a^4-a^3+a^3-a^2+4a^2-4a+6a-6=0\)

\(\Leftrightarrow a^3\left(a-1\right)+a^2\left(a-1\right)+4a\left(a-1\right)+6\left(a-1\right)=0\)

\(\Leftrightarrow\left(a-1\right)\left(a^3+a^2+4a+6\right)=0\)

\(\Leftrightarrow a-1=0\)(vì \(a\ge0\)nên \(a^3+a^2+4a+6>0\)) 

\(\Leftrightarrow a=1\)(tm).

a) \(\frac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\frac{1}{4+\sqrt{15}}=\frac{\sqrt{15}.\sqrt{5}-\sqrt{15}.\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{4-\sqrt{15}}{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}\)

\(=\sqrt{15}+4-\sqrt{15}=4\)

b) \(\left(\frac{2}{3+\sqrt{5}}+\frac{1}{2+\sqrt{5}}\right).\sqrt{6+2\sqrt{5}}=\left[\frac{2\left(3-\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}+\frac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\right].\sqrt{5+2\sqrt{5}+1}\)\(=\left(\frac{3-\sqrt{5}}{2}-2+\sqrt{5}\right).\sqrt{\left(\sqrt{5}+1\right)^2}=\frac{\sqrt{5}-1}{2}.\left(\sqrt{5}+1\right)=2\)

\(\hept{\begin{cases}\frac{x-y}{4}=\frac{x-y}{3}\\\frac{x}{4}=\frac{y}{2}+1\end{cases}}\Rightarrow\hept{\begin{cases}3\left(x-y\right)=4\left(x-y\right)\\\frac{x}{4}=\frac{y+2}{2}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}3x-3y=4x-4y\\2x=4\left(y+2\right)\end{cases}}\Leftrightarrow\hept{\begin{cases}3x-3y-4x+4y=0\\2x-4y=8\end{cases}}\Leftrightarrow\hept{\begin{cases}-x+y=0\left(1\right)\\x-2y=4\left(2\right)\end{cases}}\)

Lấy (1) + (2) theo vế

\(\Leftrightarrow\hept{\begin{cases}-y=4\\x-2y=4\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-4\\x-2y=4\end{cases}}\Leftrightarrow x=y=-4\)

Vậy hệ phương trình có nghiệm x = y = -4

\(\Leftrightarrow\hept{\begin{cases}4\left(x-y\right)=3\left(x-y\right)\\x=2y+4\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=y\left(1\right)\\2y-x+4=0\left(2\right)\end{cases}}\)

Thay (1) vào (2) ta có  

       x + 4 = 0 

<=> x       = - 4 

Thay x = -4 vào ( 1 ) ta có y = - 4 

Vậy ( x ; y ) = ( -4 ; -4 ) 

\(\Leftrightarrow\hept{\begin{cases}5x+5y=4x-3\\4x+12y=15-9y\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x+5y=-3\\4x+21y=15\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}4x+20y=-12\left(1\right)\\4x+21y=15\left(2\right)\end{cases}}\)

Lấy (2) - (1) ta có : 

y = 27

Thay y = 17 vào ( 1 ) ta có x = -138

Vậy ( x ; y ) = ( -138 ; 27 ) 

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