\(\frac{2}{x}+\frac{3}{x-1}=2\)ĐKXĐ : \(x\ne0;1\)

\(\Leftrightarrow\frac{2\left(x-1\right)}{x\left(x-1\right)}+\frac{3x}{x\left(x-1\right)}=\frac{2x\left(x-1\right)}{x\left(x-1\right)}\)

\(\Rightarrow2x-2+3x=2x^2-2x\)

\(\Leftrightarrow5x-2+2x-2x^2=0\Leftrightarrow-2x^2+7x-2=0\)

Ta có : \(\Delta=7^2-4\left(-2\right)\left(-2\right)=49-16=33\)

\(\Leftrightarrow x=\frac{-7\pm\sqrt{33}}{-4}\)

Dùng phương pháp thế là dễ nhật rồi 

a, \(\hept{\begin{cases}3x+y=5\\x-2y=-3\end{cases}}\Leftrightarrow\hept{\begin{cases}y=5-3x\left(1\right)\\x-2y=-3\left(2\right)\end{cases}}\)

Thay (1) vào (2) ta được : 

\(x-2\left(5-3x\right)=-3\Leftrightarrow x-10+6x=-3\)

\(\Leftrightarrow7x=7\Leftrightarrow x=1\)

\(\Rightarrow y=5-3x=5-3=2\)

Vậy \(x=1;y=2\)

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

Thay (1) vào (2) ta được : 

\(3x+4\left(1-2x\right)=-1\Leftrightarrow3x+4-8x=-1\)

\(\Leftrightarrow-5x=-5\Leftrightarrow x=1\)

\(\Rightarrow y=1-2x=1-2=-1\)

Vậy \(x=1;y=-1\)

tương tự nhé 

c, \(\hept{\begin{cases}2x+3y=2\\x-y=\frac{1}{6}\end{cases}}\Leftrightarrow\hept{\begin{cases}2x+3y=2\left(1\right)\\x=\frac{1}{6}+y\left(2\right)\end{cases}}\)

Thay (2) vào (1) ta được : 

\(2\left(\frac{1}{6}+y\right)+3y=2\)

\(\Leftrightarrow\frac{1}{3}+2y+3y=2\)

\(\Leftrightarrow5y=\frac{5}{3}\Leftrightarrow y=\frac{1}{3}\)

\(\Rightarrow x=\frac{1}{6}+y=\frac{1}{6}+\frac{1}{3}=\frac{1}{3}\)

Vậy \(x;y=\frac{1}{3}\)

d, \(\hept{\begin{cases}2x+5y=7\\3x-y=2\end{cases}}\Leftrightarrow\hept{\begin{cases}2x+5y=7\left(1\right)\\y=3x-2\left(2\right)\end{cases}}\)

Thay (2) vào (1) ta được : 

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

\(\Leftrightarrow2x+15x-10=7\Leftrightarrow17x=17\Leftrightarrow x=1\)

\(\Rightarrow y=3x-2=3-2=1\)

Vậy \(x;y=1\)

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OMG !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

\(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}\)

\(=\frac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\frac{8\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}\)( trục căn thức )

\(=2\sqrt{5}+\frac{8\left(1+\sqrt{5}\right)}{-4}=2\sqrt{5}-2\left(1+\sqrt{5}\right)\)

\(=2\sqrt{5}-2-2\sqrt{5}=-2\)

Vậy biểu thức nhận giá trị là -2