1)

\(\dfrac{1}{\sqrt{5}-1}-\dfrac{4}{\sqrt{5}+1}=\dfrac{1+\sqrt{5}}{4}-\dfrac{4\sqrt{5}-4}{4}\) \(=\dfrac{5-3\sqrt{5}}{4}\)

2)

HPT \(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=5\\9x+3y=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}7x=7\\y=4-3x\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(1;1\right)\) 

1: Rút gọn

Ta có: \(\dfrac{1}{\sqrt{5}-1}-\dfrac{4}{\sqrt{5}+1}\)

\(=\dfrac{\sqrt{5}+1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}-\dfrac{4\left(\sqrt{5}-1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)

\(=\dfrac{\sqrt{5}+1-4\sqrt{5}+4}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)

\(=\dfrac{5-3\sqrt{5}}{4}\)

2)

Ta có: \(\left\{{}\begin{matrix}2x+3y=5\\3x+y=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x+9y=15\\6x+2y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7y=7\\3x+y=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\3x=4-y=4-1=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\) 

\(M=\dfrac{8\left(\sqrt{5}+\sqrt{3}\right)}{2}-\dfrac{7\left(2+\sqrt{3}\right)}{4-3}+\dfrac{4\left(\sqrt{2}+1\right)}{2-1}+\dfrac{\sqrt{15}\left(\sqrt{3}-1\right)}{\sqrt{15}}\)

\(=4\left(\sqrt{5}+\sqrt{3}\right)-14-7\sqrt{3}+4\sqrt{2}+4+\sqrt{3}-1\)

\(=4\sqrt{5}+4\sqrt{3}-6\sqrt{3}+4\sqrt{2}-11\)

\(=4\sqrt{5}-2\sqrt{3}+4\sqrt{2}-11\)

\(M=\dfrac{8\left(\sqrt{5}+\sqrt{3}\right)}{5-3}+\dfrac{7\left(\sqrt{3}+2\right)}{3-4}+\dfrac{4\left(\sqrt{2}+1\right)}{2-1}+\dfrac{\sqrt{15}\left(\sqrt{3}-1\right)}{\sqrt{15}}\)

\(=4\sqrt{5}+4\sqrt{3}-7\sqrt{3}-14+4\sqrt{2}+4+\sqrt{3}-1\)

\(=4\sqrt{5}-2\sqrt{3}+4\sqrt{2}-11\)

\(\dfrac{2}{x^2-y^2}.\sqrt{\dfrac{3x^2+6xy+3y^2}{4}}\)

\(ĐK:x\ne\pm y\)

\(=\dfrac{2\left|x+y\right|}{2\left(x+y\right)\left(x-y\right)}=\dfrac{\sqrt{3}\left|x+y\right|}{\left(x+y\right)\left(x-y\right)}\)

Nếu x > -y thì x + y > 0 , ta có :\(\dfrac{\sqrt{3}}{x-y}\)

Nếu x < -y thì x + y < 0 , ta có :\(\dfrac{-\sqrt{3}}{x-y}\)

\(\dfrac{8}{\sqrt{5}-1}-\dfrac{22}{4+\sqrt{5}}+\dfrac{\sqrt{15}+2\sqrt{5}}{2+\sqrt{3}}\)
\(=\dfrac{8\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}-\dfrac{22\left(4-\sqrt{5}\right)}{\left(\sqrt{5}+4\right)\left(4-\sqrt{5}\right)}+\dfrac{\sqrt{5}\left(\sqrt{3}+2\right)}{2+\sqrt{3}}\)
\(=\dfrac{8\sqrt{5}+8}{5-1}-\dfrac{88-22\sqrt{5}}{16-5}+\sqrt{5}\)
\(=\dfrac{8\sqrt{5}+8}{4}-\dfrac{88-22\sqrt{5}}{11}+\sqrt{5}\)
\(=2\sqrt{5}+2-8+2\sqrt{5}+\sqrt{5}=5\sqrt{5}-6\)

\(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)

\(=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\)

\(=\sqrt{5}+\dfrac{\sqrt{5}}{2}\)

\(=\dfrac{2\sqrt{5}+\sqrt{5}}{2}\)

\(=\dfrac{3\sqrt{5}}{2}\)

Sửa đề: \(A=\dfrac{1-3x}{2y}\cdot\sqrt{\dfrac{36y^2}{9x^2-6x+1}}\)

\(=\dfrac{1-3x}{2y}\cdot\sqrt{\left(\dfrac{6y}{3x-1}\right)^2}\)

\(=\dfrac{1-3x}{2y}\cdot\left|\dfrac{6y}{3x-1}\right|\)

x>1/3 nên 3x-1>0

y>0 nên 6y>0

=>\(A=\dfrac{1-3x}{2y}\cdot\dfrac{6y}{3x-1}=-3y\)

`c)1/(2sqrt2)-3/2sqrt{4,5}+2/5sqrt{50}`

`=1/(2sqrt2)-3/2sqrt{9/2}+2/5sqrt{25.2}`

`=1/(2sqrt2)-9/(2sqrt2)+2sqrt2`

`=2sqrt2-8/(2sqrt2)`

`=2sqrt2-sqrt2=sqrt2`

`d)4/(3+sqrt5)-8/(1+sqrt5)+15/sqrt5`

`=(4(3-sqrt5))/(9-5)-(8(sqrt5-1))/(5-1)+3sqrt5`

`=3-sqrt5-2(sqrt5-1)+3sqrt5`

`=3+3sqrt5-3sqrt5+2=5`

\(A=\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}-\dfrac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}\)

\(=\dfrac{\sqrt{3}-\sqrt{2}}{3-2}-\sqrt{3}=\sqrt{3}-\sqrt{2}-\sqrt{3}\)

\(=-\sqrt{2}\)

\(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\sqrt{\dfrac{3}{7}}\)

\(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}=\sqrt{5}\)

\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}=\dfrac{2\left(\sqrt{8}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}=-\dfrac{2\sqrt{6}}{6}\)

`(sqrt 15 - sqrt 6)/(sqrt 35 - sqrt 14)`

`= (sqrt 3 . (sqrt 5 - sqrt 2))/(sqrt 7. (sqrt 5 - sqrt 2))`

`= sqrt3/sqrt 7`

`@ (sqrt 15 - sqrt 5)/(sqrt 3 - 1)`

`= (sqrt 5(sqrt 3 - 1))/(sqrt 3 - 1)`

`= sqrt5`

`@ (2 sqrt 8 - sqrt 12)/(sqrt18 - sqrt 48)`

`= (2(sqrt 8 - sqrt 3)/(sqrt 6(sqrt 3 - sqrt 8))`

`= (-2)/(sqrt 6) = (-2 sqrt 6)/6`