1) Ta có: \(\left\{{}\begin{matrix}3\sqrt{x}-\sqrt{y}=5\\2\sqrt{x}+3\sqrt{y}=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}9\sqrt{x}-3\sqrt{y}=15\\2\sqrt{x}+3\sqrt{y}=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}11\sqrt{x}=33\\3\sqrt{x}-\sqrt{y}=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{y}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\y=16\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=9\\y=16\end{matrix}\right.\)

2) Ta có: \(\left\{{}\begin{matrix}\sqrt{x+3}-2\sqrt{y+1}=2\\2\sqrt{x+3}+\sqrt{y+1}=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-2\sqrt{x+3}+4\sqrt{y+1}=-4\\2\sqrt{x+3}+\sqrt{y+1}=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{y+1}=0\\\sqrt{x+3}-2\sqrt{y+1}=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y+1}=0\\\sqrt{x+3}=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y+1=0\\x+3=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=1\end{matrix}\right.\)

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

4. Đk: \(x,y\ge0\)

\(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y+1}=1\\\sqrt{y}+\sqrt{x+1}=1\end{matrix}\right.\left(1\right)\)

Ta có: \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y+1}\ge0+1=1\\\sqrt{y}+\sqrt{x+1}\ge0+1=1\end{matrix}\right.\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}\sqrt{x}=0,\sqrt{x+1}=1\\\sqrt{y}=0,\sqrt{y+1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)<tmđk>

Vậy hệ pt có nghiệm \(\left(x,y\right)=\left(0;0\right)\)

\(\left\{{}\begin{matrix}\sqrt{5}x-y=\sqrt{5}\left(\sqrt{3}-1\right)\left(1\right)\\2\sqrt{3}x+3\sqrt{5}y=21\left(2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\sqrt{5}x-\sqrt{15}+\sqrt{5}\\2\sqrt{3}x+3\sqrt{5}y=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\sqrt{5}x-\sqrt{15}+\sqrt{5}\\2\sqrt{3}x+3\sqrt{5}\left(\sqrt{5}x-\sqrt{15}+\sqrt{5}\right)=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\sqrt{5}x-\sqrt{15}+\sqrt{5}\\x\left(15+2\sqrt{3}\right)=6+15\sqrt{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{3}\\y=\sqrt{5}\end{matrix}\right.\)

ĐKXD: x, y > 0.

\(Pt_{\left(1\right)}\Leftrightarrow\dfrac{y+\sqrt{x}}{x}=\dfrac{2\left(y+\sqrt{x}\right)}{y}\Leftrightarrow\left(y+\sqrt{x}\right)\left(\dfrac{1}{x}-\dfrac{2}{y}\right)=0\)

\(\Rightarrow y=2x\), thế vào Pt(2): \(2x\left(\sqrt{x^2+1}-1\right)=\sqrt{3x^2+3}\)

Đặt \(\sqrt{x^2+1}=a\) thì \(\left\{{}\begin{matrix}2x\left(a-1\right)=\sqrt{3}a\\a^2-x^2=1\end{matrix}\right.\)

Giải ra ta được \(\left(a-2\right)\left(4a^3-3a+2\right)=0\) nhưng vì \(a\ge1\) nên a=2

Do đó \(x=\pm\sqrt{3}\). Vậy (x, y)=...

a, (3 ; -3)

a, Với y >= 0 

hpt có dạng \(\left\{{}\begin{matrix}2x+y=3\\x-y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=9\\y=x-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-3\end{matrix}\right.\)(ktmđk)

Với y < 0 hpt có dạng 

\(\left\{{}\begin{matrix}2x-y=3\\x-y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-3-6=-9\end{matrix}\right.\)(tm) 

b, bạn tự làm 

c, đk : x>= 3 

\(\left\{{}\begin{matrix}2\sqrt{x+3}+\left|y-2\right|=2\\\sqrt{x+3}-3\left|y-2\right|=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x+3}+\left|y-2\right|=2\\2\sqrt{x+3}-6\left|y-2\right|=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7\left|y-2\right|=1\\2\sqrt{x+3}+\left|y-2\right|=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}y-2=\dfrac{1}{7}\\y-2=-\dfrac{1}{7}\end{matrix}\right.\\2\sqrt{x+3}+\left|y-2\right|=2\end{matrix}\right.\)

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