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 \(\left(x,y\right)=\left(0;0\right)\)

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

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

=>\(\left\{{}\begin{matrix}y=\dfrac{2+\sqrt{2}}{-4\sqrt{2}}=\dfrac{-\sqrt{2}-1}{4}\\2x=-2-y\sqrt{2}=-2+\sqrt{2}\cdot\dfrac{\sqrt{2}+1}{4}=\dfrac{-6+\sqrt{2}}{4}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=\dfrac{-\sqrt{2}-1}{4}\\x=\dfrac{-6+\sqrt{2}}{8}\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}5x\sqrt{6}+y\sqrt{2}=4\\x\sqrt{6}-y\sqrt{2}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x\cdot\sqrt{6}=6\\x\sqrt{6}-y\sqrt{2}=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{1}{\sqrt{6}}=\dfrac{\sqrt{6}}{6}\\y\sqrt{2}=x\sqrt{6}-2=1-2=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{\sqrt{6}}{6}\\y=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

2) Ta có: \(\left\{{}\begin{matrix}\sqrt{3x-1}-\sqrt{2y+1}=1\\2\sqrt{3x-1}+3\sqrt{2y+1}=12\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{3x-1}-2\sqrt{2y+1}=2\\2\sqrt{3x-1}+3\sqrt{2y+1}=12\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-5\sqrt{2y+1}=-10\\\sqrt{3x-1}-\sqrt{2y+1}=1\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}2y+1=4\\3x-1=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2y=3\\3x=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{3}{2}\\x=\dfrac{10}{3}\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{10}{3}\\y=\dfrac{3}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-2}+2\sqrt{y-3}=6\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{y-3}=10\\\sqrt{x-2}+\sqrt{y-3}=3\end{matrix}\right.\)

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

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

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=3\\y=7\end{matrix}\right.\)

Ta có: \(\left\{{}\begin{matrix}\left(\sqrt{3}-\sqrt{2}\right)x+y=\sqrt{2}\\x+\left(\sqrt{3}+\sqrt{2}\right)y=\sqrt{6}\end{matrix}\right.\)

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

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

Vậy: Hệ phương trình vô nghiệm

a,\(\left\{{}\begin{matrix}-x+2y=6\\5x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x+6y=18\left(1\right)\\10x-6y=10\left(2\right)\end{matrix}\right.\)
Cộng (1) và (2) => 7x=28
\(\Leftrightarrow\) x=4
thay x vào (1) ta có -4+2y=6
=> 2y=10
=>y=5
Vậy nghiệm của phương trình (x;y)=(4;5)

b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{3}x+\dfrac{1}{4}y=2\\5x-y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)

c: \(\Leftrightarrow\left\{{}\begin{matrix}3x=6\\5y=15\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left(x,y\right)\in\varnothing\)

a.

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\y\ge3\end{matrix}\right.\)

\(\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\5\sqrt{x-2}=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\\sqrt{x-2}=1\end{matrix}\right.\)

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

b.

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\ne-4\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{4x}{x+1}-\dfrac{10}{y+4}=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{19x}{x+1}=28\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+1}=\dfrac{28}{19}\\\dfrac{1}{y+4}=-\dfrac{4}{19}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}19x=28x+28\\4y+16=-19\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{28}{9}\\y=-\dfrac{35}{4}\end{matrix}\right.\)

Lấy phương trình trên trừ phương trình dưới thu được:

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

Thay vào phương trình dưới suy ra:

\(2\sqrt{2}x=4\sqrt{2}0\Rightarrow x=2\Rightarrow y=1\)

a)

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

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

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

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

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

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

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

Vậy hệ phương trình có tập nghiệm {1;-2}

b)

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}y=\sqrt{3}x-1\\x=\frac{3\sqrt{3}+2\sqrt{2}}{19}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\sqrt{3}.\left(\frac{3\sqrt{3}+2\sqrt{2}}{19}\right)-1\\x=\frac{3\sqrt{3}+2\sqrt{2}}{19}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{-10+2\sqrt{6}}{19}\\x=\frac{3\sqrt{3}+2\sqrt{2}}{19}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3\sqrt{3}+2\sqrt{2}}{19}\\y=\frac{-10+2\sqrt{6}}{19}\end{matrix}\right.\)

Vậy hệ phương trình có tập nghiệm \(\left\{\frac{3\sqrt{3}+2\sqrt{2}}{19};\frac{-10+2\sqrt{6}}{19}\right\}\)

c)

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

\(\Leftrightarrow\left\{{}\begin{matrix}4x+2y=10\\3x-2y=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7x=13\\4x+2y=10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{13}{7}\\4.\frac{13}{7}+2y=10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{13}{7}\\y=\frac{9}{7}\end{matrix}\right.\)

Vậy hệ phương trình có tập nghiệm \(\left\{\frac{13}{7};\frac{9}{7}\right\}\)

`{(2\sqrt{5x-1}-5/[|y+3|]=-1),(3\sqrt{5x-1}+7/[|y+3|]=13):}`     `ĐK: x >= 1/5;y ne -3`

Đặt `\sqrt{5x-1}=a;1/[|y+3|]=b` khi đó ta có:

    `{(2a-5b=-1),(3a+7b=13):}`

`<=>{(6a-15b=-3),(6a+14b=26):}`

`<=>{(29b=29),(2a-5b=-1):}`

`<=>{(b=1),(2a-5.1=-1):}`

`<=>{(a=2),(b=1):}`

  `=>{(\sqrt{5x-1}=2),(1/[|y+3|]=1):}`

`<=>{(5x-1=2),([(y+3=1),(y+3=-1):}):}`

`<=>{(x=3/5),([(y=-2),(y=-4):}):}`     (t/m)