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

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

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

\(\left\{{}\begin{matrix}4\sqrt{x}-3\sqrt{y}=4\\2\sqrt{x}+\sqrt{y}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4\sqrt{x}-3\sqrt{y}=4\\4\sqrt{x}+2\sqrt{y}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{y}=0\\2\sqrt{x}+\sqrt{y}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y}=0\\\sqrt{x}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4\sqrt{x}-3\sqrt{y}=4\\4\sqrt{x}+2\sqrt{y}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=1\end{matrix}\right.\)

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

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

Cộng từng vế của (1) và (2) ta được: \(\Rightarrow2x=3+\sqrt{2}\Leftrightarrow x=\dfrac{3+\sqrt{2}}{2}\)

Thay vào (2) ta được: \(\Rightarrow\dfrac{3+\sqrt{2}}{2}+\left(\sqrt{2}+1\right)y=1\Leftrightarrow\left(\sqrt{2}+1\right)y=1-\dfrac{3+\sqrt{2}}{2}=\dfrac{-\sqrt{2}-1}{2}\)

\(\Leftrightarrow y=\dfrac{-\sqrt{2}-1}{2\left(\sqrt{2}+1\right)}=\dfrac{-1}{2}\) Vậy...

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

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

⇔\(\left\{{}\begin{matrix}xy+2x+2y+4=xy-4y-x+4\\4xy+2x+6y+3=4xy-4x+y-1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}3x+6y=0\\6x+5y=-4\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=-\dfrac{8}{7}\\y=\dfrac{4}{7}\end{matrix}\right.\)(TM)

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

⇔\(\left\{{}\begin{matrix}5x-5y-6x-9y=12\\3x+6y-4x-8y=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}-x-14y=12\\-x-2y=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=-\dfrac{26}{3}\\y=-\dfrac{7}{12}\end{matrix}\right.\)

Vậy HPT có nghiệm (x;y) = (\(-\dfrac{26}{3};-\dfrac{7}{12}\))

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

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

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

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

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

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

Đề thiếu. Bạn coi lại đề

Đề thiếu rồi bạn

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:

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